MCQ 11 Mark
The numerical value of $\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ}$ is equal to
- ✓
$\sqrt{ } 3$
- B
$\frac{1}{\sqrt{3}}$
- C
$2 \sqrt{3}$
- D
$\frac{1}{2 \sqrt{3}}$
AnswerCorrect option: A. $\sqrt{ } 3$
(a) $\sqrt{ } 3$
Hint:
$\begin{aligned}
& \text { L.H.S. }=\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ} \\
& =\tan 20^{\circ} \tan 80^{\circ} \cot \left(90^{\circ}-40^{\circ}\right) \\
& =\tan 20^{\circ} \tan 80^{\circ} \tan 40^{\circ} \\
& =\tan 20^{\circ} \tan \left(60^{\circ}+20^{\circ}\right) \tan \left(60^{\circ}-20^{\circ}\right) \\
& =\tan 20^{\circ}\left(\frac{\tan 60^{\circ}+\tan 20^{\circ}}{1-\tan 60^{\circ} \tan 20^{\circ}}\right)\left(\frac{\tan 60^{\circ}-\tan 20^{\circ}}{1+\tan 60^{\circ} \tan 20^{\circ}}\right) \\
& =\tan 20^{\circ}\left(\frac{\sqrt{3}+\tan 20^{\circ}}{1-\sqrt{3} \tan 20^{\circ}}\right)\left(\frac{\sqrt{3}-\tan 20^{\circ}}{1+\sqrt{3} \tan 20^{\circ}}\right) \\
& \left.=\tan 20^{\circ}\right)\left(\frac{(\sqrt{3})^2-\tan 220^{\circ}}{1^2-\left(\sqrt{3} \tan 20^{\circ}\right)^2}\right] \\
& =\tan 20^{\circ}\left(\frac{3-\tan ^2 20^{\circ}}{1-3 \tan ^2 20^{\circ}}\right) \\
& =\frac{3 \tan 20^{\circ}-\tan ^3 20^{\circ}}{1-3 \tan ^2 20^{\circ}} \\
& =\tan 3\left(20^{\circ}\right) \\
& =\tan 60^{\circ} \\
& =\sqrt{3}
\end{aligned}$
View full question & answer→MCQ 21 Mark
In $\triangle A B C$ if $\cot A \cot B \cot C>0$, then the triangle is
Answer(a) acute angled
Hint:
$\cot A \cot B \cot C>0$
Case I:
$\cot A, \cot B, \cot C>0$
$\therefore \cot A>0, \cot B>0, \cot C>0$
$\therefore 0<A<\frac{\pi}{2}, 0<B<\frac{\pi}{2}, 0<C<\frac{\pi}{2}$
$\therefore \triangle \mathrm{ABC}$ is an acute angled triangle.
Case II:
Two of $\cot A, \cot B, \cot C<0$
$0<A, B, C<\pi$ and two of $\cot A, \cot B, \cot C<0$
$\therefore$ Two angles $A_t B_t C$ are in the 2nd quadrant which is not possible.
View full question & answer→MCQ 31 Mark
If $\alpha+\beta+\gamma=\pi$, then the value of $\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma$ is equal to
- A
$2 \sin \alpha$
- B
$2 \sin \alpha \cos \beta \sin \gamma$
- ✓
$2 \sin \alpha \sin \beta \cos \gamma$
- D
$2 \sin \alpha \sin \beta \sin \gamma$
AnswerCorrect option: C. $2 \sin \alpha \sin \beta \cos \gamma$
(c) $2 \sin \alpha \sin \beta \cos \gamma$
Hint:
$\begin{aligned}
& \sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma \\
& =\frac{1-\cos 2 \alpha}{2}+\frac{1-\cos 2 \beta}{2}-\sin ^2 \gamma \\
& =1-\frac{1}{2}(\cos 2 \alpha+\cos 2 \beta)-1+\cos ^2 \gamma \\
& =\frac{-1}{2} \times 2 \cos (\alpha+\beta) \cos (\alpha-\beta)+\cos ^2 \gamma \\
& =\cos \gamma \cos (\alpha-\beta)+\cos ^2 \gamma \ldots \ldots[\because \alpha+\beta+\gamma=\pi] \\
& =\cos \gamma[\cos (\alpha-\beta)+\cos \gamma] \\
& =\cos \gamma[\cos (\alpha-\beta)-\cos (\alpha+\beta)] \\
& =2 \sin \alpha \sin \beta \cos \gamma
\end{aligned}$
View full question & answer→MCQ 41 Mark
The value of $\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}$ is
- A
$\frac{1}{16}$
- ✓
$\frac{1}{64}$
- C
$\frac{1}{128}$
- D
$\frac{1}{256}$
AnswerCorrect option: B. $\frac{1}{64}$
(b) $\frac{1}{64}$
View full question & answer→MCQ 51 Mark
The value of $\cos A \cos \left(60^{\circ}-A\right) \cos \left(60^{\circ}+A\right)$ is equal to
AnswerCorrect option: C. $\frac{1}{4} \cos 3 \mathrm{~A}$
(c) $\frac{1}{4} \cos 3 \mathrm{~A}$
Hint:
$\begin{aligned}
& \cos A \cos \left(60^{\circ}-A\right) \cos \left(60^{\circ}+A\right) \\
& =(\cos A)\left(\cos 60^{\circ} \cos A+\sin 60^{\circ} \sin A\right) \\
& \quad \cdot\left(\cos 60^{\circ} \cos A-\sin 60^{\circ} \sin A\right) \\
& =(\cos A)\left(\frac{1}{2} \cos A+\frac{\sqrt{3}}{2} \sin A\right)
\end{aligned}$
$\left(\frac{1}{2} \cos \mathrm{A}-\frac{\sqrt{3}}{2} \sin \mathrm{A}\right)$
$\begin{aligned}
& =\frac{1}{4} \cos \mathrm{A}\left(\cos ^2 \mathrm{~A}-3 \sin ^2 \mathrm{~A}\right) \\
& =\frac{1}{4}\left[\cos ^3 \mathrm{~A}-3 \cos \mathrm{A}\left(1-\cos ^2 \mathrm{~A}\right)\right] \\
& =\frac{1}{4}\left(4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{A}\right)=\frac{1}{4} \cos 3 \mathrm{~A}
\end{aligned}$
View full question & answer→MCQ 61 Mark
The value of $\frac{\cos \theta}{1+\sin \theta}$ is equal to
- A
$\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)$
- B
$\tan \left(-\frac{\pi}{4}-\frac{\theta}{2}\right)$
- ✓
$\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
- D
$\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)$
AnswerCorrect option: C. $\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
(c) $\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
Hint:
$\begin{aligned}
\frac{\cos \theta}{1+\sin \theta} & =\frac{\cos ^2 \frac{\theta}{2}-\sin ^2 \frac{\theta}{2}}{\cos ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \\
& =\frac{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^2} \\
& =\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}
\end{aligned}$
Dividing numerator and denominator by $\cos \frac{\theta}{2}$, we get
$\frac{\cos \theta}{1+\sin \theta}=\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
View full question & answer→MCQ 71 Mark
If $\sin \theta=n \sin (\theta+2 \alpha)$, then $\tan (\theta+\alpha)$ is equal to
- A
$\frac{1+n}{2-n} \tan \alpha$
- B
$\frac{1-n}{1+n} \tan \alpha$
- C
$\tan \alpha$
- ✓
$\frac{1+n}{1-n} \tan \alpha$
AnswerCorrect option: D. $\frac{1+n}{1-n} \tan \alpha$
(d) $\frac{1+n}{1-n} \tan \alpha$
Hint:
$\begin{aligned}
& \sin \theta=\mathrm{n} \sin (\theta+2 \alpha) \\
& \frac{\mathrm{n}}{1}=\frac{\sin \theta}{\sin (\theta+2 \alpha)}
\end{aligned}$
By componendo-dividendo, we get
$\begin{aligned}
& \frac{\mathrm{n}+1}{\mathrm{n}-1}=\frac{\sin \theta+\sin (\theta+2 \alpha)}{\sin \theta-\sin (\theta+2 \alpha)} \\
& =\frac{2 \sin (\theta+\alpha) \cos \alpha}{-2 \cos (\theta+\alpha) \sin \alpha} \\
& \tan (\theta+\alpha)=-\left(\frac{\mathrm{n}+1}{\mathrm{n}-1}\right) \tan \alpha=\frac{1+\mathrm{n}}{1-\mathrm{n}} \tan \alpha \\
\end{aligned}$
View full question & answer→MCQ 81 Mark
If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, then $\cot (A-B)=$
- A
$\frac{1}{y}-\frac{1}{x}$
- B
$\frac{1}{x}-\frac{1}{y}$
- ✓
$\frac{1}{x}+\frac{1}{y}$
- D
$\frac{x y}{x-y}$
AnswerCorrect option: C. $\frac{1}{x}+\frac{1}{y}$
(c) $\frac{1}{x}+\frac{1}{y}$
Hint:
$\begin{aligned}
& x=\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B} \\
& =\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B}=\frac{\sin (A-B)}{\cos A \cos B} \\
& y=\frac{\cos B}{\sin B}-\frac{\cos A}{\sin A} \\
& =\frac{\sin A \cos B-\cos A \sin B}{\sin A \sin B} \\
& =\frac{\sin (A-B)}{\sin A \sin B}\\
& \frac{y}{x}=\frac{\cos \mathrm{A} \cos \mathrm{B}}{\sin \mathrm{A} \sin \mathrm{B}}=\cot \mathrm{A} \cot \mathrm{B} \\
& \cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A} \\
& =\frac{\frac{y}{x}+1}{y}=\frac{x+y}{x y} \\
& =\frac{1}{x}+\frac{1}{y} \\
\end{aligned}$
View full question & answer→MCQ 91 Mark
The value of $\sin (n+1) A \sin (n+2) A+\cos (n+1) A \cos (n+2) A$ is equal to
- A
$\sin \mathrm{A}$
- ✓
$\cos \mathrm{A}$
- C
$-\cos A$
- D
$\sin 2 \mathrm{~A}$
AnswerCorrect option: B. $\cos \mathrm{A}$
(b) $\cos \mathrm{A}$
Hint:
$\begin{aligned}
& \text { L.H.S. }=\sin [(n+1) A] \cdot \sin [(n+2) A]+\cos [(n+1) A] \cdot \cos [(n+2) A] \\
& =\cos [(n+2) A] \cdot \cos [(n+1) A]+\sin [(n+2) A] \cdot \sin [(n+1) A] \\
& \text { Let }(n+2) A=a \text { and }(n+1) A=b \ldots(i) \\
& \therefore \text { L.H.S. }=\cos a \cdot \cos b+\sin a \cdot \sin b \\
& =\cos (a-b) \\
& =\cos [(n+2) A-(n+1) A] \ldots \ldots[\text { From (i) }] \\
& =\cos [(n+2-n-1) A] \\
& =\cos A \\
& =\text { R.H.S. }
\end{aligned}$
View full question & answer→MCQ 102 Marks
If $a \cos 2 \theta+b \sin 2 \theta=c$ has $\alpha$ and $\beta$ as its roots, then the value of $\tan \alpha+\tan \beta$ is
- ✓
$\frac{2 b}{c+a}$
- B
$\frac{2 a}{b+c}$
- C
$\frac{b}{c+a}$
- D
$\frac{a}{b+c}$
AnswerCorrect option: A. $\frac{2 b}{c+a}$
(a) : Given, $a \cos 2 \theta+b \sin 2 \theta=c$
$
\begin{aligned}
& \Rightarrow a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=c \\
& \Rightarrow a\left(1-\tan ^2 \theta\right)+2 b \tan \theta=c\left(1+\tan ^2 \theta\right)
\end{aligned}
$
Since, $\alpha$ and $\beta$ are roots of the given equation.
$
\begin{aligned}
& \Rightarrow a\left(1-\tan ^2 \alpha\right)+2 b \tan \alpha=c\left(1+\tan ^2 \alpha\right) \\
& \text { and } a\left(1-\tan ^2 \beta\right)+2 b \tan \beta=c\left(1+\tan ^2 \beta\right)
\end{aligned}
$
Subtracting (ii) from (i), we get
$
\begin{aligned}
& -a\left(\tan ^2 \alpha-\tan ^2 \beta\right)+2 b(\tan \alpha-\tan \beta)=c\left(\tan ^2 \alpha-\tan ^2 \beta\right) \\
& \Rightarrow 2 b(\tan \alpha-\tan \beta)=\left(\tan ^2 \alpha-\tan ^2 \beta\right)(a+c) \\
& \Rightarrow 2 b=(\tan \alpha+\tan \beta)(a+c) \\
& \Rightarrow \tan \alpha+\tan \beta=\frac{2 b}{a+c}
\end{aligned}
$
View full question & answer→MCQ 112 Marks
If $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in H.P., then the value of $\cos 2 \theta$ is
- A
$1+4 \cos ^2 \frac{\alpha}{2}$
- ✓
$1-4 \cos ^2 \frac{\alpha}{2}$
- C
$-1-4 \cos ^2 \frac{\alpha}{2}$
- D
$-1+4 \cos ^2 \frac{\alpha}{2}$
AnswerCorrect option: B. $1-4 \cos ^2 \frac{\alpha}{2}$
(b) : $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in H.P.
$
\begin{aligned}
& \Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)} \text { are in A.P. } \\
& \Rightarrow \frac{1}{\sin (\theta-\alpha)}+\frac{1}{\sin (\theta+\alpha)}=\frac{2}{\sin \theta} \\
& \Rightarrow \frac{\sin (\theta+\alpha)+\sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)}=\frac{2}{\sin \theta} \\
& \Rightarrow \sin \theta=\frac{2 \sin (\theta-\alpha) \sin (\theta+\alpha)}{\sin (\theta+\alpha)+\sin (\theta-\alpha)} \\
& \Rightarrow \sin \theta=\frac{2\left(\sin ^2 \theta-\sin ^2 \alpha\right)}{2 \sin \theta \cos \alpha} \\
& \Rightarrow \sin ^2 \theta \cos \alpha=\sin ^2 \theta-\sin ^2 \alpha \Rightarrow \sin ^2 \theta(1-\cos \alpha)=\sin ^2 \alpha \\
& \Rightarrow \sin ^2 \theta(1-\cos \alpha)=\left(1-\cos ^2 \alpha\right) \\
& \Rightarrow \sin ^2 \theta(1-\cos \alpha)=(1-\cos \alpha)(1+\cos \alpha) \\
& \Rightarrow \sin ^2 \theta=1+\cos \alpha \Rightarrow 1-\cos ^2 \theta=1+\cos \alpha \\
& \Rightarrow \cos ^2 \theta=-\cos \alpha \Rightarrow 2 \cos ^2 \theta=-2 \cos \alpha \\
& \Rightarrow 2 \cos ^2 \theta-1=-2 \cos \alpha-1 \Rightarrow \cos 2 \theta=-2 \cos \alpha-1 \\
& =-2\left(2 \cos ^2 \frac{\alpha}{2}-1\right)-1=-4 \cos ^2 \frac{\alpha}{2}+2-1=1-4 \cos ^2 \frac{\alpha}{2} \\
&
\end{aligned}
$
View full question & answer→MCQ 122 Marks
The value of $\cos ^2 10^{\circ}-\cos 10^{\circ} \cdot \cos 50^{\circ}+\cos ^2 50^{\circ}$ is
AnswerCorrect option: B. $\frac{3}{4}$
$ \cos ^2 10^{\circ}-\cos 10^{\circ} \cdot \cos 50^{\circ}+\cos ^2 50^{\circ}$
$=\cos ^2 50^{\circ}+1-\sin ^2 10^{\circ}-\frac{1}{2}\left(2 \cos 10^{\circ} \cdot \cos 50^{\circ}\right)$
$=\cos ^2 50^{\circ}-\sin ^2 10^{\circ}+1-\frac{1}{2}\left(\cos 60^{\circ}+\cos 40^{\circ}\right)$
$[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$
$=\cos 60^{\circ} \cdot \cos 40^{\circ}+1-\frac{1}{2}(\cos 60^{\circ}+\cos 40^{\circ})$
${\left[\because \cos ^2 A-\sin ^2 B=\cos (A+B) \cos (A-B)\right]}$
$=\frac{1}{2} \cos 40^{\circ}+1-\frac{1}{2}\left(\frac{1}{2}+\cos 40^{\circ}\right)$
$=1-\frac{1}{4}$
$=\frac{3}{4}$
View full question & answer→MCQ 132 Marks
If $\theta=\frac{17 \pi}{3}$ then $\tan \theta-\cot \theta=$
- A
$\frac{1}{2 \sqrt{3}}$
- B
$\frac{-1}{2 \sqrt{3}}$
- C
$\frac{2}{\sqrt{3}}$
- ✓
$-\frac{2}{\sqrt{3}}$
AnswerCorrect option: D. $-\frac{2}{\sqrt{3}}$
Given, $\theta=\frac{17 \pi}{3}=\frac{18 \pi-\pi}{3} $
$\Rightarrow \theta=6 \pi-\frac{\pi}{3}$
Now, $\tan \theta-\cot \theta=\tan \theta-\frac{1}{\tan \theta}$
$=\tan \left(6 \pi-\frac{\pi}{3}\right)-\frac{1}{\tan \left(6 \pi-\frac{\pi}{3}\right)}$
$=\tan \left(-\frac{\pi}{3}\right)-\frac{1}{\tan \left(-\frac{\pi}{3}\right)}$
$=-\sqrt{3}-\left(\frac{1}{-\sqrt{3}}\right)$
$=\frac{-2}{\sqrt{3}}$
View full question & answer→MCQ 142 Marks
The value of $\sin 18^{\circ}$ is
- A
$\frac{\sqrt{5}+1}{4}$
- ✓
$\frac{\sqrt{5}-1}{4}$
- C
$\frac{4}{\sqrt{5}+1}$
- D
$\frac{4}{\sqrt{5}-1}$
AnswerCorrect option: B. $\frac{\sqrt{5}-1}{4}$
(b) : Let $x=18^{\circ} \Rightarrow 5 x=5 \times 18^{\circ}=90^{\circ}$
$
\begin{aligned}
& \Rightarrow 2 x+3 x=90^{\circ} \Rightarrow 2 x=90^{\circ}-3 x \\
& \Rightarrow \sin 2 x=\sin \left(90^{\circ}-3 x\right) \Rightarrow \sin 2 x=\cos 3 x \\
& \Rightarrow 2 \sin x \cos x=4 \cos ^3 x-3 \cos x \\
& \Rightarrow 2 \sin x \cos x=\cos x\left(4 \cos ^2 x-3\right) \\
& \Rightarrow 2 \sin x=4 \cos ^2 x-3 \Rightarrow 2 \sin x=4\left(1-\sin ^2 x\right)-3 \\
& \Rightarrow 2 \sin x=4-4 \sin ^2 x-3 \Rightarrow 4 \sin ^2 x+2 \sin x-1=0
\end{aligned}
$
This is a quadratic in $\sin x$, where $a=4, b=2, c=-1$
$
\begin{aligned}
\therefore \quad \sin x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-2 \pm \sqrt{4-4(4)(-1)}}{8} \\
\quad=\frac{-2 \pm \sqrt{20}}{8}=\frac{-2 \pm 2 \sqrt{5}}{8}=\frac{-1 \pm \sqrt{5}}{4}
\end{aligned}
$
But $\sin 18^{\circ}>0$ because $18^{\circ}$ lies in first quadrant.
$
\therefore \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}
$
View full question & answer→MCQ 152 Marks
If $A, B, C$ are the angles of $\triangle A B C$, then $\cot A \cdot \cot B+$ $\cot B \cdot \cot C+\cot C \cdot \cot A=$
Answer(b) : $\because A+B+C=\pi \Rightarrow A+B=\pi-C$
Now, $\cot (A+B)=\cot (\pi-C)$
$
\begin{aligned}
& \Rightarrow \frac{\cot A \cot B-1}{\cot A+\cot B}=-\cot C \\
& \Rightarrow \cot A \cot B+\cot B \cot C+\cot A \cot C=1 .
\end{aligned}
$
View full question & answer→MCQ 162 Marks
If $2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$, then $\tan \theta=$
- A
$\sqrt{3}$
- B
$-\frac{1}{\sqrt{3}}$
- C
$\frac{1}{\sqrt{3}}$
- ✓
$-\sqrt{3}$
AnswerCorrect option: D. $-\sqrt{3}$
(d) : We have, $2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$
$\begin{array}{r}\Rightarrow 2\left(\sin \theta \times \frac{1}{2}+\cos \theta \times \frac{\sqrt{3}}{2}\right)=\cos \theta \\ \times \frac{\sqrt{3}}{2}+\sin \theta \times \frac{1}{2}\end{array}$
$\begin{aligned} & \Rightarrow \frac{1}{2} \sin \theta+\frac{\sqrt{3}}{2} \cos \theta=0 \Rightarrow \frac{1}{2} \sin \theta=\frac{-\sqrt{3}}{2} \cos \theta \\ & \Rightarrow \tan \theta=-\sqrt{3}\end{aligned}$
View full question & answer→MCQ 172 Marks
$\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \ldots \cos 179^{\circ}=$
Answer(a) : $\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \ldots . \cos 179^{\circ}$ $=\cos 1^{\circ} \cdot \cos 2^{\circ} \ldots \cos 90^{\circ} \ldots \cos 179^{\circ}=0$
View full question & answer→MCQ 182 Marks
If $A , B , C$ are the angles of a triangle, then $\sum \frac{\cot A+\cot B}{\tan A+\tan B}=$
Answer(A)
$\sum \frac{\cot A+\cot B}{\tan A+\tan B}=\sum \frac{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}$
$=\sum\left(\frac{\sin B \cos A +\sin A \cos B }{\sin A \cdot \sin B }\right)$$\left(\frac{\cos A \cdot \cos B }{\sin A \cos B +\cos A \cdot \sin B }\right)$
$=\sum \cot A \cot B$
$=\cot A \cot B +\cot B \cot C +\cot C \cot A =1$$\cdots\left[\begin{array}{l}\because A+B+C=\pi, \\ \therefore \cot A \cot B+\cot B \cot C+\cot C \cot A=1\end{array}\right]$
View full question & answer→MCQ 192 Marks
If $\alpha+\beta+\gamma=2 \pi$, then
- ✓
$\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2} =\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$
- B
$\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+ \tan \frac{\gamma}{2} =-\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$
- C
$\tan \frac{\alpha}{2} \tan \frac{\beta}{2}+\tan \frac{\beta}{2} \tan \frac{\gamma}{2}+\tan \frac{\gamma}{2} \tan \frac{\alpha}{2}=1$
- D
$\tan \alpha \tan \beta+\tan \beta \tan \gamma+\tan \gamma \tan \alpha=1$
AnswerCorrect option: A. $\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2} =\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$
(A)
We have, $\alpha+\beta+\gamma=2 \pi$
$\Rightarrow \frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=\pi$
$\Rightarrow \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}\right)=\tan \pi=0$
$\Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}-\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}=0$
$\Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$
View full question & answer→MCQ 202 Marks
If $A + B + C =180^{\circ}$, then the value of $\cot \frac{ A }{2}+\cot \frac{ B }{2}+\cot \frac{ C }{2}$ will be
- A
$2 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{ C }{2}$
- B
$4 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{ C }{2}$
- ✓
$\cot \frac{ A }{2} \cot \frac{B}{2} \cot \frac{ C }{2}$
- D
$8 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{ C }{2}$
AnswerCorrect option: C. $\cot \frac{ A }{2} \cot \frac{B}{2} \cot \frac{ C }{2}$
(C)
$A+B+C=180^{\circ}$
$\Rightarrow \cot \left(\frac{ A }{2}+\frac{ B }{2}\right)=\cot \left(90^{\circ}-\frac{ C }{2}\right)$
$\Rightarrow \frac{\cot _2^{ A } \cdot \cot _2^{ B }-1}{\cot \frac{ B }{2}+\cot \frac{ A }{2}}=\tan \frac{ C }{2}=\frac{1}{\cot \frac{ C }{2}}$
$\Rightarrow\left(\cot \frac{ A }{2} \cot \frac{B}{2}-1\right) \cot \frac{ C }{2}=\cot \frac{ B }{2}+\cot \frac{ A }{2}$
$\Rightarrow \cot \frac{ A }{2} \cot \frac{B}{2} \cot \frac{ C }{2}=\cot \frac{ C }{2}+\cot \frac{ B }{2}+\cot \frac{ A }{2}$
View full question & answer→MCQ 212 Marks
If $A + B + C =180^{\circ}$, then $\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\cos A+\cos B+\cos C-1}=$
- A
$8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
- ✓
$8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{ C }{2}$
- C
$8 \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{ C }{2}$
- D
$8 \cos \frac{A}{2} \sin \frac{B}{2} \sin \frac{ C }{2}$
AnswerCorrect option: B. $8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{ C }{2}$
(B)
$\frac{\sin 2 A+\sin 2 B+\sin 2 C }{\cos A +\cos B +\cos C -1}$
$\sin 2 A+\sin 2 B+\sin 2 C =4 \sin A \sin B \sin C$
$=\frac{4 \sin A \sin B \sin C }{1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{ C }{2}-1}$
$=\frac{\left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right)\left(2 \sin \frac{B}{2} \cos \frac{B}{2}\right)\left(2 \sin \frac{ C }{2} \cos \frac{ C }{2}\right)}{\sin \frac{ A }{2} \sin \frac{B}{2} \sin \frac{ C }{2}}$$\ldots\left[\because \sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]$
$=8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{ C }{2}$
View full question & answer→MCQ 222 Marks
In a triangle ABC, the value of sin A + sin B + sin C is
- A
$4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{ C }{2}$
- ✓
$4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{ C }{2}$
- C
$4 \cos \frac{A}{2} \sin \frac{B}{2} \sin \frac{ C }{2}$
- D
$4 \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{ C }{2}$
AnswerCorrect option: B. $4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{ C }{2}$
(B)
$\sin A +\sin B +\sin C$
$=2 \sin \frac{A+ B }{2} \cos \frac{A- B }{2}+2 \sin \frac{ C }{2} \cos \frac{ C }{2}$
$=2 \sin \left(\frac{\pi}{2}-\frac{ C }{2}\right) \cos \frac{ A - B }{2}$$+2 \cos \frac{ C }{2} \sin \left(\frac{\pi}{2}-\left(\frac{ A + B }{2}\right)\right)$
$=2 \cos \frac{ C }{2} \cos \frac{A- B }{2}+2 \cos \frac{ C }{2} \cos \frac{A+ B }{2}$
$=2 \cos \frac{ C }{2}\left[\cos \frac{A- B }{2}+\cos \frac{ A + B }{2}\right]$
$=2 \cos \frac{ C }{2}\left(2 \cos \frac{A}{2} \cos \frac{B}{2}\right)$
$=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{ C }{2}$
View full question & answer→MCQ 232 Marks
If $A + B + C =\pi$, then $\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}-\cos ^2 \frac{ C }{2}$ is
- ✓
$2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}$
- B
$4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}$
- C
$1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}$
- D
$1-4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}$
AnswerCorrect option: A. $2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}$
(A)
$\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}-\cos ^2 \frac{ C }{2}$
$=\frac{1}{2}(1+\cos A )+\frac{1}{2}(1+\cos B )-\frac{1}{2}(1+\cos C )$
$=\frac{1}{2}+\frac{1}{2}(\cos A+\cos B -\cos C )$
$\cos A +\cos B - \cos C =-1+4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}$
$=\frac{1}{2}+\frac{1}{2}\left[4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}-1\right]$
$=2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{ C }{2}$
View full question & answer→MCQ 242 Marks
In a $\triangle ABC$, the value of $\cos ^2 A+\cos ^2\left(A+\frac{\pi}{3}\right)+\cos ^2\left(A-\frac{\pi}{3}\right)$ is
- A
$0$
- B
$\frac{1}{2}$
- ✓
$\frac{3}{2}$
- D
AnswerCorrect option: C. $\frac{3}{2}$
(C)
$\cos ^2 A+\cos ^2\left(A+\frac{\pi}{3}\right)+\cos ^2\left(A-\frac{\pi}{3}\right)$
$=\frac{1}{2}(1+\cos 2 A)+\frac{1}{2}\left\{1+\cos \left(2 A+\frac{2 \pi}{3}\right)\right\}$$+\frac{1}{2}\left\{1+\cos \left(2 A-\frac{2 \pi}{3}\right)\right\}$
$=\frac{3}{2}+\frac{1}{2} \cos 2 A$$+\frac{1}{2}\left\{\cos \left(2 A+\frac{2 \pi}{3}\right)+\cos \left(2 A-\frac{2 \pi}{3}\right)\right\}$
$=\frac{3}{2}+\frac{1}{2} \cos 2 A+\cos 2 A \cos \frac{2 \pi}{3}$
$\ldots[\because \cos (A+B)+\cos (A-B)$$=2 \cos A \cos B ]$
$=\frac{3}{2}+\frac{1}{2} \cos 2 A-\frac{1}{2} \cos 2 A=\frac{3}{2}$
View full question & answer→MCQ 252 Marks
If A, B, C are the angles of a triangle, then $\sin ^2 A+\sin ^2 B+\sin ^2 C-2 \cos A \cos B \cos C=$
Answer(B)
$\sin ^2 A+\sin ^2 B+\sin ^2 C$
$=1-\cos ^2 A+1-\cos ^2 B+\sin ^2 C$
$=2-\cos ^2 A-\left(\cos ^2 B-\sin ^2 C \right)$
$=2-\cos ^2 A-\cos ( B + C ) \cos ( B - C )$
$=2-\cos A [\cos A -\cos ( B - C )]$
$=2-\cos A [-\cos ( B + C )-\cos ( B - C )]$
$=2+\cos A \cdot 2 \cos B \cos C$
$\therefore \sin ^2 A+\sin ^2 B+\sin ^2 C$$-2 \cos A \cos B \cos C =2$
View full question & answer→MCQ 262 Marks
If $A + B + C =\pi$, then $\frac{\cos A}{\sin B \sin C}+\frac{\cos B}{\sin C \sin A}+\frac{\cos C}{\sin A \sin B}=$
Answer(C)
$\frac{\cos A }{\sin B \sin C }+\frac{\cos B }{\sin C \sin A }+\frac{\cos C }{\sin A \sin B }$
$=\frac{2 \sin A \cos A }{2 \sin A \sin B \sin C }+\frac{2 \sin B \cos B }{2 \sin A \sin B \sin C }$$+\frac{2 \sin C \cos C}{2 \sin A \sin B \sin C}$
$=\frac{\sin 2 A+\sin 2 B+\sin 2 C }{2 \sin A \sin B \sin C }$
$\sin 2 A+\sin 2 B+\sin 2 C =4 \sin A \sin B \sin C$
$=\frac{4 \sin A \sin B \sin C }{2 \sin A \sin B \sin C } $
$= 2$
View full question & answer→MCQ 272 Marks
If $A+B+C=\frac{3 \pi}{2}$, then
cos 2A + cos 2B + cos 2C =
Answer(D)
$\cos 2 A+\cos 2 B+\cos 2 C$
$=2 \cos (A+ B ) \cos ( A - B )+\cos 2 C$
$-2 \cos \left(\frac{3 \pi}{2}-C\right) \cos (A-B)+\cos 2 C$$\ldots\left[\because A+B=\frac{3 \pi}{2}-C\right]$
$=-2 \sin C \cos (A-B)+1-2 \sin ^2 C$
$-1-2 \sin C(\cos (A-B) \mid \sin C)$
$=1-2 \sin C\{\cos (A-B)-\cos (A+B)\}$
$=1-4 \sin A \sin B \sin C$
Trick : Check by assuming $A = B = C =\frac{\pi}{2}$
View full question & answer→MCQ 282 Marks
If $x+y+z=180^{\circ}$, then $\cos 2 x+\cos 2 y- cos 2 z$ is equal to
Answer(B)
$\cos 2 x+\cos 2 y-\cos 2 z$
$=2 \cos (x+y) \cos (x-y)-2 \cos ^2 z+1$
$=2 \cos (\pi-z) \cos (x-y)-2 \cos ^2 z+1$
$=1-2 \cos z\{\cos (x-y)-\cos (x+y)\}$
$=1-2 \cos z 2 \sin x \sin y$
$=1-4 \sin x \sin y \cos z$
View full question & answer→MCQ 292 Marks
If $A + B + C =\pi$, then $\cos 2A+\cos 2B+\cos 2C=$
- A
- B
- 1 + 4 sin A sin B cos C
- ✓
- 1 - 4 cos A cos B cos C
- D
AnswerCorrect option: C. - 1 - 4 cos A cos B cos C
(C)
\[\cos 2 A+\cos 2 B+\cos 2 C\]
$=2 \cos (A+ B ) \cos ( A - B )+\left(2 \cos ^2 C -1\right)$
$=-1-2 \cos C \cos (A-B)+2 \cos ^2 C$$\ldots[\because \cos (A+B)=-\cos C]$
$=-1-2 \cos C [\cos ( A - B )+\cos ( A + B )]$
$=-1-4 \cos A \cos B \cos C$
View full question & answer→MCQ 302 Marks
if A, B, C are the angles of a triangle, then sin 2A + sin 2B - sin 2C is equal to
Answer(D)
$\sin 2 A+\sin 2 B-\sin 2 C$
$=2 \sin A \cos A +2 \cos (B+ C ) \sin ( B - C )$
$=2 \sin A \cos A -2 \cos A \sin ( B - C )$
$=2 \cos A[\sin A -\sin ( B - C )]$
$=2 \cos A[\sin ( B + C )-\sin ( B - C )]$
$\ldots[\because \sin (B+C)=\sin A]$
$=2 \cos A(2 \cos B \sin C )$
$=4 \cos A \cos B \sin C$
Trick : Check by assuming $A = B =45^{\circ}$ and $C =90^{\circ}$
View full question & answer→MCQ 312 Marks
In triangle ABC , the value of $\sin 2 A+\sin 2 B+\sin 2 C$ is equal to
Answer(A)
In $\triangle ABC , A + B + C =\pi$
$\sin 2 A+\sin 2 B+\sin 2 C$
$=2 \sin (A+ B ) \cos ( A - B )+2 \sin C \cos C$
$=2 \sin (\pi- C ) \cos ( A - B )+2 \sin C \cos \{\pi-( A + B )\}$
$=2 \sin C \cos (A-B)-2 \sin C \cos (A+B)$
$=2 \sin C \{\cos ( A - B )-\cos ( A + B )\}$
$=2 \sin C (2 \sin A \sin B )$
$=4 \sin A \sin B \sin C$
View full question & answer→MCQ 322 Marks
If $A + B + C =180^{\circ}$, then the value of $(\cot B +\cot C )(\cot C +\cot A )(\cot A +\cot B)$ will be
Answer(B)
$\cot B +\cot C =\frac{\sin C \cos B +\sin B \cos C }{\sin B \sin C }$
$=\frac{\sin (B+C)}{\sin B \sin C}$
$=\frac{\sin \left(180^{\circ}- A \right)}{\sin B \sin C }$
$=\frac{\sin A }{\sin B \sin C }$
Similarly, $\cot C +\cot A =\frac{\sin B }{\sin C \sin A }$and $\cot A +\cot B =\frac{\sin C }{\sin A \sin B }$
$\therefore(\cot B +\cot C )(\cot C +\cot A )(\cot A +\cot B )$
$=\frac{\sin A}{\sin B \sin C} \cdot \frac{\sin B}{\sin C \sin A} \cdot \frac{\sin C}{\sin A \sin B}$
$=\operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$
View full question & answer→MCQ 332 Marks
$\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ} +\operatorname{cosec} 384^{\circ}=$
- A
- B
- ✓
$0$
- D
$\frac{\sqrt{3}}{2}$
Answer(C)
$\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ}$
$=\operatorname{cosec} 48^{\circ}+\operatorname{cosec}\left(180^{\circ}-84^{\circ}\right)$$+\operatorname{cosec}\left(180^{\circ}+12^{\circ}\right)+\operatorname{cosec}\left(360^{\circ}+24^{\circ}\right)$
$=\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 84^{\circ}-\operatorname{cosec} 12^{\circ}+\operatorname{cosec} 24^{\circ}$
$=\frac{1}{\sin 48^{\circ}}+\frac{1}{\sin 84^{\circ}}+\frac{1}{-\sin 12^{\circ}}+\frac{1}{\sin 24^{\circ}}$
$=\left(\frac{1}{\sin 48^{\circ}}-\frac{1}{\sin 12^{\circ}}\right)+\left(\frac{1}{\sin 84^{\circ}}+\frac{1}{\sin 24^{\circ}}\right)$
$=-\frac{\left(\sin 48^{\circ}-\sin 12^{\circ}\right)}{\sin 48^{\circ} \sin 12^{\circ}}, \frac{\left(\sin 84^{\circ}+\sin 24^{\circ}\right)}{\sin 84^{\circ} \sin 24^{\circ}}$
$=-\frac{2 \cos 30^{\circ} \sin 18^{\circ}}{\frac{1}{2}\left(\cos 36^{\circ}-\cos 60^{\circ}\right)}+\frac{2 \sin 54^{\circ} \cos 30^{\circ}}{\frac{1}{2}\left(\cos 60^{\circ}-\cos 108^{\circ}\right)}$
$=\frac{4 \cos 30^{\circ} \sin 18^{\circ}}{\cos 60^{\circ}-\cos 36^{\circ}}+\frac{4 \sin 54^{\circ} \cos 30^{\circ}}{\cos 60^{\circ}+\sin 18^{\circ}}$
$=4 \cos 30^{\circ}\left[\frac{\sin 18^{\circ}}{\cos 60^{\circ}-\cos 36^{\circ}}+\frac{\sin 54^{\circ}}{\cos 60^{\circ}+\sin 18^{\circ}}\right]$
$=4 \cos 30^{\circ}\left[\frac{\sin 18^{\circ}}{\cos 60^{\circ}-\cos 36^{\circ}}+\frac{\cos 36^{\circ}}{\cos 60^{\circ}+\sin 18^{\circ}}\right]$
$=4 \cos 30^{\circ}\left[\frac{\frac{\sqrt{5}-1}{4}}{\frac{1}{2}-\frac{\sqrt{5}+1}{4}}+\frac{\frac{\sqrt{5}+1}{4}}{\frac{1}{2}+\frac{\sqrt{5}-1}{4}}\right]$
$=4 \cos 30^{\circ}(-1+1)=0$
View full question & answer→MCQ 342 Marks
If $A =\tan 6^{\circ} \tan 42^{\circ}$ and $B =\cot 66^{\circ} \cot 78^{\circ}$, then
Answer(B)
$\frac{A}{B}=\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}$
$=\frac{\left(2 \sin 66^{\circ} \sin 6^{\circ}\right)\left(2 \sin 78^{\circ} \sin 42^{\circ}\right)}{\left(2 \cos 66^{\circ} \cos 6^{\circ}\right)\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)}$
$=\frac{\left(\cos 60^{\circ}-\cos 72^{\circ}\right)\left(\cos 36^{\circ}-\cos 120^{\circ}\right)}{\left(\cos 60^{\circ}+\cos 72^{\circ}\right)\left(\cos 36^{\circ}+\cos 120^{\circ}\right)}$
$=\frac{\left(\cos 60^{\circ}-\sin 18^{\circ}\right)\left(\cos 36^{\circ}+\sin 30^{\circ}\right)}{\left(\cos 60^{\circ}+\sin 18^{\circ}\right)\left(\cos 36^{\circ}-\sin 30^{\circ}\right)}$
$\ldots\left[\because \cos \left(90^{\circ}+\theta\right)=-\sin \theta\right]$
$=\frac{\left(\frac{1}{2}-\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right)}{\left(\frac{1}{2}+\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}-\frac{1}{2}\right)}$
$=\frac{9-5}{5-1}=1$
$\Rightarrow A=B$
View full question & answer→MCQ 352 Marks
$2 \sin ^2 \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta +\cos 2(\alpha+\beta)=$
- A
$\sin 2 \alpha$
- B
$\cos 2 \beta$
- ✓
$\cos 2 \alpha$
- D
$\sin 2 \beta$
AnswerCorrect option: C. $\cos 2 \alpha$
(C)
$\cos 2(\alpha+\beta)=2 \cos ^2(\alpha+\beta)-1$
and $2 \sin ^2 \beta=1-\cos 2 \beta$
Now, $2 \sin ^2 \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta$$+\cos 2(\alpha+\beta)$
$=1-\cos 2 \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta$$+2 \cos ^2(\alpha+\beta)-1$
$=2 \cos (\alpha+\beta)[2 \sin \alpha \sin \beta$$+\cos (\alpha+\beta)]-\cos 2 \beta$
$=-\cos 2 \beta+2 \cos (\alpha+\beta) \cos (\alpha-\beta)$
$=-\cos 2 \beta+\cos 2 \alpha+\cos 2 \beta$
$=\cos 2 \alpha$
View full question & answer→MCQ 362 Marks
The expression
$ \cos ^2(A-B)+\cos ^2 B-2 \cos (A-B) \cos A \cos B $ is
Answer(C)
$\cos ^2(A-B)+\cos ^2 B-2 \cos (A-B) \cos A \cos B$
$=\cos ^2(A- B )+\cos ^2 B$$-\cos ( A - B )\{\cos ( A - B )+\cos ( A + B )\}$
$=\cos ^2 B-\cos ( A - B ) \cos ( A + B )$
$=\cos ^2 B-\left(\cos ^2 A-\sin ^2 B\right)$
$=1-\cos ^2 A$
Hence, it depends on A.
View full question & answer→MCQ 372 Marks
$ \cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7} -\cos \frac{4 \pi}{7} +\cos \frac{5 \pi}{7}-\cos \frac{6 \pi}{7}=$
- A
$0$
- B
$\frac{3}{2}$
- C
$\frac{3}{4}$
- ✓
Answer(D)
$\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}-\cos \frac{4 \pi}{7}+\cos \frac{5 \pi}{7}$$-\cos \frac{6 \pi}{7}$
$=\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{3 \pi}{7}$$-\cos \frac{2 \pi}{7}+\cos \frac{\pi}{7}$$\ldots[\because \cos (\pi-\theta)=-\cos \theta]$
$=2\left(\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}\right)$
$=\frac{2 \cos \frac{\pi}{14}\left(\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}\right)}{\cos \frac{\pi}{14}}$
$=\frac{1}{\cos \frac{\pi}{14}}\left(2 \cos \frac{\pi}{7} \cos \frac{\pi}{14}-2 \cos \frac{2 \pi}{7} \cos \frac{\pi}{14}\right.$$\left.+2 \cos \frac{3 \pi}{7} \cos \frac{\pi}{14}\right)$
$=\frac{1}{\cos \frac{\pi}{14}}\left[\left(\cos \frac{3 \pi}{14}+\cos \frac{\pi}{14}\right)-\left(\cos \frac{5 \pi}{14}+\cos \frac{3 \pi}{14}\right)\right.$$\left.+\left(\cos \frac{7 \pi}{14}+\cos \frac{5 \pi}{14}\right)\right]$
$=\frac{1}{\cos \frac{\pi}{14}}\left(\cos \frac{\pi}{14}\right)=1$
View full question & answer→MCQ 382 Marks
The value of $\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{7 \pi}{7}=$
- A
- B
- C
$\frac{1}{2}$
- ✓
$\frac{-3}{2}$
AnswerCorrect option: D. $\frac{-3}{2}$
(D)
$\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{7 \pi}{7}$
$=\frac{1}{2 \sin \frac{\pi}{7}}\left\{2 \cos \frac{2 \pi}{7} \sin \frac{\pi}{7}+2 \cos \frac{4 \pi}{7} \sin \frac{\pi}{7}\right.$$\left.+2 \cos \frac{6 \pi}{7} \sin \frac{\pi}{7}\right\}+\cos \pi$
$=\frac{1}{2 \sin \frac{\pi}{7}}\left[\sin \frac{3 \pi}{7}-\sin \frac{\pi}{7}+\sin \frac{5 \pi}{7}-\sin \frac{3 \pi}{7}\right.$$\left.+\sin \pi-\sin \frac{5 \pi}{7}\right]+\cos \pi$
$=-\frac{1}{2}-1=-\frac{3}{2}$
View full question & answer→MCQ 392 Marks
$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=$
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- ✓
$\frac{1}{8}$
- D
$\frac{1}{16}$
AnswerCorrect option: C. $\frac{1}{8}$
(C)
$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)$
$=\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)$$\ldots[\because \cos (\pi-\theta)=-\cos \theta]$
$=\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right)$
$=\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8}$
$=\frac{1}{4}\left(2 \sin \frac{\pi}{8} \cdot \sin \frac{3 \pi}{8}\right)^2$
$=\frac{1}{4}\left(\cos \frac{\pi}{4}-\cos \frac{\pi}{2}\right)^2$
$=\frac{1}{8}$
View full question & answer→MCQ 402 Marks
The value of $\frac{\tan 70^{\circ}-\tan 20^{\circ}}{\tan 50^{\circ}}=$
Answer(B)
$\frac{\tan 70^{\circ}-\tan 20^{\circ}}{\tan 50^{\circ}}=\frac{\frac{\sin 70^{\circ}}{\cos 70^{\circ}}-\frac{\sin 20^{\circ}}{\cos 20^{\circ}}}{\frac{\sin 50^{\circ}}{\cos 50^{\circ}}}$
$=\frac{\frac{\sin 70^{\circ} \cos 20^{\circ}-\cos 70^{\circ} \sin 20^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}}}{\frac{\sin 50^{\circ}}{\cos 50^{\circ}}}$
$=\frac{\sin \left(70^{\circ}-20^{\circ}\right) \cos 50^{\circ}}{\cos 70^{\circ} \cos 20^{\circ} \sin 50^{\circ}}$
$=\frac{2 \sin 50^{\circ} \cos 50^{\circ}}{2 \cos 70^{\circ} \cos 20^{\circ} \sin 50^{\circ}}$
$=\frac{2 \cos 50^{\circ}}{\cos 90^{\circ}+\cos 50^{\circ}}$
$=\frac{2 \cos 50^{\circ}}{0+\cos 50^{\circ}}$
= 2
View full question & answer→MCQ 412 Marks
$\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}=$
Answer(C)
$\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}$
$=\tan 20^{\circ} \tan \left(60^{\circ}-20^{\circ}\right) \cdot \sqrt{3} \cdot \tan \left(60^{\circ}+20^{\circ}\right)$
$=\sqrt{3} \tan 20^{\circ} \cdot \tan \left(60^{\circ}-20^{\circ}\right) \cdot \tan \left(60^{\circ}+20^{\circ}\right)$
$\tan \theta \tan \left(60^{\circ}-\theta\right) \tan \left(60^{\circ}+\theta\right)=\tan 3 \theta$
$=\sqrt{3} \tan 3\left(20^{\circ}\right)$
$=\sqrt{3} \cdot \sqrt{3}=3$
View full question & answer→MCQ 422 Marks
$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=$
- A
$\frac{-3}{16}$
- B
$\frac{5}{16}$
- ✓
$\frac{3}{16}$
- D
$\frac{-5}{16}$
AnswerCorrect option: C. $\frac{3}{16}$
(C)
$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$
$=\frac{\sqrt{3}}{2} \sin 20^{\circ} \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right)$
$\sin \theta \sin \left(60^{\circ}-\theta\right) \sin \left(60^{\circ}+\theta\right)=\frac{1}{4} \sin 3 \theta$
$=\frac{\sqrt{3}}{2} \cdot \frac{1}{4} \sin 60^{\circ}$
$=\frac{\sqrt{3}}{8} \cdot \frac{\sqrt{3}}{2}$
$=\frac{3}{16}$
View full question & answer→MCQ 432 Marks
$\sin 12^{\circ} \sin 24^{\circ} \sin 48^{\circ} \sin 84^{\circ}=$
AnswerCorrect option: A. $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$
(A)
$\sin 12^{\circ} \sin 24^{\circ} \sin 48^{\circ} \sin 84^{\circ}$
$=\frac{1}{4}\left(2 \sin 12^{\circ} \sin 48^{\circ}\right)\left(2 \sin 24^{\circ} \sin 84^{\circ}\right)$
$=\frac{1}{2}\left(\cos 36^{\circ}-\cos 60^{\circ}\right)\left(\cos 60^{\circ}-\cos 108^{\circ}\right)$
$=\frac{1}{4}\left(\cos 36^{\circ}-\frac{1}{2}\right)\left(\frac{1}{2}+\sin 18^{\circ}\right)$
$=\frac{1}{4}\left\{\frac{1}{4}(\sqrt{5}+1)-\frac{1}{2}\right\}\left\{\frac{1}{2}+\frac{1}{4}(\sqrt{5}-1)\right\}=\frac{1}{16}$
Consider, $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$
$=\frac{1}{2}\left[\cos \left(60^{\circ}-20^{\circ}\right) \cos 20^{\circ} \cos \left(60^{\circ}+20^{\circ}\right)\right]$
$\cos \theta \cos \left(60^{\circ}-\theta\right) \cos \left(60^{\circ}+\theta\right)=\frac{1}{4} \cos 3 \theta$
$=\frac{1}{2}\left[\frac{1}{4} \cos 3\left(20^{\circ}\right)\right]$
$=\frac{1}{8} \cos 60^{\circ}=\frac{1}{8} \times \frac{1}{2}=\frac{1}{16}$
$\therefore$ option (A) is the correct answer.
View full question & answer→MCQ 442 Marks
The value of $\sin \frac{\pi}{16} \sin \frac{3 \pi}{16} \sin \frac{5 \pi}{16} \sin \frac{7 \pi}{16}$ is
- A
$\frac{1}{16}$
- ✓
$\frac{\sqrt{2}}{16}$
- C
$\frac{1}{8}$
- D
$\frac{\sqrt{2}}{8}$
AnswerCorrect option: B. $\frac{\sqrt{2}}{16}$
(B)
$\sin \frac{\pi}{16} \sin \frac{3 \pi}{16} \sin \frac{5 \pi}{16} \sin \frac{7 \pi}{16}$
$=\frac{1}{4}\left[2 \sin \frac{\pi}{16} \sin \frac{3 \pi}{16} \cdot 2 \sin \frac{5 \pi}{16} \sin \frac{7 \pi}{16}\right]$
$=\frac{1}{4}\left[\left(\cos \frac{\pi}{8}-\cos \frac{\pi}{4}\right)\left(\cos \frac{\pi}{8}-\cos \frac{3 \pi}{4}\right)\right]$
$=\frac{1}{4}\left[\left(\cos \frac{\pi}{8}-\frac{1}{\sqrt{2}}\right)\left(\cos \frac{\pi}{8}+\frac{1}{\sqrt{2}}\right)\right]$
$=\frac{1}{4}\left[\left(\cos ^2 \frac{\pi}{8}-\frac{1}{2}\right)\right]$
$=\frac{1}{8}\left[2 \cos ^2 \frac{\pi}{8}-1\right]=\frac{1}{8}\left[\cos \frac{\pi}{4}\right]$
$=\frac{1}{8} \times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{16}$
View full question & answer→MCQ 452 Marks
If $\cos A=\frac{3}{4}$, then the value of $2 \sin \frac{A}{2} \sin \frac{5 A}{2}=$
- A
$\frac{\sqrt{11}}{16}$
- B
$-\frac{\sqrt{11}}{16}$
- ✓
$\frac{11}{16}$
- D
$-\frac{11}{16}$
AnswerCorrect option: C. $\frac{11}{16}$
(C)
$2 \sin \frac{5 A}{2} \sin \frac{A}{2}=\cos 2 A-\cos 3 A$
$=2 \cos ^2 A-1-4 \cos ^3 A+3 \cos A$
$=2\left(\frac{9}{16}\right)-1-4\left(\frac{27}{64}\right)+3\left(\frac{3}{4}\right)$$\ldots\left[\because \cos A =\frac{3}{4}\right]$
$=\frac{9}{8}-1-\frac{27}{16}+\frac{9}{4}=\frac{11}{16}$
View full question & answer→MCQ 462 Marks
The value of $\cos \frac{\pi}{11}+\cos \frac{3 \pi}{11}+\cos \frac{5 \pi}{11}+\cos \frac{7 \pi}{11}+\cos \frac{9 \pi}{11}$ is
- A
$0$
- B
- ✓
$\frac{1}{2}$
- D
$\frac{1}{4}$
AnswerCorrect option: C. $\frac{1}{2}$
(C)
$\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)$$+\ldots \ldots+\cos \{\alpha+(n-1) \beta\}$
$=\frac{\cos \left\{\alpha+(n-1) \frac{\beta}{2}\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}}$
Here, $\alpha=\frac{\pi}{11}$ and $\beta=\frac{2 \pi}{11}$
$\therefore \cos \frac{\pi}{11}+\cos \frac{3 \pi}{11}+\cos \frac{5 \pi}{11}+\cos \frac{7 \pi}{11}+\cos \frac{9 \pi}{11}$
$=\frac{\cos \left(\frac{\pi}{11}+\frac{4 \pi}{11}\right) \sin \left(\frac{5 \pi}{11}\right)}{\sin \left(\frac{\pi}{11}\right)}$
$=\frac{\cos \frac{5 \pi}{11} \sin \frac{5 \pi}{11}}{\sin \frac{\pi}{11}}$
$=\frac{1}{2} \frac{\sin \frac{10 \pi}{11}}{\sin \frac{\pi}{11}} \quad \ldots[\because 2 \sin \theta \cos \theta=\sin 2 \theta]$
$=\frac{1}{2}$
View full question & answer→MCQ 472 Marks
$\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$
Answer(C)
$\cos \alpha+\cos 2 \alpha+\cos 3 \alpha+\ldots+\cos$ n $\alpha$
$\cos \alpha+\cos (\alpha+\beta)+ \cos (\alpha+2 \beta) +\ldots \ldots+\cos [\alpha+(n-1) \beta]$
$=\frac{\cos \left[\alpha+(n-1) \frac{\beta}{2}\right] \cdot \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}}$
If $\beta=\alpha$, then
$\cos \alpha+\cos 2 \alpha+\cos 3 \alpha+\ldots \ldots+\cos n \alpha$
$=\frac{\cos \left(\frac{ n +1}{2}\right) \alpha \cdot \sin \left(\frac{ n \alpha}{2}\right)}{\sin \left(\frac{\alpha}{2}\right)}$
Here, $n =3$ and $\alpha=\frac{2 \pi}{7}$
$\therefore \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$
$=\frac{\cos \left(\frac{3+1}{2}\right)\left(\frac{2 \pi}{7}\right) \sin \left(\frac{3 \times 2 \pi}{2 \times 7}\right)}{\sin \left(\frac{2 \pi}{7 \times 2}\right)}$
$=\frac{\cos \left(\frac{4 \pi}{7}\right) \cdot \sin \left(\frac{3 \pi}{7}\right)}{\sin \left(\frac{\pi}{7}\right)}$
Since the values of $\cos \left(\frac{4 \pi}{7}\right), \sin \left(\frac{3 \pi}{7}\right)$ and $\sin \left(\frac{\pi}{7}\right)$ are $- ve ,+ ve$ and $+ve$ respectively.
∴ option $( C )$ is the correct answer.
View full question & answer→MCQ 482 Marks
The sum $S =\sin \theta+\sin 2 \theta+\ldots .+\sin n \theta$, equals
- ✓
$\frac{\sin \frac{ n \theta}{2} \cdot \sin \frac{\theta( n +1)}{2}}{\sin \frac{\theta}{2}}$
- B
$\frac{\sin \frac{ n \theta}{2} \cdot \cos \frac{\theta( n +1)}{2}}{\sin \frac{\theta}{2}}$
- C
$\frac{\cos \frac{ n \theta}{2} \cdot \sin \frac{\theta( n +1)}{2}}{\sin \frac{\theta}{2}}$
- D
$\frac{\cos \frac{ n \theta}{2} \cdot \cos \frac{\theta( n +1)}{2}}{\sin \frac{\theta}{2}}$
AnswerCorrect option: A. $\frac{\sin \frac{ n \theta}{2} \cdot \sin \frac{\theta( n +1)}{2}}{\sin \frac{\theta}{2}}$
(A)
$S =\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots+\sin n \theta$
Since,
$\begin{aligned} \sin \theta+\sin (\theta+\beta)+ & \sin (\theta+2 \beta) \ldots+\sin [\theta+(n-1) \beta]\end{aligned}$
$=\frac{\sin \frac{ n \beta}{2}}{\sin \frac{\beta}{2}} \sin \left[\theta+\left(\frac{( n -1)}{2}\right) \beta\right]$
Here, $\beta=\theta$
$\therefore S=\frac{\sin \frac{n \theta}{2} \cdot \sin \frac{\theta(n+1)}{2}}{\sin \frac{\theta}{2}}$
View full question & answer→MCQ 492 Marks
$\frac{\sin ^2 A-\sin ^2 B}{\sin A \cos A-\sin B \cos B}=$
Answer(B)
$\frac{\sin ^2 A-\sin ^2 B}{\sin A \cos A-\sin B \cos B}$
$=\frac{2 \sin (A+B) \sin (A-B)}{\sin 2 A-\sin 2 B}$
$\ldots\left[\because \sin ^2 A-\sin ^2 B=\sin ( A + B ) \sin ( A - B )\right]$
$=\frac{2 \sin (A+B) \sin (A-B)}{2 \cos (A+B) \sin (A-B)}$
$=\tan ( A + B )$
View full question & answer→MCQ 502 Marks
If $\cos A=m \cos B$, then
- ✓
$\cot \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \tan \left(\frac{B- A }{2}\right)$
- B
$\tan \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \cot \left(\frac{B- A }{2}\right)$
- C
$\cot \left(\frac{A+B}{2}\right)=\frac{m+1}{m-1} \tan \left(\frac{A-B}{2}\right)$
- D
AnswerCorrect option: A. $\cot \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \tan \left(\frac{B- A }{2}\right)$
(A)
Given that, $\cos A=m \cos B$
$\Rightarrow \frac{m}{1}=\frac{\cos A}{\cos B}$
By componendo and dividendo, we get
$\frac{m+1}{m-1}=\frac{\cos A+\cos B}{\cos A-\cos B}$
$=\frac{2 \cos \left(\frac{A+ B }{2}\right) \cos \left(\frac{ B - A }{2}\right)}{2 \sin \left(\frac{A+ B }{2}\right) \sin \left(\frac{ B - A }{2}\right)}$
$\ldots[\because \cos (B-A)=\cos (A-B)]$
$=\cot \left(\frac{ A + B }{2}\right) \cot \left(\frac{ B - A }{2}\right)$
$\Rightarrow \cot \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \tan \left(\frac{B- A }{2}\right)$
View full question & answer→MCQ 512 Marks
If $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$, then tan A tan B tan C + tan D =
Answer(A)
$\frac{\cos ( A + B )}{\cos ( A - B )}=\frac{\sin ( C + D )}{\sin ( C - D )}$
By componendo and dividendo, we get
$\frac{\cos ( A + B )+\cos ( A - B )}{\cos ( A + B )-\cos ( A - B )}-\frac{\sin ( C + D )+\sin ( C - D )}{\sin ( C + D )-\sin ( C - D )}$
$\Rightarrow \frac{2 \cos A \cos B}{-2 \sin A \sin B}=\frac{2 \sin C \cos D}{2 \cos C \sin D}$
$\Rightarrow-\cot A \cot B =\tan C \cot D$
$\Rightarrow \tan A \tan B \tan C+\tan D=0$
View full question & answer→MCQ 522 Marks
$\frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}=$
- A
$\frac{\cos B +\sin B }{\cos B -\sin B }$
- ✓
$\frac{\cos A+\sin A}{\cos A-\sin A}$
- C
$\frac{\cos A-\sin A}{\cos A+\sin A}$
- D
$\frac{\cos B-\sin B}{\cos B+\sin B}$
AnswerCorrect option: B. $\frac{\cos A+\sin A}{\cos A-\sin A}$
(B)
$\frac{\sin ( B + A )+\cos ( B - A )}{\sin ( B - A )+\cos ( B + A )}$
$=\frac{\sin ( B + A )+\sin \left\{\left(90^{\circ}-( B - A )\right\}\right.}{\sin ( B - A )+\sin \left\{\left(90^{\circ}-( A + B )\right\}\right.}$
$=\frac{2 \sin \left(A+45^{\circ}\right) \cos \left(45^{\circ}- B \right)}{2 \sin \left(45^{\circ}- A \right) \cos \left(45^{\circ}- B \right)}$
$=\frac{\sin \left(A+45^{\circ}\right)}{\sin \left(45^{\circ}-A\right)}=\frac{\cos A+\sin A}{\cos A-\sin A}$
View full question & answer→MCQ 532 Marks
$\cos ^2 \alpha+\cos ^2\left(\alpha+120^{\circ}\right)+\cos ^2\left(\alpha-120^{\circ}\right)$ is equal to
- ✓
$\frac{3}{2}$
- B
- C
$\frac{1}{2}$
- D
$0$
AnswerCorrect option: A. $\frac{3}{2}$
(A)
$\cos ^2 \alpha+\cos ^2\left(\alpha+120^{\circ}\right)+\cos ^2\left(\alpha-120^{\circ}\right)$
$=\cos ^2 \alpha+\left\{\cos \left(\alpha+120^{\circ}\right)+\cos \left(\alpha-120^{\circ}\right)\right\}^2$$-2 \cos \left(\alpha+120^{\circ}\right) \cos \left(\alpha-120^{\circ}\right)$
$=\cos ^2 \alpha+\left\{2 \cos \alpha \cos 120^{\circ}\right\}^2$$-2\left\{\cos ^2 \alpha-\sin ^2 120^{\circ}\right\}$
$=\cos ^2 \alpha+\cos ^2 \alpha-2 \cos ^2 \alpha+2 \sin ^2 120^{\circ}$
$=2 \sin ^2 120^{\circ}$
$=2 \times \frac{3}{4}=\frac{3}{2}$
View full question & answer→MCQ 542 Marks
If $\cos x=3 \cos y$, then $2 \tan \left(\frac{y-x}{2}\right)=$
- A
$\cot \left(\frac{y-x}{2}\right)$
- B
$\cot \left(\frac{x+y}{4}\right)$
- C
$\cot \left(\frac{y-x}{4}\right)$
- ✓
$\cot \left(\frac{x+y}{2}\right)$
AnswerCorrect option: D. $\cot \left(\frac{x+y}{2}\right)$
(D)
$\cos x=3 \cos y \Rightarrow \frac{\cos x}{\cos y}=\frac{3}{1}$
Bt componendo and dividendo, we get $\frac{\cos x+\cos y}{\cos x-\cos y}=\frac{3+1}{3-1}$
$\Rightarrow \frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}=\frac{4}{2}$
$\Rightarrow-\cot \left(\frac{x+y}{2}\right) \cot \left(\frac{x-y}{2}\right)=2$
$\Rightarrow \cot \left(\frac{x+y}{2}\right) \cot \left(\frac{y-x}{2}\right)=2$
$\Rightarrow 2 \tan \left(\frac{y-x}{2}\right)=\cot \left(\frac{x+y}{2}\right)$
View full question & answer→MCQ 552 Marks
If $\sin x+\sin y=\frac{1}{2}$ and $\cos x+\cos y=1$, then $\tan (x+y)=$
- A
$\frac{-8}{3}$
- B
$\frac{8}{3}$
- ✓
$\frac{8}{3}$
- D
$-\frac{3}{4}$
AnswerCorrect option: C. $\frac{8}{3}$
(C)
$\sin x+\sin y=\frac{1}{2}$
$\Rightarrow 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{1}{2}\quad\ldots(i)$
$\cos x+\cos y=1$
$\Rightarrow 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=1\quad\ldots(ii)$
Dividing (i) by (ii), we get
$\tan \left(\frac{x+y}{2}\right)=\frac{1}{2}$
Now, $\tan (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1-\tan ^2\left(\frac{x+y}{2}\right)}$
$=\frac{2\left(\frac{1}{2}\right)}{1-\frac{1}{4}}-\frac{4}{3}$
View full question & answer→MCQ 562 Marks
If $\cos x+\cos y+\cos \alpha=0$ and $\sin x+\sin y+\sin \alpha=0$, then $\cot \left(\frac{x+y}{2}\right)=$
AnswerCorrect option: C. $\cot \alpha$
(C)
$\cos x+\cos y+\cos \alpha=0$
$\Rightarrow \cos x+\cos y=-\cos \alpha$
$ \Rightarrow 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\cos \alpha\quad\ldots(i)$
Also, $\sin x +\sin y +\sin \alpha=0$
$\Rightarrow \sin x+\sin y=-\sin \alpha$
$\Rightarrow 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=-\sin \alpha\quad\ldots(ii)$
Dividing (i) and (ii), w e get
$\frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{\cos \alpha}{\sin \alpha}$
$\Rightarrow \cot \left(\frac{x+y}{2}\right)=\cot \alpha$
View full question & answer→MCQ 572 Marks
If $\sin \theta+\sin 2 \theta+\sin 3 \theta=\sin \alpha$ and $\cos \theta+\cos 2 \theta+\cos 3 \theta=\cos \alpha$, then $\theta$ is equal to
- ✓
$\frac{\alpha}{2}$
- B
$\alpha$
- C
$2 \alpha$
- D
$\frac{\alpha}{6}$
AnswerCorrect option: A. $\frac{\alpha}{2}$
(A)
$\sin \theta+\sin 3 \theta+\sin 2 \theta=\sin \alpha$
$\Rightarrow 2 \sin 2 \theta \cos \theta+\sin 2 \theta=\sin \alpha$
$\Rightarrow \sin 2 \theta(2 \cos \theta+1)=\sin \alpha\quad\ldots(i)$
Also, $\cos \theta+\cos 3 \theta+\cos 2 \theta=\cos \alpha$
$\Rightarrow 2 \cos 20 \cos 0 + \cos 20-\cos \alpha$
$\Rightarrow \cos 2 \theta(2 \cos \theta+1)=\cos \alpha\quad\ldots(ii)$
From (i) and (ii), we get
$\tan 2 \theta=\tan \alpha$
$\Rightarrow 2 \theta=\alpha$
$\Rightarrow \theta=\frac{\alpha}{2}$
View full question & answer→MCQ 582 Marks
$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=$
Answer(C)
$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$
$=\tan 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ}+\cot 9^{\circ}$$\ldots\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$
$=\left(\tan 9^{\circ}+\cot 9^{\circ}\right)-\left(\tan 27^{\circ}+\cot 27^{\circ}\right)$
$=\frac{1}{\sin 9^{\circ} \cos 9^{\circ}}-\frac{1}{\sin 27^{\circ} \cos 27^{\circ}}$
$=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \quad \ldots[\because \sin 2 \theta=2 \sin \theta \cos \theta]$
$=2\left\{\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}\right\}$
$=2 \cdot \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}$
$=\frac{4 \cos 36^{\circ}}{\cos 36^{\circ}}$
= 4
View full question & answer→MCQ 592 Marks
The expression $\frac{\cos 6 x+6 \cos 4 x+15 \cos 2 x+10}{\cos 5 x+5 \cos 3 x+10 \cos x}$ is equal is
- A
$\cos 2 x$
- ✓
$2 \cos x$
- C
$\cos ^2 x$
- D
$1+\cos x$
AnswerCorrect option: B. $2 \cos x$
(B)
$\frac{\cos 6 x+6 \cos 4 x+15 \cos 2 x+10}{\cos 5 x+5 \cos 3 x+10 \cos x}$
$=\frac{(\cos 6 x+\cos 4 x)+5(\cos 4 x+\cos 2 x)+10(\cos 2 x+1)}{\cos 5 x+5 \cos 3 x+10 \cos x}$
$=\frac{2 \cos x \cos 5 x+10 \cos x \cos 3 x+10\left(2 \cos ^2 x-1+1\right)}{\cos 5 x+5 \cos 3 x+10 \cos x}$
$=\frac{2 \cos x(\cos 5 x+5 \cos 3 x+10 \cos x)}{\cos 5 x+5 \cos 3 x+10 \cos x}$
= 2 cos x
View full question & answer→MCQ 602 Marks
$\cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}=$
- A
$-\frac{1}{4}$
- B
$\frac{1}{2}$
- C
$0$
- ✓
$\frac{3}{4}$
AnswerCorrect option: D. $\frac{3}{4}$
(D)
$\cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}$
$=\frac{1}{2}\left [1+\cos 152^{\circ}+1+\cos 32^{\circ}-\cos 92^{\circ}-\cos 60^{\circ}\right]$
$=\frac{1}{2}\left[2-\frac{1}{2}+\cos 152^{\circ}+\cos 32^{\circ}-\cos 92^{\circ}\right]$
$=\frac{1}{2}\left[\frac{3}{2}+2 \cos 92^{\circ} \cos 60^{\circ}-\cos 92^{\circ}\right]$
$=\frac{1}{2}\left[\frac{3}{2}+\cos 92^{\circ}-\cos 92^{\circ}\right]$
$=\frac{3}{4}$
View full question & answer→MCQ 612 Marks
$2 \cos x-\cos 3 x-\cos 5 x=$
- ✓
$16 \cos ^3 x \sin ^2 x$
- B
$16 \sin ^3 x \cos ^2 x$
- C
$4 \cos ^3 x \sin ^2 x$
- D
$4 \sin ^3 x \cos ^2 x$
AnswerCorrect option: A. $16 \cos ^3 x \sin ^2 x$
(A)
$2 \cos x-\cos 3 x-\cos 5 x$
$=2 \cos x(1-\cos 4 x)$
$=2 \cos x 2 \sin ^2 2 x$
$=4 \cos x \sin ^2 2 x$
$=4 \cos x(2 \sin x \cos x)^2$
$=16 \sin ^2 x \cos ^3 x$
View full question & answer→MCQ 622 Marks
$1+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}=$
- A
$2 \cos 28^{\circ} \cos 29^{\circ} \cos 33^{\circ}$
- B
$4 \cos 28^{\circ} \cos 29^{\circ} \cos 33^{\circ}$
- ✓
$4 \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}$
- D
$2 \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}$
AnswerCorrect option: C. $4 \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}$
(C)
$1+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}$
$=2 \cos ^2 28^{\circ}+2 \sin 62^{\circ} \sin 4^{\circ}$
$=2 \cos ^2 28^{\circ}+2 \cos 28^{\circ} \cos 86^{\circ}$$\ldots\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
$=2 \cos 28^{\circ}\left(\cos 28^{\circ}+\cos 86^{\circ}\right)$
$=2 \cos 28^{\circ} .2 \cos 57^{\circ} \cos 29^{\circ}$
$=4 \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}$$\ldots\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
View full question & answer→MCQ 632 Marks
$1+\cos 10^{\circ}+\cos 20^{\circ}+\cos 30^{\circ}=$
- ✓
$4 \cos 5^{\circ} \cos 10^{\circ} \cos 15^{\circ}$
- B
$4 \cos 10^{\circ} \cos 20^{\circ} \cos 30^{\circ}$
- C
$4 \sin 5^{\circ} \sin 10^{\circ} \sin 15^{\circ}$
- D
$4 \sin 10^{\circ} \sin 20^{\circ} \sin 30^{\circ}$
AnswerCorrect option: A. $4 \cos 5^{\circ} \cos 10^{\circ} \cos 15^{\circ}$
(A)
$1+\cos 10^{\circ}+\cos 20^{\circ}+\cos 30^{\circ}$
$=2 \cos ^2 5^{\circ}+2 \cos 25^{\circ} \cos 5^{\circ}$
$=2 \cos 5^{\circ}\left(\cos 5^{\circ}+\cos 25^{\circ}\right)$
$=2 \cos 5^{\circ}\left(2 \cos 15^{\circ} \cos 10^{\circ}\right)$
$=4 \cos 5^{\circ} \cos 10^{\circ} \cos 15^{\circ}$
View full question & answer→MCQ 642 Marks
$1+\cos 2 x+\cos 4 x+\cos 6 x=$
- A
$2 \cos x \cos 2 x \cos 3 x$
- B
$4 \sin x \cos 2 x \cos 3 x$
- ✓
$4 \cos x \cos 2 x \cos 3 x$
- D
$2 \sin x \cos 2 x \cos 3 x$
AnswerCorrect option: C. $4 \cos x \cos 2 x \cos 3 x$
(C)
$1+\cos 2 x+\cos 4 x+\cos 6 x$
$=(1+\cos 6 x)+(\cos 2 x+\cos 4 x)$
$=2 \cos ^2 3 x+2 \cos 3 x \cos x$
$=2 \cos 3 x(\cos 3 x+\cos x)$
$=4 \cos x \cos 2 x \cos 3 x$
View full question & answer→MCQ 652 Marks
$\cos 10 x+\cos 8 x+3 \cos 4 x+3 \cos 2 x=$
AnswerCorrect option: D. $8 \cos x \cos ^3 3 x$
(D)
$\cos 10 x+\cos 8 x+3 \cos 4 x+3 \cos 2 x$
$=(\cos 10 x+\cos 8 x)+3(\cos 4 x+\cos 2 x)$
$=2 \cos 9 x \cos x+6 \cos 3 x \cos x$
$=2 \cos x(\cos 9 x+3 \cos 3 x)$
$=2 \cos x[\cos (3(3 x))+3 \cos 3 x]$
$=2 \cos x\left(4 \cos ^3 3 x-3 \cos 3 x+3 \cos 3 x\right)$
$=8 \cos ^3 3 x \cos x$
View full question & answer→MCQ 662 Marks
The value of $\cot 70^{\circ}+4 \cos 70^{\circ}$ is
- A
$\frac{1}{\sqrt{3}}$
- ✓
$\sqrt{3}$
- C
$2 \sqrt{3}$
- D
$\frac{1}{2}$
AnswerCorrect option: B. $\sqrt{3}$
(B)
$\cot 70^{\circ}+4 \cos 70^{\circ}$
$=\frac{\cos 70^{\circ}+4 \sin 70^{\circ} \cos 70^{\circ}}{\sin 70^{\circ}}$
$=\frac{\cos 70^{\circ}+2 \sin 140^{\circ}}{\sin 70^{\circ}}$
$=\frac{\cos \left(90^{\circ}-20^{\circ}\right)+2 \sin \left(180^{\circ}-40^{\circ}\right)}{\sin 70^{\circ}}$
$=\frac{\sin 20^{\circ}+\sin 40^{\circ}+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{2 \sin 30^{\circ} \cos 10^{\circ}+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{\cos 10^{\circ}+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{\sin 80^{\circ}+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{2 \sin 60^{\circ} \cos 20^{\circ}}{\sin 70^{\circ}}$
$=\sqrt{3}$
View full question & answer→MCQ 672 Marks
$\cos A +\cos \left(240^{\circ}+ A \right)+\cos \left(240^{\circ}- A \right)=$
- A
- ✓
$0$
- C
$\sqrt{3} \sin A$
- D
$\sqrt{3} \cos A$
Answer(B)
$\cos A +\cos \left(240^{\circ}+ A \right)+\cos \left(240^{\circ}- A \right)$
$=\cos A +2 \cos 240^{\circ} \cos A$
$=\cos A \left\{1+2 \cos \left(180^{\circ}+60^{\circ}\right)\right\}$
$=\cos A \left\{1+2\left(-\frac{1}{2}\right)\right\}$
= 0
View full question & answer→MCQ 682 Marks
The value of $\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}=$
- A
$\sin 36^{\circ}$
- B
$\cos 36^{\circ}$
- C
$\sin 7^{\circ}$
- ✓
$\cos 7^{\circ}$
AnswerCorrect option: D. $\cos 7^{\circ}$
(D)
$\sin 47^{\circ}+\sin 61^{\circ}-\left(\sin 11^{\circ}+\sin 25^{\circ}\right)$
$=2 \sin 54^{\circ} \cos 7^{\circ}-2 \sin 18^{\circ} \cos 7^{\circ}$
$=2 \cos 7^{\circ}\left(\sin 54^{\circ}-\sin 18^{\circ}\right)$
$=2 \cos 7^{\circ}\left(2 \cos 36^{\circ} \sin 18^{\circ}\right)$
$=4 \cos 7^{\circ} \cdot \frac{\sqrt{5}+1}{4} \cdot \frac{\sqrt{5}-1}{4}$
$=\cos 7^{\circ}$
View full question & answer→MCQ 692 Marks
The value of $\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}$ is
- A
$\frac{1}{2}$
- B
- ✓
$-\frac{1}{2}$
- D
$-\frac{1}{2}$
AnswerCorrect option: C. $-\frac{1}{2}$
(C)
$\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}$
$=\left(\cos 12^{\circ}+\cos 132^{\circ}\right)+\left(\cos 84^{\circ}+\cos 156^{\circ}\right)$
$=2 \cos 72^{\circ} \cos 60^{\circ}+2 \cos 120^{\circ} \cos 36^{\circ}$
$=2 \cos 72^{\circ} \times \frac{1}{2}-2 \times \frac{1}{2} \times \cos 36^{\circ}$
$=\cos 72^{\circ}-\cos 36^{\circ}$
$=\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}=\frac{-1}{2}$
View full question & answer→MCQ 702 Marks
If $\tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3$, then
- A
$\tan 2 \theta=1$
- ✓
$\tan 3 \theta=1$
- C
$\tan ^3 \theta=1$
- D
$\tan ^2 \theta=1$
AnswerCorrect option: B. $\tan 3 \theta=1$
(B)
$\tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3$
$\Rightarrow \tan \theta+\left(\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}+\frac{\tan \theta-\sqrt{3}}{1+\sqrt{3} \tan \theta}\right)=3$
$\Rightarrow \tan \theta+\frac{8 \tan \theta}{1-3 \tan ^2 \theta}=3$
$\Rightarrow \frac{9 \tan \theta-3 \tan ^3 \theta}{1-3 \tan ^2 \theta}=3$
$\Rightarrow 3 \tan 3 \theta=3$
$\Rightarrow \tan 3 \theta=1$
View full question & answer→MCQ 712 Marks
If $\sin 6 \theta=32 \cos ^5 \theta \sin \theta-32 \cos ^3 \theta \sin \theta+3 x$, then $x=$
- A
$\cos \theta$
- B
$\cos 2 \theta$
- C
$\sin \theta$
- ✓
$\sin 2 \theta$
AnswerCorrect option: D. $\sin 2 \theta$
(D)
$\sin 6 \theta=2 \sin 3 \theta \cos 3 \theta$
$=2\left(3 \sin \theta-4 \sin ^3 \theta\right)\left(4 \cos ^3 \theta-3 \cos \theta\right)$
$=24 \sin \theta \cos \theta\left(\sin ^2 \theta+\cos ^2 \theta\right)-18 \sin \theta \cos \theta$$-32 \sin ^3 \theta \cos ^3 \theta$
$=6 \sin \theta \cos \theta-32 \sin \theta \cos ^3 \theta \sin ^2 \theta$
$=3 \sin 2 \theta-32 \sin \theta \cos ^3 \theta\left(1-\cos ^2 \theta\right)$
$=32 \cos ^2 \theta \sin \theta-32 \cos ^2 \theta \sin \theta+3 \sin 2 \theta$
Given,
$\sin 6 \theta=32 \cos ^5 \theta \sin \theta-32 \cos ^3 \theta \sin \theta+3 x$
∴ On comparing, we get $x=\sin 2 \theta$
View full question & answer→MCQ 722 Marks
$\cos ^3 110^{\circ}+\cos ^3 10^{\circ}+\cos ^3 130^{\circ}=$
- A
$\frac{3}{4}$
- B
$\frac{3}{8}$
- ✓
$\frac{3 \sqrt{3}}{8}$
- D
$\frac{3 \sqrt{3}}{4}$
AnswerCorrect option: C. $\frac{3 \sqrt{3}}{8}$
(C)
$\cos ^3 \theta+\cos ^3\left(120^{\circ}-\theta\right)+\cos ^3\left(120^{\circ}+\theta\right)$$=\frac{3}{4} \cos (3 \theta)$
$\therefore \cos ^3 10^{\circ}+\cos ^3 110^{\circ}+\cos ^3 130^{\circ}$
$\begin{aligned}=\cos ^3\left(10^{\circ}\right)+\cos ^3\left(120^{\circ}-10^{\circ}\right) +\cos ^3\left(120^{\circ}+10^{\circ}\right)\end{aligned}$
$=\frac{3}{4} \cos \left(3 \times 10^{\circ}\right)=\frac{3}{4} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{8}$
View full question & answer→MCQ 732 Marks
If $\cos \theta=\frac{1}{2}\left(a+\frac{1}{a}\right)$, then the value of $\cos 3 \theta$ is
- A
$\frac{1}{8}\left(a^3+\frac{1}{a^3}\right)$
- B
$\frac{3}{2}\left(a+\frac{1}{a}\right)$
- ✓
$\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)$
- D
$\frac{1}{3}\left(a^3+\frac{1}{a^3}\right)$
AnswerCorrect option: C. $\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)$
(C)
Since $\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta$
$\therefore \quad \cos 3 \theta=4\left[\frac{1}{2^3}\left(a+\frac{1}{a}\right)^3\right]-3\left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right]$
$\Rightarrow \cos 3 \theta=\frac{1}{2}\left(a+\frac{1}{a}\right)\left[\left(a+\frac{1}{a}\right)^2-3\right]$
$\Rightarrow \cos 3 \theta=\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)$
View full question & answer→MCQ 742 Marks
If $\sin 2 A=\sin 3 A$ and $0 \leq A \leq 90^{\circ}$ then A is equal to
AnswerCorrect option: C. $0^{\circ}$ or $36^{\circ}$
(C)
$\sin 2 A=\sin 3 A$
$\Rightarrow 2 \sin A \cos A=3 \sin A-4 \sin ^3 A$
$\Rightarrow \sin A=0$ or $2 \cos A=3-4 \sin ^2 A$
$\Rightarrow A=0$ or $2 \cos A=3-4\left(1-\cos ^2 A\right)$
$\Rightarrow A=0$ or $4 \cos ^2 A-2 \cos A-1=0$
$\Rightarrow A=0$ or $\cos A=\frac{2 \pm \sqrt{4+16}}{2 \times 4}=\frac{1 \pm \sqrt{5}}{4}$
$\Rightarrow A =0^{\circ}$ or $36^{\circ} \quad \ldots\left[\because 0 \leq A \leq 90^{\circ}\right]$
View full question & answer→MCQ 752 Marks
If $3 \sin 2 \theta=2 \sin 3 \theta, 0<\theta<\pi$, then the value of $\sin \theta$ is
- A
$0$
- ✓
$\frac{\sqrt{15}}{4}$
- C
$-\frac{1}{4}$
- D
$\frac{1}{4}$
AnswerCorrect option: B. $\frac{\sqrt{15}}{4}$
(B)
$3 \sin 2 \theta=2 \sin 3 \theta$
$\Rightarrow 6 \sin \theta \cos \theta=2\left(3 \sin \theta-4 \sin ^3 \theta\right)$
Dividing by $2 \sin \theta \neq 0$, we get
$3 \cos \theta=3-4 \sin ^2 \theta$
$\Rightarrow 3 \cos \theta=3-4\left(1-\cos ^2 \theta\right)$
$\Rightarrow 4 \cos ^2 \theta-3 \cos \theta-1=0$
$\Rightarrow \cos \theta=1,-\frac{1}{4}$
But, $0<\theta<\pi$
$\therefore \cos \theta=-\frac{1}{4}$
$\Rightarrow \sin \theta=\sqrt{1-\frac{1}{16}}=\frac{\sqrt{15}}{4}$
View full question & answer→MCQ 762 Marks
For a positive integer n,let $f_n(\theta)=\left(\tan \frac{\theta}{2}\right)(1+\sec \theta)(1+\sec 2 \theta)$ $(1+\sec 4 \theta) \ldots .\left(1+\sec 2^n \theta\right)$. Then, which one of the following is incorrect?
- A
$f _2\left(\frac{\pi}{16}\right)=1$
- B
$f _3\left(\frac{\pi}{32}\right)=1$
- C
$f _4\left(\frac{\pi}{64}\right)=1$
- ✓
$f_5\left(\frac{\pi}{128}\right)=-1$
AnswerCorrect option: D. $f_5\left(\frac{\pi}{128}\right)=-1$
(D)
$f _{ n }(\theta)=\left(\tan \frac{\theta}{2}\right)(1+\sec \theta)(1+\sec 2 \theta)$$(1+\sec 4 \theta) \ldots\left(1+\sec 2^{ n } \theta\right)$
$=\left(\tan \frac{\theta}{2}\right)\left(\frac{1+\cos \theta}{\cos \theta}\right)(1+\sec 2 \theta)$$(1+\sec 4 \theta) \ldots\left(1+\sec 2^n \theta\right)$
$=\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right) \times \frac{2 \cos ^2 \frac{\theta}{2}}{\cos \theta}(1+\sec 2 \theta)$$(1+\sec 4 \theta) \ldots\left(1+\sec 2^{ n } \theta\right)$
$=\frac{\sin \theta}{\cos \theta}(1+\sec 2 \theta)(1+\sec 4 \theta)$$\ldots\left(1+\sec 2^n \theta\right)$$\ldots[\because \sin 2 A=2 \sin A \cos A ]$
$=\tan \theta\left(\frac{1+\cos 2 \theta}{\cos 2 \theta}\right)(1+\sec 4 \theta)$$\ldots\left(1+\sec 2^{ n } \theta\right)$
$=\frac{\sin \theta}{\cos \theta}\left(\frac{2 \cos ^2 \theta}{\cos 2 \theta}\right)(1+\sec 4 \theta)$$\ldots\left(1+\sec 2^{ n } \theta\right)$
$=\tan 2 \theta(1+\sec 4 \theta) \quad \ldots\left(1+\sec 2^n \theta\right)$
$=\tan 2^n \theta$
$\therefore f_2\left(\frac{\pi}{16}\right)=\tan \left(2^2 \times \frac{\pi}{16}\right)=1$
$f_3\left(\frac{\pi}{32}\right)=\tan \left(2^3 \times \frac{\pi}{32}\right)=1$
$f_4\left(\frac{\pi}{64}\right)=\tan \left(2^4 \times \frac{\pi}{64}\right)=1$
$f_5\left(\frac{\pi}{128}\right)=\tan \left(2^5 \times \frac{\pi}{128}\right)=1$
∴ option (D) is incorrect.
View full question & answer→MCQ 772 Marks
If $\sin \theta+\sin \phi= a$ and $\cos \theta+\cos \phi= b$, then $\tan \frac{\theta-\phi}{2}$ is equal to
- A
$\sqrt{\frac{a^2+b^2}{4-a^2-b^2}}$
- ✓
$\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$
- C
$\sqrt{\frac{a^2+b^2}{4+a^2+b^2}}$
- D
$\sqrt{\frac{4+a^2+b^2}{a^2+b^2}}$
AnswerCorrect option: B. $\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$
(B)
Given that $\sin \theta+\sin \phi=a \quad\ldots(i)$
and $\cos \theta+\cos \phi= b \quad\ldots(ii)$
Squaring (i) and (ii) adding, we get
$2+2(\sin \theta \sin \phi+\cos \theta \cos \phi)= a ^2+ b ^2$
$\Rightarrow 2 \cos (\theta \quad \phi)- a ^2+ b ^2 \quad 2$
$\Rightarrow \cos (\theta-\phi)=\frac{ a ^2+ b ^2-2}{2}$
$\Rightarrow \frac{1-\tan ^2 \frac{\theta-\phi}{2}}{1+\tan ^2 \frac{\theta-\phi}{2}}=\frac{a^2+b^2-2}{2}$
$\Rightarrow\left( a ^2+ b ^2\right)+\left( a ^2+ b ^2\right) \tan ^2 \frac{\theta-\phi}{2}-2$$-2 \tan ^2 \frac{\theta-\phi}{2}=2-2 \tan ^2 \frac{\theta-\phi}{2}$
$\Rightarrow \frac{4-a^2-b^2}{a^2+b^2}=\tan ^2 \frac{\theta-\phi}{2}$
$\Rightarrow \tan \left(\frac{\theta-\phi}{2}\right)=\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$
Trick : Putting $\theta=\frac{\pi}{4}=\phi$, we get $\tan \frac{\theta-\phi}{2}=0$,which is given by option (B).
View full question & answer→MCQ 782 Marks
$\frac{\sqrt{2}-\sin \alpha-\cos \alpha}{\sin \alpha-\cos \alpha}=$
- A
$\sec \left(\frac{\alpha}{2}-\frac{\pi}{8}\right)$
- B
$\cos \left(\frac{\pi}{8}-\frac{\alpha}{2}\right)$
- ✓
$\tan \left(\frac{\alpha}{2}-\frac{\pi}{8}\right)$
- D
$\cot \left(\frac{\alpha}{2}-\frac{\pi}{2}\right)$
AnswerCorrect option: C. $\tan \left(\frac{\alpha}{2}-\frac{\pi}{8}\right)$
(C)
$\frac{\sqrt{2}-\sin \alpha-\cos \alpha}{\sin \alpha-\cos \alpha}$
$=\frac{\sqrt{2}-\sqrt{2}\left\{\frac{1}{\sqrt{2}} \sin \alpha+\frac{1}{\sqrt{2}} \cos \alpha\right\}}{\sqrt{2}\left\{\frac{1}{\sqrt{2}} \sin \alpha-\frac{1}{\sqrt{2}} \cos \alpha\right\}}$
$=\frac{\sqrt{2}-\sqrt{2} \cos \left(\alpha-\frac{\pi}{4}\right)}{\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right)}$
$=\frac{\sqrt{2}(1-\cos \theta)}{\sqrt{2} \sin \theta}$, where $\theta=\alpha-\frac{\pi}{4}$
$=\frac{2 \sin ^2\left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}$
$=\tan \frac{\theta}{2}=\tan \left(\frac{\alpha}{2}-\frac{\pi}{8}\right) \quad \ldots\left[\because \theta=\alpha-\frac{\pi}{4}\right]$
View full question & answer→MCQ 792 Marks
If $\alpha=22^{\circ} 30^{\prime}$, then
$(1+\cos \alpha)(1+\cos 3 \alpha)(1+\cos 5 \alpha)(1+\cos 7 \alpha) $ equals
AnswerCorrect option: A. $\frac{1}{8}$
(A)
Since, $\sin \left(22 \frac{1}{2}\right)=\frac{1}{2} \sqrt{2-\sqrt{2}}=\cos \left(67 \frac{1}{2}\right)^{\circ}$
and $\cos \left(22 \frac{1}{2}\right)^{\circ}=\frac{1}{2} \sqrt{2+\sqrt{2}}=\sin \left(67 \frac{1}{2}\right)^{\circ}$
$\therefore\left[1+\cos \left(22 \frac{1}{2}\right)^{\circ}\right]\left[1+\cos \left(67 \frac{1}{2}\right)^{\circ}\right]$$\left[1+\cos \left(112 \frac{1}{2}\right)^{\circ}\right]\left[1+\cos \left(157 \frac{1}{2}\right)^{\circ}\right]$
$=\left[1+\cos \left(22 \frac{1}{2}\right)^{\circ}\right]\left[1+\cos \left(67 \frac{1}{2}\right)^{\circ}\right]$$\left[1-\sin \left(22 \frac{1}{2}\right)^{\circ}\right]\left[1-\sin \left(67 \frac{1}{2}\right)^{\circ}\right]$
$\ldots\left[\because \cos \left(90^{\circ}+\theta\right)=-\sin \theta\right]$
$=\left[1+\frac{1}{2} \sqrt{2+\sqrt{2}}\right]\left[1+\frac{1}{2} \sqrt{2-\sqrt{2}}\right]$$\left[1-\frac{1}{2} \sqrt{2-\sqrt{2}}\right]\left[1-\frac{1}{2} \sqrt{2+\sqrt{2}}\right]$
$=\left[1-\frac{1}{4}(2+\sqrt{2})\right]\left[1-\frac{1}{4}(2-\sqrt{2})\right]$
$=\frac{(4-2-\sqrt{2})(4-2+\sqrt{2})}{16}$
$=\frac{(2-\sqrt{2})(2+\sqrt{2})}{16}=\frac{4-2}{16}=\frac{1}{8}$
View full question & answer→MCQ 802 Marks
$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{2 \pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{4 \pi}{8}+\sin ^4 \frac{5 \pi}{8} +\sin ^4 \frac{6 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=$
- A
$\frac{3}{2}$
- B
$\frac{5}{2}$
- ✓
- D
$\frac{7}{2}$
Answer(C)
$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{2 \pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{4 \pi}{8}+\sin ^4 \frac{5 \pi}{8}$$+\sin ^4 \frac{6 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$
$=\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{\pi}{4}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{\pi}{2}+\sin ^4 \frac{3 \pi}{8}$$+\sin ^4 \frac{3 \pi}{4}+\sin ^4 \frac{\pi}{8}$
$\ldots[\because \sin (\pi-\theta)=\sin \theta]$
$=2 \sin ^4 \frac{\pi}{8}+2 \sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{\pi}{4}+\sin ^4 \frac{\pi}{2}+\sin ^4 \frac{3 \pi}{4}$
$=\frac{1}{2}\left[\left(2 \sin ^2 \frac{\pi}{8}\right)^2+\left(2 \sin ^2 \frac{3 \pi}{8}\right)^2\right]+\left(\frac{1}{\sqrt{2}}\right)^4+(1)^4$$+\sin ^4\left(\frac{\pi}{2}+\frac{\pi}{4}\right)$
$=\frac{1}{2}\left[\left(1-\cos \frac{\pi}{4}\right)^2+\left(1-\cos \frac{3 \pi}{4}\right)^2\right]+\frac{1}{4}+1+\cos ^4 \frac{\pi}{4}$
$=\frac{1}{2}\left[\left(1-\frac{1}{\sqrt{2}}\right)^2+\left(1+\frac{1}{\sqrt{2}}\right)^2\right]+\frac{5}{4}+\left(\frac{1}{\sqrt{2}}\right)^4$
$=\frac{1}{2}(3)+\frac{5}{4}+\frac{1}{4}=3$
View full question & answer→MCQ 812 Marks
The value of the expression
$\frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)} -\frac{1}{4} \sin 2 \alpha\left(\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right)$ is
AnswerCorrect option: D. $\sin ^2 \alpha$
(D)
$\frac{1+\sin 2 \alpha}{\cos (2 \alpha-2 \pi) \tan \left(\alpha-\frac{3 \pi}{4}\right)}$$-\frac{1}{4} \sin 2 \alpha\left(\cot \frac{\alpha}{2}+\cot \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)\right)$
$=\frac{1+2 \sin \alpha \cos \alpha}{\cos 2 \alpha\left(\frac{\tan \alpha+1}{1-\tan \alpha}\right)}$$-\frac{1}{4}(2 \sin \alpha \cos \alpha)\left(\frac{\cos ^2 \frac{\alpha}{2}-\sin ^2 \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right)$
$=\frac{(\cos \alpha+\sin \alpha)^2}{\cos 2 \alpha\left(\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}\right)}-\frac{1}{4}(2 \sin \alpha \cos \alpha)\left(2 \frac{\cos \alpha}{\sin \alpha}\right)$
$=1-\cos ^2 \alpha=\sin ^2 \alpha$
View full question & answer→MCQ 822 Marks
The value of $\tan \left(7 \frac{1}{2}\right)^{\circ}$ is equal to
- A
$\sqrt{6}+\sqrt{3}+\sqrt{2}-2$
- ✓
$\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
- C
$\sqrt{6}-\sqrt{3}+\sqrt{2}+2$
- D
$\sqrt{6}-\sqrt{3}-\sqrt{2}-2$
AnswerCorrect option: B. $\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
(B)
Since, $\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}$
Putting $\frac{ A }{2}=\left(7 \frac{1}{2}\right)^{\circ}$, we get
$\frac{1-\cos \theta}{\sin \theta}=\tan \frac{\theta}{2}$, where $\theta \neq(2 n+1) \pi$
$\tan \left(7 \frac{1}{2}\right)^{\circ}=\frac{1-\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{1-\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}$
$=\frac{2 \sqrt{2}-\sqrt{3}-1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
View full question & answer→MCQ 832 Marks
$\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$ is equal to
AnswerCorrect option: A. $\cot \left(7 \frac{1}{2}\right)^{\circ}$
(A)
$\cot \frac{A}{2}=\frac{1+\cos A}{\sin A}$
Putting $A =\left(7 \frac{1}{2}\right)^{\circ}$, we get
$\frac{1+\cos \theta}{\sin \theta}=\cot \frac{\theta}{2}$, where $\theta \neq 2 n \pi$
$\cot \left(7 \frac{1}{2}\right)^{\circ}=\frac{1+\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{1+\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}$
$=\frac{2 \sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
View full question & answer→MCQ 842 Marks
If $\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$, then the value of
$\sqrt{4 \cos ^4 \theta+\sin ^2 2 \theta}+4 \cot \theta \cos ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$ is
- A
$-2 \cot \theta$
- ✓
$2 \cot \theta$
- C
$2 \cos \theta$
- D
$2 \sin \theta$
AnswerCorrect option: B. $2 \cot \theta$
(B)
$\sqrt{4 \cos ^4 \theta+\sin ^2 2 \theta}+4 \cot \theta \cos ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
$=\sqrt{4 \cos ^4 \theta+4 \sin ^2 \theta \cos ^2 \theta}$$+4 \cot \theta\left[\frac{1+\cos 2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2}\right]$
$=\sqrt{4 \cos ^2 \theta\left(\cos ^2 \theta+\sin ^2 \theta\right)}+2 \cot \theta\left[1+\cos \left(\frac{\pi}{2}-\theta\right)\right]$
$=|2 \cos \theta|+2 \cot \theta+2 \cos \theta$
$=2 \cot \theta$$\ldots\left[\because \theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\right]$
View full question & answer→MCQ 852 Marks
If $\cos \theta=\frac{\cos \alpha-\cos \beta}{1-\cos \alpha \cos \beta}$, then one of the values of $\tan \left(\frac{\theta}{2}\right)$ is
- ✓
$\cot \frac{\beta}{2} \tan \frac{\alpha}{2}$
- B
$\tan \alpha \tan \frac{\beta}{2}$
- C
$\tan \frac{\beta}{2} \cot \frac{\alpha}{2}$
- D
$\tan \frac{\beta}{2} \cot \frac{\alpha}{2}$
AnswerCorrect option: A. $\cot \frac{\beta}{2} \tan \frac{\alpha}{2}$
(A)
$\cos \theta=\frac{\cos \alpha-\cos \beta}{1-\cos \alpha \cos \beta}$
By componendo - dividendo, we get
$\frac{\cos \theta+1}{\cos \theta-1}=\frac{\cos \alpha-\cos \beta+1-\cos \alpha \cos \beta}{\cos \alpha-\cos \beta-(1-\cos \alpha \cos \beta)}$
$\Rightarrow \frac{\cos \theta+1}{\cos \theta-1}=\frac{(1+\cos \alpha)(1-\cos \beta)}{-(1-\cos \alpha)(1+\cos \beta)}$
$\Rightarrow \frac{1+\cos \theta}{1-\cos \theta}=\frac{1+\cos \alpha}{1-\cos \alpha} \times \frac{1-\cos \beta}{1+\cos \beta}$
$\Rightarrow \cot ^2 \frac{\theta}{2}=\cot ^2 \frac{\alpha}{2} \tan ^2 \frac{\beta}{2}$
$\Rightarrow \tan ^2 \frac{\theta}{2}=\tan ^2 \frac{\alpha}{2} \cot ^2 \frac{\beta}{2}$
$\Rightarrow \tan \frac{\theta}{2}= \pm \tan \frac{\alpha}{2} \cot \frac{\beta}{2}$
View full question & answer→MCQ 862 Marks
$\cos 2(\theta+\phi)-4 \cos (\theta+\phi) \sin \theta \sin \phi+2 \sin ^2 \phi=$
- ✓
$\cos 2 \theta$
- B
$\cos 3 \theta$
- C
$\sin 2 \theta$
- D
$\sin 3 \theta$
AnswerCorrect option: A. $\cos 2 \theta$
(A)
Putting $\theta=\phi=\frac{\pi}{4}$ in the given expression, we get
$\cos 2\left(\frac{\pi}{2}\right)-4 \cos \left(\frac{\pi}{2}\right) \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{4}\right) +2 \sin ^2\left(\frac{\pi}{4}\right)=0$
Put $\theta=\phi=\frac{\pi}{4}$ in option (A), then
$\cos 2 \theta=\cos \frac{\pi}{2}=0$
Hence ontion (A) is the correct answer.
View full question & answer→MCQ 872 Marks
$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=$
- A
$\frac{1-\sin A}{\cos A}$
- B
$\frac{1-\cos A}{\sin A}$
- ✓
$\frac{1+\sin A}{\cos A}$
- D
$\frac{1+\cos A}{\sin A}$
AnswerCorrect option: C. $\frac{1+\sin A}{\cos A}$
(C)
$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$
$=\frac{\sin A+(1-\cos A)}{\sin A-(1-\cos A)}$
$=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}+2 \sin ^2 \frac{A}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}-2 \sin ^2 \frac{A}{2}}$
$=\frac{\cos \frac{ A }{2}+\sin \frac{ A }{2}}{\cos \frac{A}{2}-\sin \frac{ A }{2}}$
$=\frac{\left(\cos \frac{ A }{2}+\sin \frac{ A }{2}\right)^2}{\cos ^2 \frac{A}{2}-\sin ^2 \frac{A}{2}}$
$=\frac{1+\sin A}{\cos A}$
View full question & answer→MCQ 882 Marks
$(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2=$
- ✓
$4 \cos ^2\left(\frac{\alpha-\beta}{2}\right)$
- B
$4 \sin ^2\left(\frac{\alpha-\beta}{2}\right)$
- C
$4 \cos ^2\left(\frac{\alpha+\beta}{2}\right)$
- D
$4 \sin ^2\left(\frac{\alpha+\beta}{2}\right)$
AnswerCorrect option: A. $4 \cos ^2\left(\frac{\alpha-\beta}{2}\right)$
(A)
$(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2$
$\begin{array}{l}=\cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta+\sin ^2 \alpha \\ +\sin ^2 \beta+2 \sin \alpha \sin \beta\end{array}$
$=2\{1+\cos (\alpha-\beta)\}=4 \cos ^2\left(\frac{\alpha-\beta}{2}\right)$
View full question & answer→MCQ 892 Marks
If $\cos \theta=\frac{3}{5}$ and $\cos \phi=\frac{4}{5}$, where $\theta$ and $\phi$ are positive acute angles, then $\cos \frac{\theta-\phi}{2}=$
- A
$\frac{7}{\sqrt{2}}$
- ✓
$\frac{7}{5 \sqrt{2}}$
- C
$\frac{7}{\sqrt{5}}$
- D
$\frac{7}{2 \sqrt{5}}$
AnswerCorrect option: B. $\frac{7}{5 \sqrt{2}}$
(B)
Given, $\cos \theta=\frac{3}{5} \Rightarrow \sin \theta=\frac{4}{5}$
and $\cos \phi=\frac{4}{5} \Rightarrow \sin \phi=\frac{3}{5}$
$\therefore \quad \cos (\theta-\phi)=\cos \theta \cos \phi+\sin \theta \sin \phi$
$=\frac{3}{5} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}$
But, $2 \cos ^2\left(\frac{\theta-\phi}{2}\right)=1+\cos (\theta-\phi)=1+\frac{24}{25}$
$\therefore \quad \cos ^2\left(\frac{\theta-\phi}{2}\right)=\frac{49}{50}$
$\therefore \quad \cos \left(\frac{\theta-\phi}{2}\right)=\frac{7}{5 \sqrt{2}}$
View full question & answer→MCQ 902 Marks
If $\tan \left(\frac{x}{2}\right)=\operatorname{cosec} x-\sin x$, then the value of $\tan ^2\left(\frac{x}{2}\right)$ is
- A
$2-\sqrt{5}$
- ✓
$\sqrt{5}-2$
- C
$\sqrt{5}+2$
- D
$9-4 \sqrt{5}$
AnswerCorrect option: B. $\sqrt{5}-2$
(B)
$\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}$
$=\frac{2 y}{1+y^2} \quad \ldots\left[\right.$ Let $\left.y=\tan \left(\frac{x}{2}\right)\right]$
$\therefore \tan \frac{x}{2}=\operatorname{cosec} x-\sin x=\frac{1}{\sin x}-\sin x$
$\Rightarrow y=\frac{1+y^2}{2 y}-\frac{2 y}{1+y^2}$
$\Rightarrow 2 y^2\left(1+y^2\right)=1+y^4+2 y^2-4 y^2$
$\Rightarrow 1-y^4-4 v^2=0 \Rightarrow y^4+4 y^2-1=0$
$\Rightarrow y^2=\frac{-4 \pm \sqrt{16+4}}{2}=\frac{-4 \pm \sqrt{20}}{2}$
$\Rightarrow \tan ^2\left(\frac{x}{2}\right)=-2 \pm \sqrt{5}$
But, $\tan ^2\left(\frac{x}{2}\right) \neq-2-\sqrt{5}$
$\Rightarrow \tan ^2\left(\frac{x}{2}\right)=-2+\sqrt{5}$
View full question & answer→MCQ 912 Marks
$(m+2) \sin \theta+(2 m-1) \cos \theta=(2 m+1)$, if $\tan \theta$ is equal to
- ✓
$\frac{4}{3}$ or $\frac{2 m}{m^2-1}$
- B
$\frac{3}{4}$ or $\frac{2 m}{m^2+1}$
- C
$\frac{4}{3}$ or $\frac{2 m+1}{m^2}$
- D
$\frac{3}{4}$ or $\frac{m^2}{2 m+1}$
AnswerCorrect option: A. $\frac{4}{3}$ or $\frac{2 m}{m^2-1}$
(A)
Put $\tan \left(\frac{\theta}{2}\right)=t$
$\therefore (m+2)\left(\frac{2 t}{1+t^2}\right)+(2 m-1)\left(\frac{1-t^2}{1+t^2}\right)=2 m+1$
$\Rightarrow(2 m+4) t+(2 m-1)\left(1-t^2\right)$
$=(2 m+1)\left(1+ t ^2\right)$
$\Rightarrow 4 m t^2-(2 m+4) t+2=0$
$\Rightarrow 2 mt ^2 - mt - 2 t +1-0$
$\Rightarrow m t(2 t-1)-1(2 t-1)=0$
$\Rightarrow(2 t -1)( mt -1)=0$
$\Rightarrow t =\frac{1}{2} \quad$ or $\quad t =\frac{1}{m}$
If $t =\frac{1}{2}$, then
$\tan \theta=\frac{2 t}{1-t^2}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$
If $t=\frac{1}{m}$, then
$\tan \theta=\frac{\frac{2}{m}}{1-\left(\frac{1}{m^2}\right)}=\frac{2 m}{m^2-1}$
View full question & answer→MCQ 922 Marks
$\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$ is equal to
- A
$\sec \theta$
- ✓
$2 \sec \theta$
- C
$\sec \frac{\theta}{2}$
- D
$\sin \theta$
AnswerCorrect option: B. $2 \sec \theta$
(B)
$\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
$=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}+\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}$
$=\frac{\left(1+\tan \frac{\theta}{2}\right)^2+\left(1-\tan \frac{\theta}{2}\right)^2}{1-\tan ^2 \frac{\theta}{2}}$
$=2\left(\frac{1+\tan ^2 \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}\right)$
$=\frac{2}{\cos \theta}$
$=2 \sec \theta$
View full question & answer→MCQ 932 Marks
If $\sec \theta=1 \frac{1}{4}$, then $\tan \frac{\theta}{2}=$
- ✓
$\frac{1}{3}$
- B
$\frac{3}{4}$
- C
$\frac{1}{4}$
- D
$\frac{5}{4}$
AnswerCorrect option: A. $\frac{1}{3}$
(A)
Given that, $\sec \theta=\frac{5}{4}$
Since, $\sec \theta=\frac{1+\tan ^2\left(\frac{\theta}{2}\right)}{1-\tan ^2\left(\frac{\theta}{2}\right)}$
$\Rightarrow \frac{5}{4}=\frac{1+\tan ^2\left(\frac{\theta}{2}\right)}{1-\tan ^2\left(\frac{\theta}{2}\right)}$
$\Rightarrow 5-5 \tan ^2\left(\frac{\theta}{2}\right)=4+4 \tan ^2\left(\frac{\theta}{2}\right)$
$\Rightarrow 9 \tan ^2\left(\frac{\theta}{2}\right)=1$
$\Rightarrow \tan \left(\frac{\theta}{2}\right)=\frac{1}{3}$
View full question & answer→MCQ 942 Marks
If $\theta$ is an acute angle and $\sin \frac{\theta}{2}=\sqrt{\frac{x-1}{2 x}}$, then $\tan \theta$ is equal to
- A
$x^2-1$
- ✓
$\sqrt{x^2-1}$
- C
$\sqrt{x^2+1}$
- D
$x^2+1$
AnswerCorrect option: B. $\sqrt{x^2-1}$
(B)
Given, $\sin \frac{\theta}{2}=\sqrt{\frac{x-1}{2 x}}$
$\therefore \quad \cos \frac{\theta}{2}=\sqrt{1-\sin ^2 \frac{\theta}{2}}=\sqrt{\frac{x+1}{2 x}}$and $\tan \frac{\theta}{2}=\frac{\sqrt{x-1}}{\sqrt{x+1}}$
Since, $\tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}$
$\therefore \quad \tan \theta=\sqrt{x^2-1}$
View full question & answer→MCQ 952 Marks
If $90^{\circ}< A <180^{\circ}$ and $\sin A =\frac{4}{5}$, then $\tan \frac{ A }{2}$ is equal to
- A
$\frac{1}{2}$
- B
$\frac{3}{5}$
- C
$\frac{3}{2}$
- ✓
Answer(D)
$\sin A =\frac{4}{5}$
$\Rightarrow \tan A=-\frac{4}{3}$$\ldots\left[\because 90^{\circ}< A <180^{\circ}\right]$
Now, $\tan A =\frac{2 \tan \frac{\Lambda}{2}}{1-\tan ^2 \frac{A}{2}}\left(\right.$ Let $\left.\tan \frac{ A }{2}= P \right)$
$\Rightarrow-\frac{4}{3}=\frac{2 P}{1-P^2}$
$\Rightarrow 4 P ^2-6 P -4=0$
$\Rightarrow P =\frac{-1}{2}$ or $P =2$
$P =\frac{-1}{2}$ is not possible
$\therefore \quad P=2 \Rightarrow \tan \frac{\Lambda}{2}=2$
View full question & answer→MCQ 962 Marks
If $\sin \theta=\frac{-4}{5}$ and $\theta$ lies in the third quadrant, then $\cos \frac{\theta}{2}=$
- A
$\frac{1}{\sqrt{5}}$
- ✓
$-\frac{1}{\sqrt{5}}$
- C
$\sqrt{\frac{2}{5}}$
- D
$-\sqrt{\frac{2}{5}}$
AnswerCorrect option: B. $-\frac{1}{\sqrt{5}}$
(B)
$\sin \theta=-\frac{4}{5}$
$\Rightarrow \cos \theta=\sqrt{1-\frac{16}{25}}= \pm \frac{3}{5}$
$\rightarrow \cos \theta=\frac{-3}{5} \ldots\left[\because \theta\right.$ lies in the $3^{\text {rd }}$ quadrant $]$
Since, $\cos \frac{\theta}{2}= \pm \sqrt{\frac{1+\cos \theta}{2}}$
$= \pm \sqrt{\frac{1-3 / 5}{2}}$
$= \pm \sqrt{\frac{1}{5}}$
$\therefore \quad \cos \frac{\theta}{2}=-\frac{1}{\sqrt{5}} \quad \ldots\left[\because \frac{\theta}{2}\right.$ lies in the $2^{\text {nd }}$ quadrant $]$
View full question & answer→MCQ 972 Marks
If $0<\theta<\frac{\pi}{2}$ and $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$, then $y$ is equal is
AnswerCorrect option: B. $\tan \frac{\theta}{2}$
(B)
$\frac{y+1}{1-y}=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$
$\Rightarrow \frac{y+1}{1-y}-\sqrt{\frac{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^2}{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^2}}$
$\Rightarrow \frac{1+y}{1-y}=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\left|\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right|}$
$\Rightarrow \frac{1+y}{1-y}=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}$
$\ldots\left[\begin{array}{rl}\because 0<\theta & <\frac{\pi}{2} \Rightarrow 0<\frac{\theta}{2}<\frac{\pi}{4} \\ & \Rightarrow \cos \frac{\theta}{2}>\sin \frac{\theta}{2}\end{array}\right]$
$\Rightarrow \frac{1+y}{1-y}=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$
$\Rightarrow y=\tan \frac{\theta}{2}$
View full question & answer→MCQ 982 Marks
If $\operatorname{cosec} \theta=\frac{p+q}{p-q}$, then $\cot \left(\frac{\pi}{4}+\frac{\theta}{2}\right)=$
- A
$\sqrt{\frac{p}{q}}$
- ✓
$\sqrt{\frac{q}{p}}$
- C
$\sqrt{ pq }$
- D
AnswerCorrect option: B. $\sqrt{\frac{q}{p}}$
(B)
Given, $\operatorname{cosec} \theta=\frac{p+q}{p-q} \Rightarrow \frac{1}{\sin \theta}=\frac{p+q}{p-q}$
By componrndo - dividendo ,we get
$\frac{1+\sin \theta}{1-\sin \theta}=\frac{p+q+p-q}{p+q-p+q}$
$\Rightarrow\left\{\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right\}^2=\frac{p}{q}$
$\Rightarrow\left\{\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}\right\}^2=\frac{p}{q} \Rightarrow\left\{\frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}\right\}=\frac{p}{q}$
$\Rightarrow \tan ^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\frac{ p }{ q } \Rightarrow \cot ^2\left(\frac{\pi}{2}+\frac{\theta}{2}\right)=\frac{ q }{ p }$
$\Rightarrow \cot \left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\sqrt{\frac{q}{p}}$
View full question & answer→MCQ 992 Marks
If $A =133^{\circ}$, then $2 \cos \frac{A}{2}$ is equal to
- A
$-\sqrt{1+\sin A}-\sqrt{1-\sin A}$
- B
$-\sqrt{1+\sin A }+\sqrt{1-\sin A }$
- ✓
$\sqrt{1+\sin A }-\sqrt{1-\sin A }$
- D
$\sqrt{1+\sin A }+\sqrt{1-\sin A }$
AnswerCorrect option: C. $\sqrt{1+\sin A }-\sqrt{1-\sin A }$
(C)
For $A =133^{\circ}, \frac{ A }{2}=66.5^{\circ}$
$\Rightarrow \sin \frac{ A }{2}>\cos \frac{ A }{2}>0$
$\sqrt{ 1+\sin A} =\sin \frac{ A }{2}+\cos \frac{ A }{2} \quad\ldots(i)$
and $\sqrt{1-\sin A }=\sin \frac{ A }{2}-\cos \frac{ A }{2}\quad\ldots(ii)$
Subtracting (ii) from (i), we get
$2 \cos \frac{A}{2}=\sqrt{1+\sin A }-\sqrt{1-\sin A }$
View full question & answer→MCQ 1002 Marks
If $\tan x=\frac{2 b}{a-c}(a \neq c)$,
$\begin{array}{l}y= a \cos ^2 x+2 b \sin x \cos x+ c \sin ^2 x \text { and } \\ z = a \sin ^2 x-2 b \sin x \cos x+ c \cos ^2 x, \text { then }\end{array}$
- A
$y= z$
- ✓
$y+z=a+c$
- C
$y-z=a+c$
- D
$y-z=(a-c)^2+4 b^2$
AnswerCorrect option: B. $y+z=a+c$
(B)
$\begin{aligned} y- z = a \left(\cos ^2 x-\sin ^2 x\right)+ & 4 b \sin x \cos x - c \left(\cos ^2 x-\sin ^2 x\right)\end{aligned}$
$=( a - c ) \cos 2 x +2 b \sin 2 x$
$=( a - c )\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)+2 b\left(\frac{2 \tan x}{1+\tan ^2 x}\right)$
$=(a-c)\left\{\frac{1-4 b^2 /(a-c)^2}{1+4 b^2 /(a-c)^2}\right\}+2 b\left\{\frac{2.2 b /(a-c)}{1+4 b^2 /(a-c)^2}\right\}$
$\ldots\left[\because \tan x=\frac{2 b}{a-c}\right.$ (given) $]$
$=\frac{(a-c)\left\{(a-c)^2-4 b^2\right\}+8 b^2(a-c)}{(a-c)^2+4 b^2}$
$\therefore y-z=\frac{(a-c)\left\{(a-c)^2+4 b^2\right\}}{(a-c)^2+4 b^2}=a-c$
$\Rightarrow y \neq z \quad \ldots[\because a \neq c]$
$y+ z = a \left(\cos ^2 x+\sin ^2 x\right)+ c \left(\sin ^2 x+\cos ^2 x\right)$
= a + c
View full question & answer→MCQ 1012 Marks
If $0 < x<\pi$ and $\cos x+\sin x=\frac{1}{2}$, then $\tan x$ i equal to
- A
$\frac{1-\sqrt{7}}{4}$
- B
$\frac{4-\sqrt{7}}{3}$
- ✓
$-\frac{4+\sqrt{7}}{3}$
- D
$\frac{\sqrt{7}+1}{4}$
AnswerCorrect option: C. $-\frac{4+\sqrt{7}}{3}$
(C)
$\cos x+\sin x=\frac{1}{2}$
$\Rightarrow(\cos x+\sin x)^2=\frac{1}{4}$
$\Rightarrow 1+\sin 2 x=\frac{1}{4}$
$\Rightarrow \sin 2 x=-\frac{3}{4}$
$\Rightarrow \frac{2 \tan x}{1+\tan ^2 x}=-\frac{3}{4}$
$\Rightarrow 3 \tan ^2 x+8 \tan x+3=0$
$\Rightarrow \tan x=\frac{-4 \pm \sqrt{7}}{3}$
View full question & answer→MCQ 1022 Marks
If $\sin x+\cos x=\frac{1}{5}$ and $0 \leq x \leq \pi$, then $\tan x$ equal to
- ✓
$-\frac{4}{3}$
- B
$-\frac{1}{2}$
- C
$-\frac{2}{3}$
- D
$\frac{3}{2}$
AnswerCorrect option: A. $-\frac{4}{3}$
(A)
$\sin x+\cos x=\frac{1}{5}$
Squaring on both sides, we get
$1+\sin 2 x=\frac{1}{25}$
$\Rightarrow \sin 2 x=\frac{-24}{25}$
$\Rightarrow \frac{2 \tan x}{1+\tan ^2 x}=\frac{-24}{25}$
$\Rightarrow 24 \tan ^2 x+50 \tan x+24=0$
$\Rightarrow 12 \tan ^2 x+25 \tan x+12=0$
$\Rightarrow(3 \tan x+4)(4 \tan x+3)=0$
$\Rightarrow \tan x=\frac{-4}{3}$ or $\frac{-3}{4}$
View full question & answer→MCQ 1032 Marks
If $2 \tan A=3 \tan B$, then $\frac{\sin 2 B}{5-\cos 2 B}$ is equal to
- A
$\tan A-\tan B$
- ✓
$\tan (A-B)$
- C
$\tan (A+B)$
- D
$\tan (A+2 B)$
AnswerCorrect option: B. $\tan (A-B)$
(B)
$2 \tan A=3 \tan B$
$\Rightarrow \tan A =\frac{3}{2} \tan B=\frac{3}{2} t \quad[$ Let $\tan B =t]$
$\sin 2 B=\frac{2 t }{1+ t ^2}, \cos 2 B=\frac{1- t ^2}{1+ t ^2}$
$\therefore \frac{\sin 2 B}{5-\cos 2 B}=\frac{\left(\frac{2 t}{1+t^2}\right)}{5-\left(\frac{1-t^2}{1+t^2}\right)}=\frac{2 t}{4+6 t^2}$
$=\frac{t}{2+3 t^2}=\frac{\frac{3 t}{2}-t}{1+\frac{3 t^2}{2}}$
$=\tan ( A - B )$
View full question & answer→MCQ 1042 Marks
If $\tan \alpha=\frac{1}{7}$ and $\tan \beta=\frac{1}{3}$, then $\cos 2 \alpha=$
- A
$\sin 2 \beta$
- ✓
$\sin 4 \beta$
- C
$\sin 3 \beta$
- D
$\sin \beta$
AnswerCorrect option: B. $\sin 4 \beta$
(B)
$\cos 2 \alpha=\frac{1-t^2}{1+t^2}=\frac{24}{25}($ Let $t=\tan \alpha)$
$\sin 2 \beta-\frac{2 T}{1+T^2}-\frac{3}{5}(\operatorname{Let} T-\tan \beta)$
$\Rightarrow \cos 2 \beta=\frac{4}{5}$
Now, $\sin 4 \beta=2 \sin 2 \beta \cos 2 \beta$
$=2 \cdot \frac{3}{5}. \frac{4}{5}$
$=\frac{24}{25}$
$=\cos 2 \alpha$
View full question & answer→MCQ 1052 Marks
If $2 \sin ^2\left[\left(\frac{\pi}{2}\right) \cos ^2 x\right]=1-\cos (\pi \sin 2 x)$,
$x \neq(2 n+1) \frac{\pi}{2}, n \in I$, then $\cos 2 x$ is equal to
- A
$\frac{1}{5}$
- ✓
$\frac{3}{5}$
- C
$\frac{4}{5}$
- D
AnswerCorrect option: B. $\frac{3}{5}$
(B)
$2 \sin ^2\left[\left(\frac{\pi}{2}\right) \cos ^2 x\right]=1-\cos (\pi \sin 2 x)$
$\Rightarrow 2 \sin ^2\left[\left(\frac{\pi}{2}\right) \cos ^2 x\right]=2 \sin ^2\left[\frac{\pi \sin 2 x}{2}\right]$
$\Rightarrow \cos ^2 x=\sin 2 x \Rightarrow \cos ^2 x=2 \sin x \cos x$
$\Rightarrow \tan x=\frac{1}{2}$
Now, $\cos 2 x=\frac{1-\tan ^2 x}{1+\tan ^2 x}=\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}$
View full question & answer→MCQ 1062 Marks
If $\cot \frac{2 x}{3}+\tan \frac{x}{3}=\operatorname{cosec} \frac{ kx }{3}$, then the value of k is
Answer(B)
$\cot 2 \theta+\tan \theta=\frac{1}{\tan 2 \theta}+\tan \theta$
$=\frac{1-\tan ^2 \theta}{2 \tan \theta}+\tan \theta$
$=\frac{1+\tan ^2 \theta}{2 \tan \theta}=\frac{1}{\sin 2 \theta}$
Now, $\cot \frac{2 x}{3}+\tan \frac{x}{3}=\operatorname{cosec} \frac{ k x}{3}$
$\Rightarrow \operatorname{cosec} \frac{2 x}{3}=\operatorname{cosec} \frac{ k x}{3}$
$\Rightarrow k =2$
View full question & answer→MCQ 1072 Marks
$\tan \left(\frac{\pi}{4}+\theta\right)-\tan \left(\frac{\pi}{4}-\theta\right)=$
- ✓
$2 \tan 2 \theta$
- B
$2 \cot 2 \theta$
- C
$\tan 2 \theta$
- D
$\cot 2 \theta$
AnswerCorrect option: A. $2 \tan 2 \theta$
(A)
$\tan \left(\frac{\pi}{4}+\theta\right)-\tan \left(\frac{\pi}{4}-\theta\right)$
$\begin{aligned}i. \tan \left(45^{\circ}+\theta\right) & =\frac{1+\tan \theta}{1-\tan \theta} \\ & =\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\end{aligned}$
$\begin{aligned}ii. \tan \left(45^{\circ}-\theta\right) & =\frac{1-\tan \theta}{1+\tan \theta} \\ & =\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\end{aligned}$
$ =\frac{1+\tan \theta}{1-\tan \theta}-\frac{1-\tan \theta}{1+\tan \theta} $
$=\frac{4 \tan \theta}{1-\tan ^2 \theta}$
$=2\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)=2 \tan 2 \theta$
View full question & answer→MCQ 1082 Marks
If $0 < x<\frac{\pi}{4}$, then $\sec 2 x-\tan 2 x=$
- A
$\tan \left(x-\frac{\pi}{4}\right)$
- ✓
$\tan \left(\frac{\pi}{4}-x\right)$
- C
$\tan \left(x+\frac{\pi}{4}\right)$
- D
$\tan ^2\left(x+\frac{\pi}{4}\right)$
AnswerCorrect option: B. $\tan \left(\frac{\pi}{4}-x\right)$
(B)
$\sec 2 x-\tan 2 x=\frac{1-\sin 2 x}{\cos 2 x}=\frac{(\cos x-\sin x)^2}{\left(\cos ^2 x-\sin ^2 x\right)}$
$=\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}$
$=\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \binom{\pi}{4} \tan x}=\tan \left(\frac{\pi}{4}-x\right)$
View full question & answer→MCQ 1092 Marks
$\frac{\sec 8 A-1}{\sec 4 A-1}=$
- A
$\frac{\tan 2 A}{\tan 8 A}$
- B
$\frac{\tan 8 A}{\tan 2 A}$
- C
$\frac{\cot 8 A}{\cot 2 A}$
- ✓
$\frac{\tan 6 A}{\tan 2 A}$
AnswerCorrect option: D. $\frac{\tan 6 A}{\tan 2 A}$
(D)
$\frac{\sec 8 A-1}{\sec 4 A-1}=\frac{1-\cos 8 A}{\cos 8 A} \cdot \frac{\cos 4 A}{1-\cos 4 A}$
$=\frac{2 \sin ^2 4 A}{\cos 8 A} \cdot \frac{\cos 4 A}{2 \sin ^2 2 A}$
$=\frac{2 \sin 4 A \cos 4 A}{\cos 8 A} \cdot \frac{\sin 4 A}{2 \sin ^2 2 A}$
$=\tan 8 A \frac{2 \sin 2 A \cos 2 A}{2 \sin ^2 2 A}$
$=\frac{\tan 8 A}{\tan 2 A}$
View full question & answer→MCQ 1102 Marks
$\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$, then $\tan \alpha=$
AnswerCorrect option: A. $\sqrt{2} \tan \beta$
(A)
$\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$
By componendo - dividendo, we get
$\frac{\cos 2 \alpha+1}{\cos 2 \alpha-1}=\frac{3 \cos 2 \beta-1+3-\cos 2 \beta}{3 \cos 2 \beta-1-(3-\cos 2 \beta)}$
$\Rightarrow \frac{2 \cos ^2 \alpha}{-2 \sin ^2 \alpha}=\frac{2+2 \cos 2 \beta}{4 \cos 2 \beta-4}$
$\Rightarrow \frac{-\cos ^2 \alpha}{\sin ^2 \alpha}=\frac{1+\cos 2 \beta}{2(\cos 2 \beta-1)}=\frac{2 \cos ^2 \beta}{-4 \sin ^2 \beta}$
$\Rightarrow \frac{\sin ^2 \alpha}{\cos ^2 \alpha}=\frac{2 \sin ^2 \beta}{\cos ^2 \beta} \Rightarrow \tan ^2 \alpha=2 \tan ^2 \beta$
$\Rightarrow \tan \alpha=\sqrt{2} \tan \beta$
View full question & answer→MCQ 1112 Marks
If $5\left(\tan ^2 x-\cos ^2 x\right)=2 \cos 2 x+9$, then the value of $\cos 4 x$ is
- ✓
$-\frac{7}{9}$
- B
$-\frac{3}{5}$
- C
$\frac{1}{3}$
- D
$\frac{2}{9}$
AnswerCorrect option: A. $-\frac{7}{9}$
(A)
$5\left(\tan ^2 x-\cos ^2 x\right)=2 \cos 2 x+9$
Put $\cos ^2 x= t$
$\therefore \quad 5\left(\frac{1-t}{t}-t\right)=2(2 t-1)+9$
$\Rightarrow 5\left(1-t-t^2\right)=4 t^2-2 t+9 t$
$\Rightarrow 9 t ^2+12 t -5=0$
$\Rightarrow 9 t^2+15 t-3 t-5=0$
$\Rightarrow 3 l(3 l+5)-1(3 l+5)=0$
$\Rightarrow(3 t+5)(3 t-1)=0$
$\Rightarrow t=\frac{1}{3}$ or $t=\frac{-5}{3}$
but t cannot be negative
$\therefore \quad t=\frac{1}{3}$
$\therefore \quad \cos ^2 x=\frac{1}{3}$
$\Rightarrow \cos 2 x=2 \cos ^2 x-1$
$=\frac{2}{3}-1=\frac{-1}{3}$
$\Rightarrow \cos 4 x=2 \cos ^2 2 x-1$
$=2\left(\frac{-1}{3}\right)^2-1$
$=\frac{-7}{9}$
View full question & answer→MCQ 1122 Marks
If $8 \cos 2 \theta+8 \sec 2 \theta=65,0<\theta<\frac{\pi}{2}$, then the value of $4 \cos 4 \theta$ is equal to
- A
$-\frac{33}{8}$
- ✓
$-\frac{31}{8}$
- C
$-\frac{31}{32}$
- D
$-\frac{33}{32}$
AnswerCorrect option: B. $-\frac{31}{8}$
(B)
$8 \cos 2 \theta+8 \sec 2 \theta=65$
$\Rightarrow 8 \cos ^2 2 \theta+8=65 \cos 2 \theta$
$\Rightarrow 8 \cos ^2 2 \theta-65 \cos 2 \theta+8=0$
$\Rightarrow(\cos 2 \theta-8)(8 \cos 2 \theta-1)=0$
Since $\cos 2 \theta \in[-1,1]$
$\therefore \quad \cos 2 \theta=\frac{1}{8}$
Now, $4 \cos 4 \theta=4\left(2 \cos ^2 2 \theta-1\right)$
$=4\left[2\left(\frac{1}{8}\right)^2-1\right]=-\frac{31}{8}$
View full question & answer→MCQ 1132 Marks
If $\sin 2 \theta+\sin 2 \phi=\frac{1}{2}$ and
$\cos 2 \theta+\cos 2 \phi=\frac{3}{2}$, then $\cos ^2(\theta-\phi)=$
- A
$\frac{3}{8}$
- ✓
$\frac{5}{8}$
- C
$\frac{3}{4}$
- D
$\frac{5}{4}$
AnswerCorrect option: B. $\frac{5}{8}$
(B)
$\sin 2 \theta+\sin 2 \phi=\frac{1}{2}$
and $\cos 2 \theta+\cos 2 \phi=\frac{3}{2}$
Squaring and adding (i) and (ii), we get
$\left(\sin ^2 2 \theta+\cos ^2 2 \theta\right)+\left(\sin ^2 2 \phi+\cos ^2 2 \phi\right)$
$+2(\sin 2 \theta \sin 2 \phi+\cos 2 \theta \cos 2 \phi)=\frac{1}{4}+\frac{9}{4}$
$\Rightarrow \cos 2 \theta \cos 2 \phi+\sin 2 \theta \sin 2 \phi=\frac{1}{4}$
$\Rightarrow \cos (2 \theta-2 \phi)=\frac{1}{4}$
$\Rightarrow 2 \cos ^2(\theta-\phi)-1=\frac{1}{4}$
$\Rightarrow \cos ^2(\theta-\phi)=\frac{5}{8}$
View full question & answer→MCQ 1142 Marks
If $\theta$ and $\phi$ are angles in the $1^{\text {st }}$ quadrant such that $\tan \theta=\frac{1}{7}$ and $\sin \phi=\frac{1}{\sqrt{10}}$, then
- A
$\theta+2 \phi=90^{\circ}$
- B
$\theta+2 \phi=60^{\circ}$
- C
$\theta+2 \phi=30^{\circ}$
- ✓
$\theta+2 \phi=45^{\circ}$
AnswerCorrect option: D. $\theta+2 \phi=45^{\circ}$
(D)
Given, $\tan \theta=\frac{1}{7}, \sin \phi=\frac{1}{\sqrt{10}}$
$\Rightarrow \sin \theta=\frac{1}{\sqrt{50}}, \cos \theta=\frac{7}{\sqrt{50}}, \cos \phi=\frac{3}{\sqrt{10}}$
$\therefore \quad \cos 2 \phi=2 \cos ^2 \phi-1=2\left(\frac{9}{10}\right)-1=\frac{4}{5}$
$\sin 2 \phi=2 \sin \phi \cos \phi=2\left(\frac{1}{\sqrt{10}}\right)\left(\frac{3}{\sqrt{10}}\right)=\frac{3}{5}$
$\therefore \quad \cos (\theta+2 \phi)=\cos \theta \cos 2 \phi-\sin \theta \sin 2 \phi$
$=\frac{7}{\sqrt{50}} \cdot \frac{4}{5}-\frac{1}{\sqrt{50}} \cdot \frac{3}{5}=\frac{28}{5 \sqrt{50}}-\frac{3}{5 \sqrt{50}}$
$=\frac{25}{5 \sqrt{50}}=\frac{1}{\sqrt{2}}$
$\therefore \quad \theta+2 \phi=45^{\circ}$
View full question & answer→MCQ 1152 Marks
If $\sin (\theta+\alpha)=a$ and $\sin (\theta+\beta)=b$, then $\cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta)$ is equal to
- A
$1-a^2-b^2$
- ✓
$1-2 a^2-2 b^2$
- C
$2+a^2+b^2$
- D
$2-a^2-b^2$
AnswerCorrect option: B. $1-2 a^2-2 b^2$
(B)
$\cos (\alpha-\beta)=\cos [\theta+\alpha-(\theta+\beta)]$
$=\cos (\theta+\alpha) \cos (\theta+\beta)$$+\sin (\theta+\alpha) \sin (\theta+\beta)$
$=\sqrt{1-a^2} \sqrt{1-b^2}+a b$
Now, $\cos 2(\alpha-\beta)-4 ab \cos (\alpha-\beta)$
$=2 \cos ^2(\alpha-\beta)-1-4 a b \cos (\alpha-\beta)$
$=2\left(\sqrt{1-a^2} \sqrt{1-b^2}+\right. a b)^2 -4 a b\left(\sqrt{1-a^2} \sqrt{1-b^2}+a b\right)-1$
$=2\left\{\left(1-a^2\right)\left(1-b^2\right)+a^2 b^2\right.$$\left.+2 a b \sqrt{1-a^2} \sqrt{1-b^2}\right\}$$-4 a b\left(\sqrt{1-a^2} \sqrt{1-b^2}+a b\right)-1$
$=2\left(1-b^2-a^2+a^2 b^2\right)+2 a^2 b^2-4 a^2 b^2-1$
$=2\left(1-a^2-b^2\right)-1$
$=1-2 a^2-2 b^2$
View full question & answer→MCQ 1162 Marks
The value of $\frac{1}{8}(3-4 \cos 2 \theta+\cos 4 \theta)$ is
- A
$\cos 4 \theta$
- B
$\sin 4 \theta$
- ✓
$\sin ^4 \theta$
- D
$\cos ^4 \theta$
AnswerCorrect option: C. $\sin ^4 \theta$
(C)
$\frac{1}{8}(3-4 \cos 2 \theta+\cos 4 \theta)$
$=\frac{1}{8}\left(3-4 \cos 2 \theta+2 \cos ^2 2 \theta-1\right)$
$=\frac{1}{8}\left(2 \cos ^2 2 \theta-4 \cos 2 \theta+2\right)$
$=\frac{1}{4}\left(\cos ^2 2 \theta-2 \cos 2 \theta+1\right)$
$=\frac{1}{4}(\cos 2 \theta-1)^2$
$=\frac{1}{4}\left(-2 \sin ^2 \theta\right)^2 \quad\ldots[\because \cos 2 \theta=1-2 \sin ^2 \theta]$
$=\frac{1}{4}\left(4 \sin ^4 \theta\right)=\sin ^4 \theta$
View full question & answer→MCQ 1172 Marks
If $\tan ^2 \theta=2 \tan ^2 \phi+1$, then $\cos 2 \theta+\sin ^2 \phi$ equals
Answer(B)
$\tan ^2 \theta=2 \tan ^2 \phi+1$
$\Rightarrow 1+\tan ^2 \theta=2\left(1+\tan ^2 \phi\right)$
$\Rightarrow \sec ^2 \theta=2 \sec ^2 \phi$
$\Rightarrow \cos ^2 \phi=2 \cos ^2 \theta$
$\Rightarrow \cos ^2 \phi=1+\cos 2 \theta$
$\Rightarrow \sin ^2 \phi+\cos 2 \theta=0 \ldots\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
View full question & answer→MCQ 1182 Marks
If $\sin A +\sin 2 A=x$ and $\cos A +\cos 2 A=y$, then $\left(x^2+y^2\right)\left(x^2+y^2-3\right)=$
Answer(A)
Squaring and adding the given expressions, we get
$x^2+y^2=1+1+2 \cos (2 A- A )$
$\therefore \frac{x^2+y^2-2}{2}=\cos A \quad\ldots(i)$
Also, $\cos A +2 \cos ^2 A-1=y$
$\Rightarrow(\cos A +1)(2 \cos A-1)=y$
putting the value of cos A from (i), we get
$\left(x^2+y^2\right)\left(x^2+y^2-3\right)=2 y$
View full question & answer→MCQ 1192 Marks
$\tan \left(60^{\circ}+ A \right) \tan \left(60^{\circ}- A \right)$ is equal to
- ✓
$\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$
- B
$\frac{2 \cos 2 A-1}{2 \cos 2 A+1}$
- C
$\frac{\cos 2 A+21}{\cos 2 A-1}$
- D
$\frac{\cos 2 A-1}{\cos 2 A+21}$
AnswerCorrect option: A. $\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$
(A)
$\tan \left(60^{\circ}+ A \right) \tan \left(60^{\circ}- A \right)$
$=\frac{\sin ^2 60^{\circ}-\sin ^2 A}{\cos ^2 60^{\circ}-\sin ^2 A}$
$=\frac{\frac{3}{4}-\left(\frac{1-\cos 2 A}{2}\right)}{\frac{1}{4}-\left(\frac{1-\cos 2 A}{2}\right)}=\frac{3-2+2 \cos 2 A}{1-2+2 \cos 2 A}$
$=\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$
View full question & answer→MCQ 1202 Marks
If $\tan x=\frac{b}{a}$, then $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=$
- A
$\frac{2 \sin x}{\sqrt{\sin 2 x}}$
- ✓
$\frac{2 \cos x}{\sqrt{\cos 2 x}}$
- C
$\frac{2 \cos x}{\sqrt{\sin 2 x}}$
- D
$\frac{2 \sin x}{\sqrt{\cos 2 x}}$
AnswerCorrect option: B. $\frac{2 \cos x}{\sqrt{\cos 2 x}}$
(B)
Given that, $\tan x=\frac{ b }{ a }$
$\therefore \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\sqrt{\frac{1+\frac{b}{a}}{1-\frac{b}{a}}}+\sqrt{\frac{1-\frac{b}{a}}{1+\frac{b}{a}}}$
$=\frac{2}{\sqrt{1-\frac{b^2}{a^2}}}=\frac{2}{\sqrt{1-\tan ^2 x}}$
$=\frac{2}{\sqrt{1-\frac{\sin ^2 x}{\cos ^2 x}}}=\frac{2 \cos x}{\sqrt{\cos 2 x}}$
View full question & answer→MCQ 1212 Marks
If $\sec 2 \theta=p+\tan 2 \theta$, then the value of $\sin ^2 \theta$ in terms of $p$ is
- ✓
$\frac{(p-1)^2}{2\left(p^2+1\right)}$
- B
$\frac{1}{2}\left(\frac{p-1}{p+1}\right)^2$
- C
$\frac{p^2-1}{2\left(p^2+1\right)}$
- D
$\frac{ p ^2-1}{2( p +1)^2}$
AnswerCorrect option: A. $\frac{(p-1)^2}{2\left(p^2+1\right)}$
(A)
$\sec 2 \theta= p +\tan 2 \theta$
$\Rightarrow \sec 2 \theta-\tan 2 \theta=p \quad\ldots(i)$
$ \Rightarrow \sec 2 \theta+\tan 2 \theta=\frac{1}{p} \quad\ldots(ii)$
$\ldots\left[\because \sec ^2 2 \theta-\tan ^2 2 \theta=1\right]$
Adding (i) and (ii), we get
$2 \sec 2 \theta=p+\frac{1}{p}$
$\Rightarrow \cos 2 \theta=\frac{2 p}{p^2+1} \Rightarrow 1-2 \sin ^2 \theta=\frac{2 p}{p^2+1}$
$\Rightarrow 2 \sin ^2 \theta=1-\frac{2 p}{p^2+1} \Rightarrow 2 \sin ^2 \theta=\frac{(p-1)^2}{p^2+1}$
$\Rightarrow \sin ^2 \theta=\frac{(p-1)^2}{2\left(p^2+1\right)}$
View full question & answer→MCQ 1222 Marks
If $x+\frac{1}{x}=2 \cos \alpha$, then $x^n+\frac{1}{x^n}=$
- A
$2^n \cos \alpha$
- B
$2^n \cos n \alpha$
- C
$2 i \sin n \alpha$
- ✓
$2 \cos n \alpha$
AnswerCorrect option: D. $2 \cos n \alpha$
(D)
$x+\frac{1}{x}=2 \cos \alpha$
Squaring on both sides, we get
$x^2+\frac{1}{x^2}+2=4 \cos ^2 \alpha$
$\Rightarrow x^2+\frac{1}{x^2}=4 \cos ^2 \alpha-2$
$\Rightarrow x^2+\frac{1}{x^2}=2\left(2 \cos ^2 \alpha-1\right)$
$=2 \cos 2 \alpha$
Similarly, $x^{ n }+\frac{1}{x^{ n }}=2 \cos n \alpha$
View full question & answer→MCQ 1232 Marks
If $\cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right)$, then $\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)=$
- A
$\sin 2 \theta$
- ✓
$\cos 2 \theta$
- C
$\tan 2 \theta$
- D
$\sec 2 \theta$
AnswerCorrect option: B. $\cos 2 \theta$
(B)
Given that, $\cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right)$
$\Rightarrow x+\frac{1}{x}=2 \cos \theta$
Now, $x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2$
$=(2 \cos \theta)^2-2$
$=4 \cos ^2 \theta-2=2 \cos 2 \theta$
$\therefore \quad \frac{1}{2}\left(x^2+\frac{1}{x^2}\right)=\frac{1}{2} \times 2 \cos 2 \theta=\cos 2 \theta$
View full question & answer→MCQ 1242 Marks
Let $B =2 \sin ^2 x-\cos 2 x$, then
- ✓
$-1 \leq B \leq 3$
- B
$0 \leq B \leq 2$
- C
$-1 \leq B \leq 1$
- D
$-2 \leq B \leq 2$
AnswerCorrect option: A. $-1 \leq B \leq 3$
(A)
$2 \sin ^2 x-\cos 2 x=4 \sin ^2 x-1$ and $0 \leq \sin ^2 x \leq 1 \Leftrightarrow 0 \leq 4 \sin ^2 x \leq 4$
$\Rightarrow-1 \leq 4 \sin ^2 x-1 \leq 3$
View full question & answer→MCQ 1252 Marks
$2 \cos ^2 \theta-2 \sin ^2 \theta=1$, then $\theta=$
- A
$15^{\circ}$
- ✓
$30^{\circ}$
- C
$45^{\circ}$
- D
$60^{\circ}$
AnswerCorrect option: B. $30^{\circ}$
(B)
$2 \cos ^2 \theta-2 \sin ^2 \theta=1$
$\Rightarrow 2 \cos 2 \theta=1 \Rightarrow \cos 2 \theta=\frac{1}{2}=\cos 60^{\circ}$
$\Rightarrow 2 \theta=60^{\circ} \Rightarrow \theta=30^{\circ}$
View full question & answer→MCQ 1262 Marks
$\cos 15^{\circ}=$
- ✓
$\sqrt{\frac{1+\cos 30^{\circ}}{2}}$
- B
$\sqrt{\frac{1-\cos 30^{\circ}}{2}}$
- C
$\pm \sqrt{\frac{1+\cos 30^{\circ}}{2}}$
- D
$\pm \sqrt{\frac{1-\cos 30^{\circ}}{2}}$
AnswerCorrect option: A. $\sqrt{\frac{1+\cos 30^{\circ}}{2}}$
(A)
$\cos 15^{\circ}=\sqrt{\frac{1+\cos \left(2 \times 15^{\circ}\right)}{2}}=\sqrt{\frac{1+\cos 30^{\circ}}{2}}$
View full question & answer→MCQ 1272 Marks
The value of $\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}$ is equal to
- A
$\frac{1}{8}$
- B
$\frac{1}{16}$
- C
$\frac{1}{32}$
- ✓
$\frac{1}{64}$
AnswerCorrect option: D. $\frac{1}{64}$
(D)
$\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}$
$\begin{aligned}=\sin \frac{\pi}{14} \sin & \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \times 1 \times \sin \left(\pi-\frac{5 \pi}{14}\right) \sin \left(\pi-\frac{3 \pi}{14}\right) \sin \left(\pi-\frac{\pi}{14}\right)\end{aligned}$
$=\left(\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14}\right)^2$
$=\left(\cos \frac{6 \pi}{14} \cos \frac{4 \pi}{14} \cos \frac{2 \pi}{14}\right)^2 \cdots\left[\because \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta\right]$
$=\left(\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7}\right)^2$
$\ldots\left[\because \cos \frac{3 \pi}{7}=\cos \left(\pi-\frac{4 \pi}{7}\right)=-\cos \frac{4 \pi}{7}\right]$
$\cos \alpha \cdot \cos 2 \alpha \cdot \cos 2^2 \alpha \cdot \cos 2^3 \alpha \ldots \cos 2^{n-1} \alpha$
$=\frac{\sin 2^n \alpha}{2^n \sin \alpha}, \text { if } \alpha \neq n \pi$
$=1, \text { if } \alpha=2 n \pi$
$=-1, \text { if } \alpha=(2 n+1) \pi$
$=\left(-\frac{\sin \frac{2^3 \pi}{7}}{2^3 \sin \frac{\pi}{7}}\right)^2=\frac{1}{64}\left(-\frac{\sin \frac{8 \pi}{7}}{\sin \frac{\pi}{7}}\right)^2$
$=\frac{1}{64} \quad \ldots\left[\because \sin \frac{8 \pi}{7}=\sin \left(\pi+\frac{\pi}{7}\right)=-\sin \frac{\pi}{7}\right]$
View full question & answer→MCQ 1282 Marks
If $k=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}$, then the numerical value of $k$ is
- A
$\frac{1}{4}$
- ✓
$\frac{1}{8}$
- C
$\frac{1}{16}$
- D
$\frac{1}{32}$
AnswerCorrect option: B. $\frac{1}{8}$
(B)
$k=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}$
$=\cos \left(\frac{\pi}{2}-\frac{\pi}{18}\right) \cos \left(\frac{\pi}{2}-\frac{5 \pi}{18}\right) \cos \left(\frac{\pi}{2}-\frac{7 \pi}{18}\right)$
$\cos \alpha \cdot \cos 2 \alpha \cdot \cos 2^2 \alpha \cdot \cos 2^3 \alpha \ldots . . \cos 2^{n-1} \alpha$
$=\frac{\sin 2^n \alpha}{2^n \sin \alpha}, \text { if } \alpha \neq n \pi$
$=1, \text { if } \alpha=2 n \pi$
$=-1, \text { if } \alpha=(2 n+1) \pi$
$=\cos \frac{\pi}{9} \cos \frac{2 \pi}{9} \cos \frac{4 \pi}{9}=\frac{\sin 2^3 \frac{\pi}{9}}{2^3 \sin \frac{\pi}{9}}$
$=\frac{\sin \frac{8 \pi}{9}}{8 \sin \frac{\pi}{9}}=\frac{\sin \left(\pi-\frac{\pi}{9}\right)}{8 \sin \frac{\pi}{9}}$
$=\frac{1}{8}$
View full question & answer→MCQ 1292 Marks
$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=$
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
$\frac{1}{8}$
- ✓
$\frac{1}{16}$
AnswerCorrect option: D. $\frac{1}{16}$
(D)
$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$
$\cos \alpha \cdot \cos 2 \alpha \cdot \cos 2^2 \alpha \cdot \cos 2^3 \alpha \ldots \cos 2^{n-1} \alpha$
$=\frac{\sin 2^n \alpha}{2^n \sin \alpha}, \text { if } \alpha \neq n \pi$
$=1, \text { if } \alpha=2 n \pi$
$=-1, \text { if } \alpha=(2 n+1) \pi$
$=\frac{\sin 2^4 \frac{2 \pi}{15}}{2^4 \sin \frac{2 \pi}{15}}=\frac{\sin \frac{32 \pi}{15}}{16 \sin \frac{2 \pi}{15}}$
$=\frac{\sin \left(2 \pi+\frac{2 \pi}{15}\right)}{16 \sin \frac{2 \pi}{15}}=\frac{\sin \frac{2 \pi}{15}}{16 \sin \frac{2 \pi}{15}}$
$=\frac{1}{16}$
View full question & answer→MCQ 1302 Marks
$\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}=$
- A
$\frac{1}{16}$
- B
$0$
- C
$\frac{-1}{8}$
- ✓
$\frac{-1}{16}$
AnswerCorrect option: D. $\frac{-1}{16}$
(D)
$\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}$
$\cos \alpha \cdot \cos 2 \alpha \cdot \cos 2^2 \alpha \cdot \cos 2^3 \alpha \ldots . . \cos 2^{n-1} \alpha$
$=\frac{\sin 2^n \alpha}{2^n \sin \alpha}, \text { if } \alpha \neq n \pi$
$=1, \text { if } \alpha=2 n \pi$
$=-1, \text { if } \alpha=(2 n+1) \pi$
$ =\frac{\sin \frac{2^4 \pi}{5}}{2^4 \sin \frac{\pi}{5}}=\frac{\sin \frac{16 \pi}{5}}{16 \sin \frac{\pi}{5}} $
$=\frac{\sin \left(3 \pi+\frac{\pi}{5}\right)}{16 \sin \frac{\pi}{5}}$
$=\frac{-\sin \frac{\pi}{5}}{16 \sin \frac{\pi}{5}}=-\frac{1}{16}$
View full question & answer→MCQ 1312 Marks
$\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7}=$
- A
$0$
- B
$\frac{1}{2}$
- C
$\frac{1}{4}$
- ✓
$-\frac{1}{8}$
AnswerCorrect option: D. $-\frac{1}{8}$
(D)
$\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7}$
$\cos \alpha \cdot \cos 2 \alpha \cdot \cos 2^2 \alpha \cdot \cos 2^3 \alpha \ldots . . \cos 2^{n-1} \alpha$
$=\frac{\sin 2^n \alpha}{2^n \sin \alpha}, \text { if } \alpha \neq n \pi$
$=1, \text { if } \alpha=2 n \pi$
$=-1, \text { if } \alpha=(2 n+1) \pi$
$=\left[\frac{\sin \left(2^3 \cdot \frac{\pi}{7}\right)}{2^3 \sin \left(\frac{\pi}{7}\right)}\right]$
$=\frac{\sin \frac{8 \pi}{7}}{8 \sin \frac{\pi}{7}}$
$=-\frac{1}{8} \quad \ldots\left[\because \sin \frac{8 \pi}{7}=\sin \left(\pi+\frac{\pi}{7}\right)=-\sin \frac{\pi}{7}\right]$
View full question & answer→MCQ 1322 Marks
If $\theta=\frac{\pi}{2^n+1}$, then
$\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{n-1} \theta$ is equal to
- ✓
$\frac{1}{2^n}$
- B
$\cos \theta$
- C
- D
$2^n$
AnswerCorrect option: A. $\frac{1}{2^n}$
(A)
$\cos \alpha \cdot \cos 2 \alpha \cdot \cos 2^2 \alpha \cdot \cos 2^3 \alpha \ldots . . \cos 2^{n-1} \alpha$
$=\frac{\sin 2^n \alpha}{2^n \sin \alpha}, \text { if } \alpha \neq n \pi$
$=1, \text { if } \alpha=2 n \pi$
$=-1, \text { if } \alpha=(2 n+1) \pi$
$\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^n \sin \theta}$
$=\frac{\sin (\pi-\theta)}{2^n \sin \theta} \ldots\left[\begin{array}{r}\because \theta=\frac{n}{2^n+1}(\text { given }) \\ \Rightarrow 2^n \theta+\theta=\pi \\ \Rightarrow 2^n \theta=\pi-\theta\end{array}\right]$
$=\frac{1}{2^n}$
View full question & answer→MCQ 1332 Marks
Maximum and minimum values of $\sin ^4 \theta+\cos ^4 \theta$ are
- A
$0$,2
- ✓
$1, \frac{1}{2}$
- C
$-1,1$
- D
$1,-\frac{1}{2}$
AnswerCorrect option: B. $1, \frac{1}{2}$
(B)
$\sin ^4 \theta+\cos ^4 \theta=\left(\sin ^2 \theta+\cos ^2 \theta\right)^2$$-2 \sin ^2 \theta \cos ^2 \theta$
$=1-\frac{1}{2}(\sin 2 \theta)^2$
Since $0 \leq \sin ^2 2 \theta \leq 1$
$\therefore \quad 0 \geq-\frac{1}{2} \sin ^2 2 \theta \geq-\frac{1}{2}$
$\Rightarrow 1+0>1-\frac{1}{2} \sin ^2 2 \theta>1-\frac{1}{2}$
$\Rightarrow 1 \geq \sin ^4 \theta+\cos ^4 \theta \geq \frac{1}{2}$
View full question & answer→MCQ 1342 Marks
If $|\tan A|<1$ and $|A|$ is acute, then
$\frac{\sqrt{1+\sin 2 A}+\sqrt{1-\sin 2 A}}{\sqrt{1+\sin 2 A}-\sqrt{1-\sin 2 A}}$ is equal to
Answer(C)
$|\tan A |<1$ and $| A |$ is acute.
$\therefore \quad-\frac{\pi}{4}< A <\frac{\pi}{4} \Rightarrow \cos A>\sin A$
$\therefore \frac{\sqrt{1+\sin 2 A}+\sqrt{1-\sin 2 A}}{\sqrt{1+\sin 2 A}-\sqrt{1-\sin 2 A}}$
$=\frac{\sqrt{(\cos A+\sin A)^2}+\sqrt{(\cos A-\sin A)^2}}{\sqrt{(\cos A+\sin A)^2}-\sqrt{(\cos A-\sin A)^2}}$
$=\frac{|\cos A+\sin A|+|\cos A-\sin A|}{|\cos A+\sin A|-|\cos A-\sin A|}$
$=\frac{(\cos A+\sin A)+(\cos A-\sin A)}{(\cos A+\sin A)-(\cos A-\sin A)}=\cot A$
View full question & answer→MCQ 1352 Marks
$\frac{\sqrt{3}}{\sin \left(20^{\circ}\right)}-\frac{1}{\cos \left(20^{\circ}\right)}=$
Answer(C)
$\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$
$=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
$=\frac{2\left[\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right]}{\frac{2}{2} \sin 20^{\circ} \cos 20^{\circ}}$
$=\frac{4 \cos \left(20^{\circ}+30^{\circ}\right)}{\sin 40^{\circ}}=\frac{4 \cos 50^{\circ}}{\sin 40^{\circ}}$
$=\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}}$
= 4
View full question & answer→MCQ 1362 Marks
The value of $\tan \left(1^{\circ}\right)+\tan \left(89^{\circ}\right)$ is_________
- A
$\frac{2}{\sin \left(1^{\circ}\right)}$
- B
$\frac{1}{\sin \left(1^{\circ}\right)}$
- C
$\frac{1}{\sin \left(2^{\circ}\right)}$
- ✓
$\frac{2}{\sin \left(2^{\circ}\right)}$
AnswerCorrect option: D. $\frac{2}{\sin \left(2^{\circ}\right)}$
(D)
$\tan \left(1^{\circ}\right)+\tan \left(89^{\circ}\right)$
$=\tan 1^{\circ}+\cot 1^{\circ} \quad \ldots\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$
$=\frac{\tan ^2 1^{\circ}+1}{\tan 1^{\circ}}$
$=\frac{\sec ^2 1^{\circ}}{\tan 1^{\circ}}=\frac{1}{\sin 1^{\circ} \cos 1^{\circ}}=\frac{2}{\sin 2^{\circ}}$
View full question & answer→MCQ 1372 Marks
If $x=\cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}$, then the value of $x$ is
- A
$\frac{1}{4} \tan 10^{\circ}$
- B
$\frac{1}{8} \cot 10^{\circ}$
- C
$\frac{1}{8} \operatorname{cosec} 10^{\circ}$
- ✓
$\frac{1}{8} \sec 10^{\circ}$
AnswerCorrect option: D. $\frac{1}{8} \sec 10^{\circ}$
(D)
$A-\cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}$
$=\frac{1}{2 \sin 10^{\circ}}\left(2 \sin 10^{\circ} \cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right)$
$=\frac{1}{2.2 \sin 10^{\circ}}\left(2 \sin 20^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right)$
$=\frac{1}{2.4 \sin 10^{\circ}}\left(2 \sin 40^{\circ} \cos 40^{\circ}\right)$
$=\frac{1}{8 \sin 10^{\circ}}\left(\sin 80^{\circ}\right)$
$=\frac{1}{8 \sin 10^{\circ}}\left(\cos 10^{\circ}\right)=\frac{1}{8} \cot 10^{\circ}$
View full question & answer→MCQ 1382 Marks
$8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$ is equal to
- A
$8 sin x$
- ✓
$sin x$
- C
$cos x$
- D
$8 cos x$
AnswerCorrect option: B. $sin x$
(B)
$8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$
$=4\left(2 \sin \frac{x}{8} \cos \frac{x}{8}\right) \cos \frac{x}{2} \cos \frac{x}{4}$
$=4\left(\sin \frac{x}{4} \cos \frac{x}{2} \cos \frac{x}{4}\right)$
$\ldots[\because 2 \sin A \cos A=\sin 2 A]$
$=2\left(2 \sin \frac{x}{4} \cos \frac{x}{4}\right) \cos \frac{x}{2}$
$=2 \sin \frac{x}{2} \cos \frac{x}{2}=\sin x$
View full question & answer→MCQ 1392 Marks
If $2 \sec 2 \alpha=\tan \beta+\cot \beta$, then one of the values of $\alpha+\beta$ is
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{2}$
- C
$\pi$
- D
$2 \pi$
AnswerCorrect option: A. $\frac{\pi}{4}$
(A)
$2 \sec 2 \alpha=\tan \beta+\cot \beta$
$\Rightarrow \frac{2}{\cos 2 \alpha}=\frac{\sin \beta}{\cos \beta}+\frac{\cos \beta}{\sin \beta}$
$=\frac{\sin ^2 \beta+\cos ^2 \beta}{\cos \beta \sin \beta}=\frac{1}{\cos \beta \cdot \sin \beta}$
$\Rightarrow \cos 2 \alpha=\sin 2 \beta \Rightarrow \cos 2 \alpha=\cos \left(\frac{\pi}{2}-2 \beta\right)$
$\Rightarrow 2 \alpha=\frac{\pi}{2}-2 \beta \Rightarrow 2 \alpha+2 \beta=\frac{\pi}{2}$
$\Rightarrow \alpha+\beta=\frac{\pi}{4}$
View full question & answer→MCQ 1402 Marks
If $\tan \theta-\cot \theta=a$ and $\sin \theta+\cos \theta=b$, then $\left(b^2-1\right)^2\left(a^2+4\right)$ is equal to
Answer(D)
$\tan \theta-\cot \theta= a$ and $\sin \theta+\cos \theta= b$
$\therefore \quad\left(b^2-1\right)^2\left(a^2+4\right)$
$=\left\{(\sin \theta+\cos \theta)^2-1\right\}^2\left\{(\tan \theta-\cot \theta)^2+4\right\}$
$=(1+\sin 2 \theta-1)^2\left(\tan ^2 \theta+\cot ^2 \theta-2+4\right)$
$-\sin ^2 2 \theta\left(\operatorname{cosec}^2 \theta+\operatorname{scc}^2 \theta\right)$
$=4 \sin ^2 \theta \cos ^2 \theta\left(\frac{1}{\sin ^2 \theta}+\frac{1}{\cos ^2 \theta}\right)=4$
View full question & answer→MCQ 1412 Marks
If $\sin 2 \theta=\frac{3}{4}$, then $\sin ^3 \theta+\cos ^3 \theta=$
- A
$\frac{\sqrt{5}}{8}$
- B
$\frac{\sqrt{7}}{8}$
- C
$\frac{\sqrt{11}}{8}$
- ✓
$\frac{5 \sqrt{7}}{16}$
AnswerCorrect option: D. $\frac{5 \sqrt{7}}{16}$
(D)
$\sin ^3 \theta+\cos ^3 \theta$
$=(\sin \theta+\cos \theta)\left(\cos ^2 \theta+\sin ^2 \theta-\frac{\sin 2 \theta}{2}\right)$
$=\sqrt{(\sin \theta+\cos \theta)^2}\left(1-\frac{\sin 2 \theta}{2}\right)$
$\Rightarrow \sin ^3 \theta+\cos ^3 \theta=\sqrt{1+\frac{3}{4}}\left(1-\frac{3}{8}\right)$
$=\frac{\sqrt{7}}{2} \times \frac{5}{8}=\frac{5 \sqrt{7}}{16}$
View full question & answer→MCQ 1422 Marks
$3(\sin x-\cos x)^4+6(\sin x +\cos x)^2 +4\left(\sin ^6 x+\cos ^6 x\right)=$
Answer(D)
$3(\sin x-\cos x)^4+6(\sin x+\cos x)^2$$+4\left(\sin ^6 x+\cos ^6 x\right)$
$=3\left(\sin ^2 x+\cos ^2 x-2 \sin \ x \ \cos \ x\right)^2+6\left(\sin ^2 x+\cos ^2 x+2 \sin x \cos x\right)+4\left(\sin ^6 x+\cos ^6 x\right)$
$=3(1-\sin 2 x)^2+6(1+\sin 2 x)$$+4\left(1-3 \sin ^2 x \cos ^2 x\right)$
$=3+3 \sin ^2 2 x-6 \sin 2 x+6 +6 \sin 2 x$$+4-3 \sin ^2 2 x$
$=9+4+3 \sin ^2 2 x-3 \sin ^2 2 x=13$
View full question & answer→MCQ 1432 Marks
Which of the following numbers is/are rational?
AnswerCorrect option: C. $\sin 15^{\circ} \cos 15^{\circ}$
(C)
$\sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}}=$ irrational
$\cos 15^{\circ}=\cos \left(45^{\circ}-30^{\circ}\right)=\frac{\sqrt{3}+1}{2 \sqrt{2}}=$ irrational
$\sin 15^{\circ} \cos 15^{\circ}=\frac{1}{2}\left(2 \sin 15^{\circ} \cos 15^{\circ}\right)$
$=\frac{1}{2} \sin 30^{\circ}=\frac{1}{2} \cdot \frac{1}{2}$
$=\frac{1}{4}=$ rational
$\sin 15^{\circ} \cos 75^{\circ}=\sin 15^{\circ} \sin 15^{\circ}$
$=\sin ^2 15^{\circ}$
$=\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2$
$=\frac{4-2 \sqrt{3}}{8}=$ irrational
View full question & answer→MCQ 1442 Marks
If $A$ lies in the third quadrant and $3 \tan A-4=0,$ then $5 \sin \ 2 A+3 \ \sin A+4 \ \cos A=$
- ✓
$0$
- B
$\frac{-24}{5}$
- C
$\frac{24}{5}$
- D
$\frac{48}{5}$
Answer(A)
$3 \tan A-4=0$
$\Rightarrow \tan A=\frac{4}{3}$
$\Rightarrow \sin A=-\frac{4}{5}, \cos A=-\frac{3}{5}$
$\therefore 5 \sin 2 A+3 \sin A+4 \cos A$
$=10 \sin A \cos A +3 \sin A+4 \cos A$
$=10\left(\frac{12}{25}\right)-\frac{12}{5}-\frac{12}{5}=0$
View full question & answer→MCQ 1452 Marks
$\sin^2\left(3^{\circ}\right)+\sin^2\left(6^{\circ}\right)+\sin^2\left(9^{\circ}\right)+\ldots+\sin^2\left(84^{\circ}\right)+\sin^2\left(87^{\circ}\right)+\sin^2\left(90^{\circ}\right)=$
- ✓
$\frac{31}{2}$
- B
$\frac{39}{2}$
- C
$\frac{59}{2}$
- D
AnswerCorrect option: A. $\frac{31}{2}$
(A)
$\sin^2\left(3^{\circ}\right)+\sin^2\left(6^{\circ}\right)+\sin^2\left(9^{\circ}\right)+\ldots+\sin^2\left(84^{\circ}\right)$$+\sin^2\left(87^{\circ}\right)+\sin^2\left(90^{\circ}\right)$
$=\sin^2\left(3^{\circ}\right)+\sin^2\left(6^{\circ}\right)+\sin^2\left(9^{\circ}\right)+\ldots$$+\cos^2\left(6^{\circ}\right)+\cos^2\left(3^{\circ}\right)+\sin^2\left(90^{\circ}\right)$
$\ldots\left[\because\sin\left(90^{\circ}-\theta\right)=\cos\theta\right]$
$=\left[\sin^2\left(3^{\circ}\right)+\cos^2\left(3^{\circ}\right)\right]+\ldots$$+\left[\sin^2\left(42^{\circ}\right)+\cos^2\left(42^{\circ}\right)\right]+\sin^2\left(30^{\circ}\right)$$+\sin^2\left(45^{\circ}\right)+\sin^2\left(60^{\circ}\right)+\sin^2\left(90^{\circ}\right)$
$=(1+1+\ldots+1)+\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2+(1)^2$
$=13+\frac{1}{4}+\frac{1}{2}+\frac{3}{4}+1$
$=\frac{31}{2}$
View full question & answer→MCQ 1462 Marks
$\cos^2\frac{\pi}{12}+\cos^2\frac{\pi}{4}+\cos^2\frac{5\pi}{12}=$
- A
$\frac{2}{3+\sqrt{3}}$
- B
$\frac{2}{3}$
- C
$\frac{3+\sqrt{3}}{2}$
- ✓
$\frac{3}{2}$
AnswerCorrect option: D. $\frac{3}{2}$
(D)
$\cos^7\frac{\pi}{12}+\cos^7\frac{\pi}{4}+\cos^7\frac{5\pi}{12}$
$=\cos^215^{\circ}+\cos^245^{\circ}+\cos^275^{\circ}$
$=\cos^215^{\circ}+\cos^275^{\circ}+\cos^245^{\circ}$
$=\cos^215^{\circ}+\sin^215^{\circ}+\left(\frac{1}{\sqrt{2}}\right)^2$
$\ldots\left[\because\cos^2\theta=\sin^2\left(90^{\circ}-\theta\right)\right]$
$=1+\frac{1}{2}$
$=\frac{3}{2}$
View full question & answer→MCQ 1472 Marks
$\cos^2\left(\frac{\pi}{4}-\beta\right)-\sin^2\left(\alpha-\frac{\pi}{4}\right)=$
- A
$\sin(\alpha+\beta)\sin(\alpha-\beta)$
- B
$\cos(\alpha+\beta)\cos(\alpha-\beta)$
- C
$\sin(\alpha-\beta)\cos(\alpha+\beta)$
- ✓
$\sin(\alpha+\beta)\cos(\alpha-\beta)$
AnswerCorrect option: D. $\sin(\alpha+\beta)\cos(\alpha-\beta)$
(D)
$\cos^2\left(\frac{\pi}{4}-\beta\right)-\sin^2\left(\alpha-\frac{\pi}{4}\right)$
$=\cos\left(\frac{\pi}{4}-\beta+\alpha-\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}-\beta-\alpha+\frac{\pi}{4}\right)$
$=\cos(\alpha-\beta)\cos\left(\frac{\pi}{2}-(\alpha+\beta)\right)$
$=\cos(\alpha-\beta)\sin(\alpha+\beta)$
View full question & answer→MCQ 1482 Marks
$\sec50^{\circ}+\tan50^{\circ}$is equal to
- A
$\tan20^{\circ}+\tan50^{\circ}$
- B
$2\tan20^{\circ}+\tan50^{\circ}$
- ✓
$\tan20^{\circ}+2\tan50^{\circ}$
- D
$2\tan20^{\circ}+2\tan50^{\circ}$
AnswerCorrect option: C. $\tan20^{\circ}+2\tan50^{\circ}$
(C)
$\sec50^{\circ}+\tan50^{\circ}$
$=\frac{1}{\cos50^{\circ}}+\tan50^{\circ}$
$=\frac{\cos20^{\circ}}{\cos20^{\circ}\cos50^{\circ}}+\tan50^{\circ}$
$=\frac{\sin70^{\circ}}{\cos20^{\circ}\cos50^{\circ}}+\tan50^{\circ}$
$\ldots\left[\because\cos\theta=\sin\left(90^{\circ}-\theta\right)\right]$
$=\frac{\sin\left(50^{\circ}+20^{\circ}\right)}{\cos20^{\circ}\cos50^{\circ}}+\tan50^{\circ}$
$=\frac{\sin50^{\circ}\cos20^{\circ}+\cos50^{\circ}\sin20^{\circ}}{\cos20^{\circ}\cos50^{\circ}}+\tan50^{\circ}$
$=\frac{\sin50^{\circ}\cos20^{\circ}}{\cos20^{\circ}\cos50^{\circ}}+\frac{\cos50^{\circ}\sin20^{\circ}}{\cos20^{\circ}\cos50^{\circ}}+\tan50^{\circ}$
$=\tan50^{\circ}+\tan20^{\circ}+\tan50^{\circ}$
$=2\tan50^{\circ}+\tan20^{\circ}$
View full question & answer→MCQ 1492 Marks
$\tan70^{\circ}$ is equal to
- A
$\tan20^{\circ}+\tan50^{\circ}$
- B
$2\tan20^{\circ}+\tan50^{\circ}$
- ✓
$\tan20^{\circ}+2\tan50^{\circ}$
- D
$2\tan20^{\circ}+2\tan50^{\circ}$
AnswerCorrect option: C. $\tan20^{\circ}+2\tan50^{\circ}$
(C)
$\tan50^{\circ}=\tan\left(70^{\circ}-20^{\circ}\right)=\frac{\tan70^{\circ}-\tan20^{\circ}}{1+\tan70^{\circ}\tan20^{\circ}}$
$\Rightarrow\tan50^{\circ}+\tan70^{\circ}\tan20^{\circ}\tan50^{\circ}$
$\Rightarrow\tan70^{\circ}-\tan20^{\circ}$
$\Rightarrow\tan50^{\circ}+\tan50^{\circ}=\tan70^{\circ}-\tan20^{\circ}$
$\ldots\left[\because\tan70^{\circ}=\cot20^{\circ}\right]$
$\Rightarrow2\tan50^{\circ}+\tan20^{\circ}=\tan70^{\circ}$
View full question & answer→MCQ 1502 Marks
If $\alpha+\beta=\frac{\pi}{2}$and$\beta+\gamma=\alpha$, then $\tan\alpha$ equals
AnswerCorrect option: C. $\tan\beta+2\tan\gamma$
(C)
$\beta+\gamma=\alpha$
$\Rightarrow\gamma=\alpha-\beta$
$\Rightarrow\tan\gamma=\tan(\alpha-\beta)$
$\Rightarrow\tan\gamma=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$
$\Rightarrow\tan\gamma=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\cot\alpha}$
$\ldots\left[\because\alpha+\beta=\frac{\pi}{2},\therefore\beta=\frac{\pi}{2}-\alpha\right]$
$\Rightarrow\tan\gamma=\frac{1}{2}(\tan\alpha-\tan\beta)$
$\Rightarrow\tan\alpha=\tan\beta+2\tan\gamma$
View full question & answer→MCQ 1512 Marks
If $\tan20^{\circ}=\lambda$, then$\frac{\tan160^{\circ}-\tan110^{\circ}}{1+\left(\tan160^{\circ}\right)\left(\tan110^{\circ}\right)}=$
- A
$\frac{1+\lambda^2}{2\lambda}$
- B
$\frac{1+\lambda^2}{\lambda}$
- C
$\frac{1-\lambda^2}{\lambda}$
- ✓
$\frac{1-\lambda^2}{2\lambda}$
AnswerCorrect option: D. $\frac{1-\lambda^2}{2\lambda}$
(D)
$\frac{\tan160^{\circ}-\tan110^{\circ}}{1+\left(\tan160^{\circ}\right)\left(\tan110^{\circ}\right)}$
$=\frac{\tan\left(180^{\circ}-160^{\circ}\right)-\cot\left(90^{\circ}-110^{\circ}\right)}{1+\left[\tan\left(180^{\circ}-160^{\circ}\right)\cot\left(90^{\circ}-110^{\circ}\right)\right]}$
$=\frac{-\tan20^{\circ}+\cot20^{\circ}}{1+\left(-\tan20^{\circ}\right)\left(-\cot20^{\circ}\right)}$
$=\frac{-\lambda+\frac{1}{\lambda}}{1+1}=\frac{1-\lambda^2}{2\lambda}$
View full question & answer→MCQ 1522 Marks
$\frac{1-\tan2^{\circ}\cot62^{\circ}}{\tan152^{\circ}-\cot88^{\circ}}=$
- A
$\sqrt{3}$
- ✓
$-\sqrt{3}$
- C
$\sqrt{2}-1$
- D
$1-\sqrt{2}$
AnswerCorrect option: B. $-\sqrt{3}$
(B)
$\frac{1-\tan2^{\circ}\cot62^{\circ}}{\tan152^{\circ}-\cot88^{\circ}}$
$=\frac{1-\tan2^{\circ}\cot62^{\circ}}{\tan\left(90^{\circ}+62^{\circ}\right)-\cot\left(90^{\circ}-2^{\circ}\right)}$
$=\frac{1-\tan2^{\circ}\cot62^{\circ}}{\cot62^{\circ}\tan2^{\circ}}$
$=\frac{\tan62^{\circ}-\tan2^{\circ}}{-\left(1+\tan2^{\circ}\tan62^{\circ}\right)}$
$=-\tan\left(62^{\circ}-2^{\circ}\right)$
$=-\tan60^{\circ}=-\sqrt{3}$
View full question & answer→MCQ 1532 Marks
If$A+B=225^{\circ}$,then$\frac{cotA}{1+cotA}\cdot\frac{cotB}{1+cotB}=$
AnswerCorrect option: D. $\frac{1}{2}$
(D)
$\frac{\cot A }{1+\cot A } \cdot \frac{\cot B }{1+\cot B }$
$=\frac{1}{(1+\tan A)(1+\tan B)}$
$=\frac{1}{\tan A+\tan B+1+\tan A \tan B}$
$=\frac{1}{1-\tan A \tan B+1+\tan A \tan B}$
$\cdots\left[\begin{array}{l}\because \tan (A+B)=\tan 225^{\circ} \\ \Rightarrow \tan A+\tan B=1-\tan A \tan B\end{array}\right]$
$=\frac{1}{2}$
View full question & answer→MCQ 1542 Marks
$\tan100^{\circ}+\tan125^{\circ}+\tan100^{\circ}\tan125^{\circ}=$
Answer(D)
$\tan\left(100^{\circ}+125^{\circ}\right)=\frac{\tan100^{\circ}+\tan125^{\circ}}{1-\tan100^{\circ}\tan125^{\circ}}$
$\Rightarrow\tan225^{\circ}=\frac{\tan100^{\circ}+\tan125^{\circ}}{1-\tan100^{\circ}\tan125^{\circ}}$
$\Rightarrow1=\frac{\tan100^{\circ}+\tan125^{\circ}}{1-\tan100^{\circ}\tan125^{\circ}}$
$\Rightarrow1=\frac{\tan100^{\circ}+\tan125^{\circ}}{1-\tan100^{\circ}\tan125^{\circ}}$
$\ldots\left[\because\tan\left(180^{\circ}+45^{\circ}\right)=\tan45^{\circ}=1\right]$
$\Rightarrow\tan100^{\circ}+\tan125^{\circ}+\tan100^{\circ}\tan125^{\circ}=1$
View full question & answer→MCQ 1552 Marks
$\frac{\cos12^{\circ}-\sin12^{\circ}}{\cos12^{\circ}+\sin12^{\circ}}+\frac{\sin147^{\circ}}{\cos147^{\circ}}=$
Answer(C)
$\frac{\cos12^{\circ} - \sin12^{\circ}}{\cos12^{\circ}+\sin12^{\circ}}+\frac{\sin147^{\circ}}{\cos147^{\circ}}\\=\frac{1-\tan12^{\circ}}{1+\tan12^{\circ}}+\tan147^{\circ}$
$=\tan\left(45^{\circ}-12^{\circ}\right)+\tan\left(180^{\circ}-33^{\circ}\right)$
$=\tan33^{\circ}-\tan33^{\circ}=0$
View full question & answer→MCQ 1562 Marks
$\begin{aligned}3\left[\sin^4\left(\frac{3\pi}{2}-\alpha\right)+\right.&\left.\sin^4(3\pi+\alpha)\right]-2\left[\sin^6\left(\frac{\pi}{2}+\alpha\right)+\sin^6(5\pi-\alpha)\right]=\end{aligned}$
Answer(B)
$3\left\{\sin^4\left(\frac{3\pi}{2}-\alpha\right)+\sin^4(3\pi+\alpha)\right\}$
$-2\left\{\sin^6\left(\frac{\pi}{2}+\alpha\right)+\sin^6(5\pi-\alpha)\right\}$
$=3\left\{(-\cos\alpha)^4+(-\sin\alpha)^4\right\}-2\left(\cos^6\alpha+\sin^6\alpha\right)$
$=3\left(1-2\sin^2\alpha\cos^2\alpha\right)-2\left(1-3\sin^2\alpha\cos^2\alpha\right)$
$=3-6\sin^2\alpha\cos^2\alpha-2+6\sin^2\alpha\cos^2\alpha$
$=3-2=1$
View full question & answer→MCQ 1572 Marks
$\frac{\sin\left(-660^{\circ}\right)\tan\left(1050^{\circ}\right)\sec\left(-420^{\circ}\right)}{\cos\left(225^{\circ}\right)\operatorname{cosec}\left(315^{\circ}\right)\cos\left(510^{\circ}\right)}=$
- A
$\frac{\sqrt{3}}{4}$
- B
$\frac{\sqrt{3}}{2}$
- ✓
$\frac{2}{\sqrt{3}}$
- D
$\frac{4}{\sqrt{3}}$
AnswerCorrect option: C. $\frac{2}{\sqrt{3}}$
(C)
Givenexpression$=\frac{-\sin\left(660^{\circ}\right)\tan\left(1050^{\circ}\right)\sec\left(420^{\circ}\right)}{\cos\left(180^{\circ}+45^{\prime}\right)\operatorname{cosec}\left(360^{\circ}-45^{\prime}\right)\cos\left(360^{\circ}+150^{\circ}\right)}$
$=\frac{-\sin\left(7\times90+30^{\circ}\right)\tan\left(3\times360^{\circ}-30^{\circ}\right)\sec\left(360^{\circ}+60^{\circ}\right)}{\left(-\cos45^{\circ}\right)\left(-\operatorname{cosec}45^{\circ}\right)\cos150^{\circ}}$
$=\frac{\cos\left(30^{\circ}\right)\left(-\tan30^{\circ}\right)\sec60^{\circ}}{\left(-\cos45^{\circ}\right)\left(-\operatorname{cosec}45^{\circ}\right)\left(-\cos30^{\circ}\right)}$
$=\frac{\frac{1}{\sqrt{3}}\cdot2}{\frac{1}{\sqrt{2}}\cdot\sqrt{2}}$
$=\frac{2}{\sqrt{3}}$
View full question & answer→MCQ 1582 Marks
The value of $\sin600^{\circ}\cos330^{\circ}+\cos120^{\circ}\sin150^{\circ}$ is
- ✓
- B
- C
$\frac{1}{\sqrt{2}}$
- D
$\frac{\sqrt{3}}{2}$
Answer(A)
$\sin600^{\circ}\cos330^{\circ}+\cos120^{\circ}\sin150^{\circ}$
$=-\sin60^{\circ}\cos30^{\circ}-\sin30^{\circ}\cos60^{\circ}$
$=-\left\{\sin\left(60^{\circ}+30^{\circ}\right)\right\}$
$=-1$
View full question & answer→MCQ 1592 Marks
The value of $\cos105^{\circ}+\sin105^{\circ}$ is
- A
$\frac{1}{2}$
- B
- C
$\sqrt{2}$
- ✓
$\frac{1}{\sqrt{2}}$
AnswerCorrect option: D. $\frac{1}{\sqrt{2}}$
(D)
$\cos105^{\circ}+\sin105^{\circ}$
$=\cos\left(90^{\circ}+15^{\circ}\right)+\sin\left(90^{\circ}+15^{\circ}\right)$
$=\cos15^{\circ}-\sin15^{\circ}$
$=\cos\left(45^{\circ}-30^{\circ}\right)-\sin\left(45^{\circ}-30^{\circ}\right)$
$=\frac{\sqrt{3+1}}{2\sqrt{2}}-\frac{\sqrt{3-1}}{2\sqrt{2}}$
$=\frac{2}{2\sqrt{2}}$
$=\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 1602 Marks
$\tan75^{\circ}-\cot75^{\circ}=$
- ✓
$2\sqrt{3}$
- B
$2+\sqrt{3}$
- C
$2-\sqrt{3}$
- D
$-2\sqrt{3}$
AnswerCorrect option: A. $2\sqrt{3}$
(A)
$\tan75^{\circ}-\cot75^{\circ}$
$=\tan\left(90^{\circ}-15^{\circ}\right)-\cot75^{\circ}$
$=\cot15^{\circ}-\cot75^{\circ}$
$=(2+\sqrt{3})-(2-\sqrt{3})$
$=2\sqrt{3}$
View full question & answer→MCQ 1612 Marks
$\tan\left(\frac{\pi}{4}+\theta\right)\tan\left(\frac{3\pi}{4}+\theta\right)$is equal to
Answer(B)
$\tan\left(\frac{\pi}{4}+\theta\right)\tan\left(\frac{3\pi}{4}+\theta\right)$
$=\tan\left(\frac{\pi}{4}+\theta\right)\tan\left(\frac{\pi}{2}+\left(\frac{\pi}{4}+\theta\right)\right)$
$=\tan\left(\frac{\pi}{4}+\theta\right)\left\{-\cot\left(\frac{\pi}{4}+\theta\right)\right\}$
$\ldots\left[\because\tan\left(\frac{\pi}{2}+\alpha\right)=-\cot\alpha\right]$
$=-1$
View full question & answer→MCQ 1622 Marks
$\cot\left(45^{\circ}+\theta\right)\cot\left(45^{\circ}-\theta\right)=$
Answer(C)
$\cot\left(45^{\circ}+\theta\right)\cot\left(45^{\circ}-\theta\right)$
$=\tan\left(90^{\circ}-45^{\circ}-\theta\right)\cot\left(45^{\circ} \theta\right)$
$\ldots\left[\because\tan\left(90^{\circ}-\theta\right)=\cot\theta\right]$
$=\tan\left(45^{\circ}-\theta\right)\cot\left(45^{\circ}-\theta\right)=1$
View full question & answer→MCQ 1632 Marks
If $\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, then $\frac{\sin (\alpha+\beta+\gamma)}{\sin \alpha+\sin \beta+\sin \gamma}$ is
Answer(A)
$\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)$
$=\sin \alpha+\sin \beta+\sin \gamma-\sin \alpha \cos \beta \cos \gamma$$-\cos \alpha \sin \beta \cos \gamma-\cos \alpha \cos \beta \sin \gamma$$+\sin \alpha \sin \beta \sin \gamma$
$=\sin \alpha(1-\cos \beta \cos \gamma)+\sin \beta(1-\cos \alpha \cos \gamma)$$+\sin \gamma(1-\cos \alpha \cos \beta)+\sin \alpha \sin \beta \sin \gamma>0$
$\therefore \quad \sin \alpha+\sin \beta+\sin \gamma>\sin (\alpha+\beta+\gamma)$
$\Rightarrow \frac{\sin (\alpha+\beta+\gamma)}{\sin \alpha + \sin \beta + \sin \gamma}<1$
View full question & answer→MCQ 1642 Marks
If $\tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \sin ^2 \alpha}$, then $\tan (\alpha-\beta)$ is equal to
AnswerCorrect option: B. $(1-n) \tan \alpha$
(B)
$\tan \beta=\frac{ n \sin \alpha \cos \alpha}{1- n \sin ^2 \alpha}$
$\Rightarrow \tan \beta=\frac{ n \tan \alpha}{\sec ^2 \alpha- n \tan ^2 \alpha}$
$=\frac{ n \tan \alpha}{1+\tan ^2 \alpha- n \tan ^2 \alpha}$
Now, $\tan (\alpha-\beta)$
$=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}$
$=\frac{\tan \alpha-\frac{ n \tan \alpha}{1+\tan ^2 \alpha- n \tan ^2 \alpha}}{1+\tan \alpha\left(\frac{ n \tan \alpha}{1+\tan ^2 \alpha- n \tan ^2 \alpha}\right)}$
$=\frac{\tan \alpha+\tan ^3 \alpha- n \tan ^3 \alpha- n \tan \alpha}{1+\tan ^2 \alpha- n \tan ^2 \alpha+ n \tan ^2 \alpha}$
$=\frac{\tan \alpha\left(1+\tan ^2 \alpha\right)- n \tan \alpha\left(1+\tan ^2 \alpha\right)}{\left(1+\tan ^2 \alpha\right)}$
$=(1-n)(\tan \alpha)$
View full question & answer→MCQ 1652 Marks
$\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}=$
Answer(D)
$\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}$
$=\frac{1}{\tan 3 A-\tan A}+\frac{\tan A \tan 3 A}{\tan 3 A-\tan A}$
$=\frac{1}{\frac{\tan 3 A-\tan A }{1+\tan 3 A \cdot \tan A }}=\frac{1}{\tan 2 A}=\cot 2 A$
View full question & answer→MCQ 1662 Marks
If $\tan A=2 \tan B+\cot B$, then $2 \tan (A-B)$ is equal to
Answer(C)
$2 \tan (A-B)=2\left(\frac{\tan A-\tan B}{1+\tan A \tan B}\right)$
$=2\left[\frac{2 \tan B+\cot B-\tan B}{1+(2 \tan B+\cot B) \tan B}\right]$
$\ldots[\because \tan A=2 \tan B+\cot B]$
$=2\left[\frac{\tan B+\cot B}{2\left(1+\tan ^2 B\right)}\right]$
$=\frac{\cot B \left(\tan ^2 B+1\right)}{1+\tan ^2 B}=\cot B$
View full question & answer→MCQ 1672 Marks
$\tan \frac{2 \pi}{5}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{2 \pi}{5} \tan \frac{\pi}{15}$ is equal to
- A
$-\sqrt{3}$
- B
$\frac{1}{\sqrt{3}}$
- C
- ✓
$\sqrt{3}$
AnswerCorrect option: D. $\sqrt{3}$
(D)
$\tan \left(\frac{6 \pi}{15}-\frac{\pi}{15}\right)-\tan \frac{\pi}{3}$
$\Rightarrow \frac{\tan \frac{6 \pi}{15}-\tan \frac{\pi}{15}}{1+\tan \frac{6 \pi}{15} \tan \frac{\pi}{15}}=\tan \frac{\pi}{3}$
$\Rightarrow \tan \frac{6 \pi}{15}-\tan \frac{\pi}{15}=\sqrt{3}+\sqrt{3} \tan \frac{6 \pi}{15} \tan \frac{\pi}{15}$
$\Rightarrow \tan \frac{6 \pi}{15}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{6 \pi}{15} \tan \frac{\pi}{15}=\sqrt{3}$
$\Rightarrow \tan \frac{2 \pi}{5}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{2 \pi}{5} \tan \frac{\pi}{15}=\sqrt{3}$
View full question & answer→MCQ 1682 Marks
$\tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}=$
- A
$\frac{1}{\sqrt{3}}$
- ✓
$\sqrt{3}$
- C
$-\frac{1}{\sqrt{3}}$
- D
$-\sqrt{3}$
AnswerCorrect option: B. $\sqrt{3}$
(B)
$\tan \left(20^{\circ}+40^{\circ}\right)=\frac{\tan 20^{\circ}+\tan 40^{\circ}}{1-\tan 20^{\circ} \tan 40^{\circ}}$
$\Rightarrow \sqrt{3}=\frac{\tan 20^{\circ}+\tan 40^{\circ}}{1-\tan 20^{\circ} \tan 40^{\circ}}$
$\Rightarrow \sqrt{3}-\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}=\tan 20^{\circ}+\tan 40^{\circ}$
$\Rightarrow \tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}=\sqrt{3}$
View full question & answer→MCQ 1692 Marks
tan 3A - tan 2A - tan A =
- ✓
- B
- C
tan A tan 2A - tan 2A tan 3A
- D
tan 2A tan 3A - tan A tan 2A
Answer(A)
Since $\tan 3 A=\frac{\tan A+\tan 2 A}{1-\tan A \tan 2 A}$
$\Rightarrow \tan 3 A-\tan 2 A-\tan A$
$=\tan 3 A \tan 2 A \tan A$
View full question & answer→MCQ 1702 Marks
if x, y, z are any three real numbers, then tan (x - y) + tan (y - z) + tan (z - x) is equal to
- A
- B
$0$
- ✓
tan(x - y) tan(y - z) tan(z - x)
- D
tan(y - x) tan(z - y) tan(x - z)
AnswerCorrect option: C. tan(x - y) tan(y - z) tan(z - x)
(C)
Let $x-y=\alpha, y-z=\beta$ and $z-x=\gamma$,
then $\alpha+\beta+\gamma=0$
$\Rightarrow \alpha+\beta=-\gamma$
$\Rightarrow \tan (\alpha+\beta)=\tan (-\gamma)$
$\Rightarrow \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}--\tan \gamma$
$\Rightarrow \tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$
View full question & answer→MCQ 1712 Marks
If $\cos (\alpha+\beta)=\frac{3}{5}, \sin (\alpha-\beta)=\frac{5}{13}$ and $0<\alpha, \beta<\frac{\pi}{4}$, then $\tan (2 \alpha)$ is equal to
- A
$\frac{63}{52}$
- ✓
$\frac{63}{16}$
- C
$\frac{21}{16}$
- D
$\frac{33}{52}$
AnswerCorrect option: B. $\frac{63}{16}$
(B)
$0<\alpha, \beta<\frac{\pi}{4}$
$\Rightarrow 0<\alpha+\beta<\frac{\pi}{2}$ and $\frac{-\pi}{4}<\alpha-\beta<\frac{\pi}{4}$
$\cos (\alpha+\beta)=\frac{3}{5} \Rightarrow \tan (\alpha+\beta)=\frac{4}{3}$
$\sin (\alpha-\beta)=\frac{3}{13} \Rightarrow \tan (\alpha-\beta)=\frac{5}{12}$
$\therefore \tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)]$
$=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)}$
$=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}=\frac{63}{16}$
View full question & answer→MCQ 1722 Marks
If $A + B =45^{\circ}$, then $(\cot A -1)(\cot B -1)=$
Answer(B)
$A + B =45^{\circ}$
$\Rightarrow \tan (A+B)=1$
$\Rightarrow \tan A+\tan B=1-\tan A \tan B$
$\Rightarrow \frac{1}{\cot A}+\frac{1}{\cot B}=1-\frac{1}{\cot A \cot B }$
$\Rightarrow \cot A +\cot B =\cot A \cot B -1$
$\Rightarrow \cot A \cot B -\cot A -\cot B =1$
$\Rightarrow \cot A \cot B -\cot A -\cot B +1=2$
$\Rightarrow(\cot A -1)(\cot B -1)=2$
View full question & answer→MCQ 1732 Marks
If $y=(1+\tan A)(1-\tan B)$, where $A-B=\frac{\pi}{4}$,then $(y+1)^{y+1}$ is equal to
Answer(C)
$(1+\tan A )(1-\tan B )=2$
$\Rightarrow y=2$
$\therefore(y+1)^{y+1}-(2+1)^{2+1}-(3)^3-27$
View full question & answer→MCQ 1742 Marks
If $A+B=\frac{\pi}{4}$, then $(1+\tan A)(1+\tan B)=$
Answer(B)
$A + B =\frac{\pi}{4}$
$\Rightarrow \tan ( A + B )=\tan \frac{\pi}{4}$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}-1$
$\Rightarrow \tan A +\tan B +\tan A \tan B =1$
$\Rightarrow(1+\tan A)(1+\tan B)=2$
View full question & answer→MCQ 1752 Marks
If $\tan \theta=\frac{1}{2}$ and $\tan \phi=\frac{1}{3}$, then $\tan (2 \theta+\phi)$ is equal to
Answer(C)
$\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}=1$
$\tan (2 \theta+\phi)=\tan [\theta+(\theta+\phi)]$
$=\frac{\tan \theta+\tan (\theta+\phi)}{1-\tan \theta \tan (\theta+\phi)}=\frac{\frac{1}{2}+1}{1-\frac{1}{2} \cdot 1}=3$
View full question & answer→MCQ 1762 Marks
If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$, then $\sin \alpha+\cos \alpha$ $\sin \alpha-\cos \alpha$ is equal to
- ✓
$\sqrt{2} \cos \theta, \sqrt{2} \sin \theta$
- B
$-\sqrt{2} \sin \theta,-\sqrt{2} \cos \theta$
- C
$\sqrt{2} \sin \theta, \sqrt{2} \sin \theta$
- D
$\sqrt{2} \cos \theta, \sqrt{2} \cos \theta$
AnswerCorrect option: A. $\sqrt{2} \cos \theta, \sqrt{2} \sin \theta$
(A)
We have, $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$
$\Rightarrow \tan \theta=\frac{\frac{1}{\sqrt{2}}(\sin \alpha-\cos \alpha)}{\frac{1}{\sqrt{2}}(\sin \alpha+\cos \alpha)}$
$\Rightarrow \tan \theta=\frac{\sin \alpha \cos \frac{\pi}{4}-\cos \alpha \sin \frac{\pi}{4}}{\sin \alpha \sin \frac{\pi}{4}+\cos \alpha \cos \frac{\pi}{4}}$
$\Rightarrow \tan \theta=\frac{\sin \left(\alpha-\frac{\pi}{4}\right)}{\cos \left(\alpha-\frac{\pi}{4}\right)}$
$\Rightarrow \tan \theta=\tan \left(\alpha-\frac{\pi}{4}\right)$
$\Rightarrow \theta=\alpha-\frac{\pi}{4} \Rightarrow \alpha=\theta+\frac{\pi}{4}$
$\therefore \sin \alpha+\cos \alpha=\sin \left(\theta+\frac{\pi}{4}\right)+\cos \left(\theta+\frac{\pi}{4}\right)$
$=\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta$
$=\frac{2}{\sqrt{2}} \cos \theta=\sqrt{2} \cos \theta$
and $\sin \alpha-\cos \alpha=\sin \left(\theta+\frac{\pi}{4}\right)-\cos \left(\theta+\frac{\pi}{4}\right)$
$=\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta$
$=\frac{2}{\sqrt{2}} \sin \theta=\sqrt{2} \sin \theta$
View full question & answer→MCQ 1772 Marks
If $\cos (\theta-\alpha)=a, \sin (\theta-\beta)=b$, then $\cos ^2(\alpha-\beta)+2 ab \ \sin (\alpha-\beta)$ is equal to
- A
$4 a^2 b^2$
- B
$a^2-b^2$
- ✓
$a^2+b^2$
- D
$-a^2 b^2$
AnswerCorrect option: C. $a^2+b^2$
(C)
$\sin (\alpha-\beta)=\sin [(\theta-\beta)-(\theta-\alpha)]$
$=\sin (\theta-\beta) \cos (\theta-\alpha)-\cos (\theta-\beta) \sin (\theta-\alpha)$
$=b a-\sqrt{1-b^2} \sqrt{1-a^2}$
and $\cos (\alpha-\beta)=\cos [(\theta-\beta)-(\theta-\alpha)]$
$=\cos (\theta-\beta) \cos (\theta-\alpha)+\sin (\theta-\beta) \sin (\theta-\alpha)$
$=a \sqrt{1-b^2}+b \sqrt{1-a^2}$
$\therefore \quad \cos ^2(\alpha-\beta)+2 ab \sin (\alpha-\beta)$
$=\left(a \sqrt{1-b^2}+b \sqrt{1-a^2}\right)^2+2 a b\left(a b-\sqrt{1-a^2} \sqrt{1-b^2}\right)$
$=a^2+b^2$
View full question & answer→MCQ 1782 Marks
If $\sin \theta=3 \sin (\theta+2 \alpha)$, then the value $\tan (\theta+\alpha)+2 \tan \alpha$ is
Answer(D)
$\sin \theta=3 \sin (\theta+2 \alpha)$
$\Rightarrow \sin (\theta+\alpha-\alpha)=3 \sin (\theta+\alpha+\alpha)$
$\Rightarrow \sin (\theta+\alpha) \cos \alpha-\cos (\theta+\alpha) \sin \alpha$
$=3 \sin (\theta+\alpha) \cos \alpha+3 \cos (\theta+\alpha) \sin \alpha$
$\Rightarrow-2 \sin (\theta+\alpha) \cos \alpha=4 \cos (\theta+\alpha) \sin \alpha$
$\Rightarrow \frac{-\sin (\theta+\alpha)}{\cos (\theta+\alpha)}=\frac{2 \sin \alpha}{\cos \alpha}$
$\Rightarrow \tan (\theta+\alpha)+2 \tan \alpha=0$
View full question & answer→MCQ 1792 Marks
If $\sin \alpha=\frac{1}{\sqrt{5}}$ and $\sin \beta=\frac{3}{5}$, then $\beta-\alpha$ lies the interval
- ✓
$\left(0, \frac{\pi}{4}\right)$
- B
$\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$
- C
$[0, \pi]$
- D
$\left(\pi, \frac{5 \pi}{4}\right)$
AnswerCorrect option: A. $\left(0, \frac{\pi}{4}\right)$
(A)
We have, $\sin \alpha=\frac{1}{\sqrt{5}}$
$\therefore \cos \alpha=\sqrt{1-\left(\frac{1}{\sqrt{5}}\right)^2}=\frac{2}{\sqrt{5}}$and $\sin \beta=\frac{3}{5}$
$\therefore \cos \beta=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}$
$\therefore \sin (\beta-\alpha)=\sin \beta \cos \alpha-\cos \beta \sin \alpha$
$=\frac{3}{5} \times \frac{2}{\sqrt{5}}-\frac{4}{5} \times \frac{1}{\sqrt{5}}=\frac{2}{5 \sqrt{5}}$
$=0.1789$
Now, $\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}=0.7071$
Since, $0<0.1789<0.7071$
$\therefore \sin 0<\sin (\beta-\alpha)<\sin \frac{\pi}{4}$
$\Rightarrow 0<(\beta-\alpha)<\frac{\pi}{4}$
View full question & answer→MCQ 1802 Marks
If $\sin A=\frac{1}{\sqrt{10}}$ and $\sin B=\frac{1}{\sqrt{5}}$, where $A$ and are positive acute angles, then A + B =
- A
$\pi$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
(D)
$\sin ( A + B )=\sin A \cos B +\cos A \sin B$
$=\frac{1}{\sqrt{10}} \sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}} \sqrt{1-\frac{1}{10}}$
$=\frac{1}{\sqrt{10}} \sqrt{\frac{4}{5}}+\frac{1}{\sqrt{5}} \sqrt{\frac{9}{10}}$
$=\frac{1}{\sqrt{50}}(2+3)$
$=\frac{5}{\sqrt{50}}$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow \sin (A+B)=\sin \frac{\pi}{4}$
$\Rightarrow A+B=\frac{\pi}{4}$
View full question & answer→MCQ 1812 Marks
If $\sin \theta=\frac{12}{13},\left(0<\theta<\frac{\pi}{2}\right)$ and $\cos \phi=-\frac{3}{5}$ $\left(\pi<\phi<\frac{3 \pi}{2}\right)$, then $\sin (\theta+\phi)$ will be
- A
$\frac{-56}{61}$
- ✓
$\frac{-56}{65}$
- C
$\frac{1}{65}$
- D
$-56$
AnswerCorrect option: B. $\frac{-56}{65}$
(B)
We have, $\sin \theta=\frac{12}{13}$
$\therefore \quad \cos \theta=\sqrt{1-\sin ^2 \theta}$
$=\sqrt{1-\left(\frac{12}{13}\right)^2=\frac{5}{13}} \quad \ldots\left[\because 0<\theta<\frac{\pi}{2}\right]$
and $\cos \phi=\frac{-3}{5}$
$\therefore \quad \sin \phi=\sqrt{1-\frac{9}{25}}=\frac{-4}{5} \quad \ldots\left[\because \pi<\phi<\frac{3 \pi}{2}\right]$
$\therefore \quad \sin (\theta+\phi)=\sin \theta \cdot \cos \phi+\cos \theta \cdot \sin \phi$
$=\left(\frac{12}{13}\right)\left(\frac{-3}{5}\right)+\left(\frac{5}{13}\right)\left(\frac{-4}{5}\right)$
$=\frac{-36}{65}-\frac{20}{65}$
$=\frac{-56}{65}$
View full question & answer→MCQ 1822 Marks
If $\tan \theta_1=k \cot \theta_2$, then $\frac{\cos \left(\theta_1+\theta_2\right)}{\cos \left(\theta_1-\theta_2\right)}=$
- A
$\frac{1+k}{1-k}$
- ✓
$\frac{1-k}{1+k}$
- C
$\frac{k+1}{k-1}$
- D
$\frac{k-1}{k+1}$
AnswerCorrect option: B. $\frac{1-k}{1+k}$
(B)
$\tan \theta_1= k \cot \theta_2$
$\Rightarrow \frac{\tan \theta_1}{\cot \theta_2}= k$
$\Rightarrow \frac{\tan \theta_1+\cot \theta_2}{\tan \theta_1-\cot \theta_2}=\frac{ k +1}{ k -1}$
$\Rightarrow \frac{\sin \theta_1 \sin \theta_2+\cos \theta_1 \cos \theta_2}{\sin \theta_1 \sin \theta_2-\cos \theta_1 \cos \theta_2}=\frac{k+1}{k-1}$
$\Rightarrow \frac{\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2-\sin \theta_1 \sin \theta_2}=\frac{ k +1}{1- k }$
$\Rightarrow \frac{\cos \left(\theta_1-\theta_2\right)}{\cos \left(\theta_1+\theta_2\right)}=\frac{ k +1}{1- k }$
$\Rightarrow \frac{\cos \left(\theta_1+\theta_2\right)}{\cos \left(\theta_1-\theta_2\right)}=\frac{1- k }{1+ k }$
View full question & answer→MCQ 1832 Marks
If $\frac{-\pi}{2}<\theta<\frac{\pi}{2}$ and $\theta \neq \pm \frac{\pi}{4}$, then the value of $\cot \left(\frac{\pi}{4}+\theta\right) \cot \left(\frac{\pi}{4}-\theta\right)$ is
Answer(C)
Since $\cos \left(\frac{\pi}{2}\right)=0$
$\therefore \cos \left[\left(\frac{\pi}{4}+\theta\right)+\left(\frac{\pi}{4}-\theta\right)\right]=0$
$\Rightarrow \cos \left(\frac{\pi}{4}+\theta\right) \cos \left(\frac{\pi}{4}-\theta\right)-\sin \left(\frac{\pi}{4}+\theta\right) \sin \left(\frac{\pi}{4}-\theta\right)=0$
$\Rightarrow \cos \left(\frac{\pi}{4}+\theta\right) \cos \left(\frac{\pi}{4}-\theta\right)=\sin \left(\frac{\pi}{4}+\theta\right) \sin \left(\frac{\pi}{4}-\theta\right)$
$\Rightarrow \cot \left(\frac{\pi}{4}+\theta\right) \cot \left(\frac{\pi}{4}-\theta\right)=1$
View full question & answer→MCQ 1842 Marks
The maximum value of $\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)$ in the interval $\left(0, \frac{\pi}{2}\right)$ is attained at
- A
$x=\frac{\pi}{3}$
- ✓
$x=\frac{\pi}{12}$
- C
$x=\frac{\pi}{6}$
- D
$x=\frac{\pi}{2}$
AnswerCorrect option: B. $x=\frac{\pi}{12}$
(B)
Let $f (x)=\sin \left(x+\frac{\pi}{x}\right)+\cos \left(x+\frac{\pi}{x}\right)$. Then,
$f(x)=\sqrt{2}\left[\cos \left(x+\frac{\pi}{6}\right) \frac{1}{\sqrt{2}}+\sin \left(x+\frac{\pi}{6}\right) \frac{1}{\sqrt{2}}\right]$
$=\sqrt{2}\left[\cos \left(x+\frac{\pi}{6}\right) \cos \frac{\pi}{4}+\sin \left(x+\frac{\pi}{6}\right) \sin \frac{\pi}{4}\right]$
$=\sqrt{2} \cos \left(x+\frac{\pi}{6}-\frac{\pi}{4}\right)$
$\begin{aligned} \ldots[\because \cos (A-B) & =\cos A \cos B +\sin A \sin B]\end{aligned}$
$=\sqrt{2} \cos \left(x-\frac{\pi}{12}\right)$
Since, $-1 \leq \cos \left(x-\frac{\pi}{12}\right) \leq 1$
$\therefore f (x)$ is maximum, if $x-\frac{\pi}{12}=0$ i.e., if $x=\frac{\pi}{12}$
View full question & answer→MCQ 1852 Marks
If $\cos (A-B)=\frac{3}{5}$ and $\tan A \ \tan B=2$, then
- ✓
$\cos A \ \cos B =\frac{1}{5}$
- B
$\sin A \ \sin B=-\frac{2}{5}$
- C
$\cos A \ \cos B =-\frac{1}{5}$
- D
$\sin A \ \sin B=-\frac{1}{5}$
AnswerCorrect option: A. $\cos A \ \cos B =\frac{1}{5}$
(A)
Given, $\cos ( A - B )-\frac{3}{5}$
$\therefore 5 \cos A \cos B+5 \sin A \sin B=3 \quad\ldots(i)$
Also, $\tan A \tan B =2$
$\therefore \sin A \sin B=2 \cos A \cos B \quad\ldots(ii)$
From (i) and (ii), we get
$\cos A \cos B =\frac{1}{5}$ and $\sin A \sin B =\frac{2}{5}$
View full question & answer→MCQ 1862 Marks
If $\cos \theta=\frac{8}{17}$ and $\theta$ lies in the $1^{\text {st }}$ quadrant, then the value of
$\cos \left(30^{\circ}+\theta\right)+\cos \left(45^{\circ}-\theta\right)+\cos \left(120^{\circ}-\theta\right)$ is
- ✓
$\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$
- B
$\frac{23}{17}\left(\frac{\sqrt{3}+1}{2}+\frac{1}{\sqrt{2}}\right)$
- C
$\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}-\frac{1}{\sqrt{2}}\right)$
- D
$\frac{23}{17}\left(\frac{\sqrt{3}+1}{2}-\frac{1}{\sqrt{2}}\right)$
AnswerCorrect option: A. $\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$
(A)
$\cos \theta=\frac{8}{17}$ and $0<\theta<\frac{\pi}{2}$
$\Rightarrow \sin \theta=\sqrt{1-\left(\frac{8}{17}\right)^2}=\frac{15}{17}$
$\begin{array}{l}\cos \left(30^{\circ}+\theta\right)+\cos \left(45^{\circ}-\theta\right)+\cos \left(120^{\circ}-\theta\right) \\ =\cos 30^{\circ} \cos \theta-\sin 30^{\circ} \sin \theta+\cos 45^{\circ} \cos \theta \\ +\sin 45^{\circ} \sin \theta+\cos 120^{\circ} \cos \theta+\sin 120^{\circ} \sin \theta\end{array}$
$=\cos \theta\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right)-\sin \theta\left(\frac{1}{2}-\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\right)$
$=\frac{8}{17}\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right)+\frac{15}{17}\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right)$
$=\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)$
View full question & answer→MCQ 1872 Marks
If $\cos P=\frac{1}{7}$ and $\cos Q=\frac{13}{14}$, where P and Q both are acute angles, then the value of P - Q is
- A
$30^{\circ}$
- ✓
$60^{\circ}$
- C
$45^{\circ}$
- D
$75^{\circ}$
AnswerCorrect option: B. $60^{\circ}$
(B)
$\cos P =\frac{1}{7} \Rightarrow \sin P =\frac{\sqrt{48}}{7}$
$\cos Q=\frac{13}{14} \Rightarrow \sin Q=\frac{\sqrt{27}}{14}$
$\therefore \quad \cos (P-Q)=\cos P \cos Q+\sin P \sin Q$
$=\frac{1}{7} \cdot \frac{13}{14}+\frac{\sqrt{48}}{7} \cdot \frac{\sqrt{27}}{14}$
$=\frac{13+36}{98}=\frac{1}{2}=\cos 60^{\circ}$
$\therefore \quad P-Q=60^{\circ}$
View full question & answer→MCQ 1882 Marks
If $\sin A=\frac{4}{5}$ and $\cos B=-\frac{12}{13}$, where A and B lie in first and third quadrant respectively, then $\cos (A+B)=$
- A
$\frac{56}{65}$
- B
$-\frac{56}{65}$
- C
$\frac{16}{65}$
- ✓
$-\frac{16}{65}$
AnswerCorrect option: D. $-\frac{16}{65}$
(D)
$\cos (A+B)=\cos A \cos B-\sin A \sin B$
$=\sqrt{1-\frac{16}{25}}\left(-\frac{12}{13}\right)-\frac{4}{5} \sqrt{1-\frac{144}{169}}$
$=\frac{3}{5}\left(-\frac{12}{13}\right)-\frac{4}{5}\left(-\frac{5}{13}\right)$
$\ldots[\because$ A lies in first quadrant and B lies in third quadrant]
$=-\frac{16}{65}$
View full question & answer→MCQ 1892 Marks
If $A , B$ and C are the angles of a plain triangle and $\tan \frac{A}{2}=\frac{1}{3}, \tan \frac{B}{2}=\frac{2}{3}$, then $\tan \frac{C}{2}$ is equal to
- ✓
$\frac{7}{9}$
- B
$\frac{2}{9}$
- C
$\frac{1}{3}$
- D
$\frac{2}{3}$
AnswerCorrect option: A. $\frac{7}{9}$
(A)
$A + B + C =\pi$
$\therefore \tan \left(\frac{ A + B }{2}\right)=\tan \left(\frac{\pi}{2}-\frac{ C }{2}\right)$
$\Rightarrow \frac{\tan \frac{ A }{2}+\tan \frac{ B }{2}}{1-\tan \frac{ A }{2} \cdot \tan \frac{B}{2}}=\cot \frac{ C }{2}$
$\Rightarrow \frac{\frac{1}{3}+\frac{2}{3}}{1-\frac{1}{3} \cdot \frac{2}{3}}=\cot \frac{ C }{2}$
$\Rightarrow \frac{9}{7}=\cot \frac{C}{2}$
$\Rightarrow \tan \frac{C}{2}=\frac{7}{9}$
View full question & answer→MCQ 1902 Marks
If $A + B + C =180^{\circ}$, then $\tan A +\tan B +\tan C$ is equal to
AnswerCorrect option: A. $\tan A \tan B \tan C$
(A)
$\tan ( A + B )=\tan \left(180^{\circ}- C \right)$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C$
$\Rightarrow \tan A+\tan B=-\tan C(1-\tan A \tan B)$
$\Rightarrow \tan A+\tan B=-\tan C+\tan A \tan B \tan C$
$\Rightarrow \tan A+\tan B+\tan C=\tan A \tan B \tan C$
View full question & answer→MCQ 1912 Marks
If $A + B + C =\pi$, then $\cos ^2 A+\cos ^2 B-\cos ^2 C$ is equal to
- A
$1-4 \sin A \cos B \sin C$
- B
$1-2 \sin A \sin B \sin C$
- ✓
$1-2 \sin A \sin B \cos C$
- D
$1-4 \sin A \sin B \cos C$
AnswerCorrect option: C. $1-2 \sin A \sin B \cos C$
(C)
$\cos ^2 A+\cos ^2 B-\cos ^2 C$
$=\frac{1}{2}(1+\cos 2 A)+\frac{1}{2}(1+\cos 2 B)$$-\frac{1}{2}(1+\cos 2 C )$
$=\frac{1}{2}+\frac{1}{2}(\cos 2 A+\cos 2 B-\cos 2 C )$
$=\frac{1}{2}+\frac{1}{2}(1-4 \sin A \sin B \cos C )$
$=1-2 \sin A \sin B \cos C$
View full question & answer→MCQ 1922 Marks
If $\alpha+\beta+\gamma=\pi$, then the value of
$\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma$ is equal to
- A
$2 \sin \alpha$
- B
$2 \sin \alpha \cos \beta \sin \gamma$
- ✓
$2 \sin \alpha \sin \beta \cos \gamma$
- D
$2 \sin \alpha \sin \beta \sin \gamma$
AnswerCorrect option: C. $2 \sin \alpha \sin \beta \cos \gamma$
(C)
$\sin ^2 \alpha \mid \sin ^2 \beta-\sin ^2 \gamma$
$=\sin ^2 \alpha+\sin (\beta-\gamma) \sin (\beta+\gamma)$
$=\sin ^2 \alpha+\sin (\pi-\alpha) \sin (\beta-\gamma)$
$=\sin \alpha[\sin \alpha+\sin (\beta-\gamma)]$
$=\sin \alpha[\sin (\beta+\gamma)+\sin (\beta-\gamma)]$
$=2 \sin \alpha \sin \beta \cos \gamma$
View full question & answer→MCQ 1932 Marks
If $A + B + C =\pi$ and $\cos A =\cos B \cos C$, then $\tan B \tan C$ is equal to
- A
$\frac{1}{2}$
- ✓
- C
- D
$-\frac{1}{2}$
Answer(B)
$\cos A =\cos B \cos C$
$\Rightarrow \cos [\pi-( B + C )]=\cos B \cos C$
$\Rightarrow-\cos ( B + C )=\cos B \cos C$
$\Rightarrow-[\cos B \cos C -\sin B \sin C ]=\cos B \cos C$
$\Rightarrow \sin B \sin C =2 \cos B \cos C$
$\Rightarrow \tan B \tan C=2$
View full question & answer→MCQ 1942 Marks
In a $\triangle A B C, \operatorname{cosec} A(\sin B \cos C+\cos B \sin C)$ is equal to
Answer(A)
$\operatorname{cosec} A(\sin B \cos C+\cos B \sin C)$
$=\operatorname{cosec} A \sin ( B + C )$
$=\operatorname{cosec} A \sin \left(180^{\circ}- A \right)$
$=\operatorname{cosec} A \sin A$
= 1
View full question & answer→MCQ 1952 Marks
In $\triangle ABC , \angle A =\frac{\pi}{2}$, then $\cos ^2 B+\cos ^2 C =$
Answer(D)
Given, $\angle A =\frac{\pi}{2}$
In $\triangle ABC , \angle A +\angle B +\angle C =\pi$
$\therefore\angle B +\angle C =\frac{\pi}{2}$
$\Rightarrow B =\frac{\pi}{2}- C$
$\Rightarrow \cos ^2 B=\cos ^2\left(\frac{\pi}{2}- C \right)=\sin ^2 C$
$\therefore \cos ^2 B+\cos ^2 C =\sin ^2 C +\cos ^2 C =1$
View full question & answer→MCQ 1962 Marks
If $A + B + C =\pi$, then $\sin ( A + B )=$
Answer(D)
$\sin (A+B)=\sin (\pi-C)=\sin C$
View full question & answer→MCQ 1972 Marks
The value of $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$ is equal to
- ✓
$\frac{1}{16}$
- B
$\frac{3}{16}$
- C
$\frac{\sqrt{3}}{16}$
- D
$\frac{\sqrt{3}}{32}$
AnswerCorrect option: A. $\frac{1}{16}$
(A)
$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$
$=\frac{1}{2} \cdot \frac{1}{2}\left(2 \cos 40^{\circ} \cos 20^{\circ}\right) \cos 80^{\circ}$
$=\frac{1}{4}\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \cos 80^{\circ}$
$=\frac{1}{8}\left(\cos 80^{\circ}+2 \cos 20^{\circ} \cos 80^{\circ}\right)$
$=\frac{1}{8}\left(\cos 80^{\circ}+\cos 100^{\circ}+\cos 60^{\circ}\right)$
$=\frac{1}{16} \quad \ldots\left[\because \cos \left(180^{\circ}-A\right)=-\cos A\right]$
View full question & answer→MCQ 1982 Marks
The value of $10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}$ is equal to
- A
$\frac{1}{8}$
- ✓
$\frac{1}{16}$
- C
$\frac{\sqrt{3}}{8}$
- D
$\frac{\sqrt{3}}{16}$
AnswerCorrect option: B. $\frac{1}{16}$
(B)
$\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}$
$=\frac{1}{2} \cdot \frac{1}{2}\left(2 \sin 10^{\circ} \sin 50^{\circ}\right) \sin 70^{\circ}$
$=\frac{1}{4}\left(\cos 40^{\circ}-\cos 60^{\circ}\right) \sin 70^{\circ}$
$=\frac{1}{8}\left(2 \sin 70^{\circ} \cos 40^{\circ}-\sin 70^{\circ}\right)$
$=\frac{1}{8}\left(\sin 110^{\circ}+\sin 30^{\circ}-\sin 70^{\circ}\right)$
$=\frac{1}{8}\left(\sin 70^{\circ}+\frac{1}{2}-\sin 70^{\circ}\right)$
$\ldots\left[\because \sin \left(180^{\circ}-A\right)=\sin A\right]$
$=\frac{1}{16}$
View full question & answer→MCQ 1992 Marks
$\sin 18^{\circ} \sin 70^{\circ}+\sin 16^{\circ} \sin 36^{\circ}=$
- ✓
$\sin 54^{\circ} \sin 34^{\circ}$
- B
$\sin 54^{\circ} \cos 34^{\circ}$
- C
$\cos 54^{\circ} \sin 34^{\circ}$
- D
$\cos 54^{\circ} \cos 34^{\circ}$
AnswerCorrect option: A. $\sin 54^{\circ} \sin 34^{\circ}$
(A)
$\sin 18^{\circ} \sin 70^{\circ}+\sin 16^{\circ} \sin 36^{\circ}$
$=\frac{1}{2}\left[2 \sin 18^{\circ} \sin 70^{\circ}+2 \sin 16^{\circ} \sin 36^{\circ}\right]$
$=\frac{1}{2}\left[\cos 52^{\circ}-\cos 88^{\circ}+\cos 20^{\circ}-\cos 52^{\circ}\right]$
$=\frac{1}{2}\left[\cos 20^{\circ}-\cos 88^{\circ}\right]$
$=\frac{1}{2}\left[2 \sin 54^{\circ} \sin 34^{\circ}\right]$
$=\sin 54^{\circ} \sin 34^{\circ}$
View full question & answer→MCQ 2002 Marks
$4 \sin \left(\frac{\pi}{3}+\theta\right) \sin \left(\frac{\pi}{3}-\theta\right)=$
- A
$1+\cos 2 \theta$
- B
$1-2 \cos 2 \theta$
- C
$2 \cos 2 \theta-1$
- ✓
$1+2 \cos 2 \theta$
AnswerCorrect option: D. $1+2 \cos 2 \theta$
(D)
$4 \sin \left(\frac{\pi}{3}+\theta\right) \sin \left(\frac{\pi}{3}-\theta\right)$
$=2\left[2 \sin \left(\frac{\pi}{3}+\theta\right) \sin \left(\frac{\pi}{3}-\theta\right)\right]$
$=2\left[\cos \left(\frac{\pi}{3}+\theta-\frac{\pi}{3}+\theta\right)-\cos \left(\frac{\pi}{3}+\theta+\frac{\pi}{3}-\theta\right)\right]$
$=2 [\cos 2 \theta-\cos \left(\frac{2 \pi}{3}\right)]$
$=2 \cos 2 \theta+1$
View full question & answer→MCQ 2012 Marks
$\sin \left(45^{\circ}+A\right) \sin \left(45^{\circ}-A\right)=$
- A
$\cos A$
- ✓
$\frac{1}{2} \cos 2 A$
- C
$\cos 2 A$
- D
$\frac{1}{2} \cos A$
AnswerCorrect option: B. $\frac{1}{2} \cos 2 A$
(B)
$\sin \left(45^{\circ}+ A \right) \sin \left(45^{\circ}- A \right)$
$=\frac{1}{2}\left[2 \sin \left(45^{\circ}+ A \right) \sin \left(45^{\circ}- A \right)\right]$
$=\frac{1}{2}\left[\cos \left(45^{\circ}+ A -45^{\circ}+ A \right)\right.$$\left.-\cos \left(45^{\circ}+ A +45^{\circ}- A \right)\right]$
$=\frac{1}{2}\left[\cos 2 A-\cos 90^{\circ}\right]$
$=\frac{1}{2} \cos 2 A$
View full question & answer→MCQ 2022 Marks
The value of $\cos 75^{\circ} \cos 15^{\circ}$ is equal to
- A
$\frac{1}{2}$
- B
$\frac{1}{\sqrt{2}}$
- ✓
$\frac{1}{4}$
- D
AnswerCorrect option: C. $\frac{1}{4}$
(C)
$\cos 75^{\circ} \cos 15^{\circ}$
$=\frac{1}{2}\left[2 \cos 75^{\circ} \cos 15^{\circ}\right]$
$=\frac{1}{2}\left[\cos \left(75^{\circ}+15^{\circ}\right)+\cos \left(75^{\circ}-15^{\circ}\right)\right]$
$=\frac{1}{2}\left[\cos 90^{\circ}+\cos 60^{\circ}\right]$
$=\frac{1}{4}$
View full question & answer→MCQ 2032 Marks
$2 \sin \frac{5 \pi}{12} \cos \frac{\pi}{12}=$
- ✓
$\frac{2+\sqrt{3}}{2}$
- B
$\frac{2-\sqrt{3}}{2}$
- C
$\frac{1}{2}$
- D
$\frac{3}{2}$
AnswerCorrect option: A. $\frac{2+\sqrt{3}}{2}$
(A)
$2 \sin \frac{5 \pi}{12} \cos \frac{\pi}{12}=\sin \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)+\sin \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)$
$=\sin \frac{\pi}{2}+\sin \frac{\pi}{3}$
$=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}$
View full question & answer→MCQ 2042 Marks
The value of $2 \sin 3 x \cos 2 x$ is equal to
- ✓
$\sin 5 x+\sin x$
- B
$\sin 3 x+\sin x$
- C
$\sin 7 x+\sin x$
- D
$\sin 4 x+\sin x$
AnswerCorrect option: A. $\sin 5 x+\sin x$
(A)
$2 \sin 3 x \cos 2 x=\sin (3 x+2 x)+\sin (3 x-2 x)$
$=\sin 5 x+\sin x$
View full question & answer→MCQ 2052 Marks
If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{ a + b }{ a - b }$, then $\frac{\tan x}{\tan y}$ is equal to
- A
$\frac{ b }{ a }$
- ✓
$\frac{a}{b}$
- C
- D
$a-b$
AnswerCorrect option: B. $\frac{a}{b}$
(B)
$\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
By componendo and dividendo, we get
$\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2 \sin x \cos y}{2 \cos x \sin y}=\frac{2 a }{2 b}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{ a }{ b }$
View full question & answer→MCQ 2062 Marks
$\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}=$
- A
$\tan 3 \theta$
- B
$\cot 3 \theta$
- ✓
$\tan 6 \theta$
- D
$\cot 6 \theta$
AnswerCorrect option: C. $\tan 6 \theta$
(C)
$\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}$
$=\frac{(\sin 3 \theta+\sin 9 \theta)+(\sin 5 \theta+\sin 7 \theta)}{(\cos 3 \theta+\cos 9 \theta)+(\cos 5 \theta+\cos 7 \theta)}$
$=\frac{2 \sin 6 \theta \cos 3 \theta+2 \sin 6 \theta \cos \theta}{2 \cos 6 \theta \cos 3 \theta+2 \cos 6 \theta \cos \theta}$
$=\frac{2 \sin 6 \theta(\cos 3 \theta+\cos \theta)}{2 \cos 6 \theta(\cos 3 \theta+\cos \theta)}$
$=\tan 6 \theta$
View full question & answer→MCQ 2072 Marks
$\frac{\sin 3 A-\cos \left(\frac{\pi}{2}-A\right)}{\cos A+\cos (\pi+3 A)}=$
Answer(D)
$\frac{\sin 3 A-\cos \left(\frac{\pi}{2}-A\right)}{\cos A+\cos (\pi+3 A)}=\frac{\sin 3 A-\sin A}{\cos A - \cos 3 A}$
$=\frac{2 \cos 2 A \sin A }{2 \sin 2 A \sin A }$
$= cot 2A$
View full question & answer→MCQ 2082 Marks
$\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}=$
- A
$\frac{1}{\sqrt{3}}$
- ✓
$\sqrt{3}$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: B. $\sqrt{3}$
(B)
$\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}=\frac{\sin 70^{\circ}+\sin 50^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}}$
$=\frac{2 \sin 60^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}$
$=\tan 60^{\circ}$
$=\sqrt{3}$
View full question & answer→MCQ 2092 Marks
The value of $\frac{\sin 55^{\circ}-\cos 55^{\circ}}{\sin 10^{\circ}}$ is
- A
$\frac{1}{\sqrt{2}}$
- B
- C
- ✓
$\sqrt{2}$
AnswerCorrect option: D. $\sqrt{2}$
(D)
$\frac{\sin 55^{\circ}-\cos 55^{\circ}}{\sin 10^{\circ}}=\frac{\sin 55^{\circ}-\sin 35^{\circ}}{\sin 10^{\circ}}$
$=\frac{2 \cos 45^{\circ} \sin 10^{\circ}}{\sin 10^{\circ}}=\sqrt{2}$
View full question & answer→MCQ 2102 Marks
$\frac{\sin 85^{\circ}-\sin 35^{\circ}}{\cos 65^{\circ}}=$
Answer(C)
$\frac{\sin 85^{\circ}-\sin 35^{\circ}}{\cos 65^{\circ}}=\frac{2 \cos 60^{\circ} \sin 25^{\circ}}{\sin 25^{\circ}}$
$=2 \times \frac{1}{2}$
= 1
View full question & answer→MCQ 2112 Marks
$\frac{\sin A-\sin B}{\cos A+\cos B}$ is equal to
- A
$\sin \left(\frac{A+B}{2}\right)$
- B
$2 \tan ( A + B )$
- C
$\cot \left(\frac{ A - B }{2}\right)$
- D
$\tan \left(\frac{A-B}{2}\right)$
View full question & answer→MCQ 2122 Marks
$\frac{\sin 3 x-\sin x}{\cos 2 x}=$
- A
$\sin x$
- B
$\cos x$
- C
$2 \sin x$
- D
$2 \cos x$
View full question & answer→MCQ 2132 Marks
If $\sin 4 A-\cos 2 A=\cos 4 A-\sin 2 A \left(0 < A<\frac{\pi}{4}\right)$, then the value of $\tan 4 A=$
View full question & answer→MCQ 2142 Marks
$\sin (\beta+\gamma-\alpha)+\sin (\gamma+\alpha-\beta) +\sin (\alpha+\beta-\gamma) -\sin (\alpha+\beta+\gamma)=$
- A
$2 \sin \alpha \sin \beta \sin \gamma$
- B
$4 \sin \alpha \sin \beta \sin \gamma$
- C
$8 \sin \alpha \sin \beta \sin \gamma$
- D
$\sin \alpha \sin \beta \sin \gamma$
View full question & answer→MCQ 2152 Marks
The expression $\cos \frac{10 \pi}{13}+\cos \frac{8 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$ is equal to
View full question & answer→MCQ 2162 Marks
The value of $\cos 52^{\circ}+\cos 68^{\circ}+\cos 172^{\circ}$ is
View full question & answer→MCQ 2172 Marks
$\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}$ is equal to
View full question & answer→MCQ 2182 Marks
$\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=$
- A
$-\sqrt{2} \sin x$
- B
$\sqrt{2} \sin x$
- C
$\cos x$
- D
$-\sqrt{2} \cos x$
View full question & answer→MCQ 2192 Marks
$\cos 18^{\circ}-\sin 18^{\circ}=$
View full question & answer→MCQ 2202 Marks
The value of $\cos 57^{\circ}+\sin 27^{\circ}$ is equal to
- A
$\cos 30^{\circ}$
- B
$\cos 3^{\circ}$
- C
$\sin 3^{\circ}$
- D
$\sin 30^{\circ}$
View full question & answer→MCQ 2212 Marks
If $x+\frac{1}{x}=2 \cos \theta$, then $x^3+\frac{1}{x^3}=$
View full question & answer→MCQ 2222 Marks
If $\cos 3 \theta=\alpha \cos \theta+\beta \cos ^3 \theta$, then $(\alpha, \beta)=$
- A
$(3,4)$
- B
$(4,3)$
- C
$(-3,4)$
- D
$(3,-4)$
View full question & answer→MCQ 2232 Marks
If $\tan A=\frac{1}{2}$, then $\tan 3 A=$
- A
$\frac{9}{2}$
- B
$\frac{11}{2}$
- C
$\frac{7}{2}$
- D
$-\frac{1}{2}$
View full question & answer→MCQ 2242 Marks
If $\tan \beta=\cos \theta \tan \alpha$, then $\tan ^2 \frac{\theta}{2}=$
- A
$\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}$
- B
$\frac{\cos (\alpha-\beta)}{\cos (\alpha+\beta)}$
- ✓
$\frac{\sin (\alpha-\beta)}{\sin (\alpha+\beta)}$
- D
$\frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}$
AnswerCorrect option: C. $\frac{\sin (\alpha-\beta)}{\sin (\alpha+\beta)}$
(C)
$\tan ^2 \frac{\theta}{2}=\frac{1-\cos \theta}{1+\cos \theta}$
$=\frac{\tan \alpha-\tan \beta}{\tan \alpha+\tan \beta} \ldots\left[\because \cos \theta=\frac{\tan \beta}{\tan \alpha}(\right.$ given $\left.)\right]$
$=\frac{\frac{\sin \alpha}{\cos \alpha}-\frac{\sin \beta}{\cos \beta}}{\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}}$
$=\frac{\sin \alpha \cos \beta-\sin \beta \cos \alpha}{\sin \alpha \cos \beta+\sin \beta \cos \alpha}$
$=\frac{\sin (\alpha-\beta)}{\sin (\alpha+\beta)}$
View full question & answer→MCQ 2252 Marks
$\frac{1+\sin A-\cos A}{1+\sin A+\cos A}=$
- A
$\sin \frac{A}{2}$
- B
$\cos \frac{ A }{2}$
- ✓
$\tan \frac{A}{2}$
- D
$\cot \frac{ A }{2}$
AnswerCorrect option: C. $\tan \frac{A}{2}$
(C)
$\frac{1+\sin A -\cos A }{1+\sin A +\cos A }=\frac{2 \sin ^2 \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos ^2 \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}}$
$=\frac{2 \sin \frac{A}{2}\left(\sin \frac{A}{2}+\cos \frac{ A }{2}\right)}{2 \cos \frac{A}{2}\left(\cos \frac{A}{2}+\sin \frac{ A }{2}\right)}$
$=\tan \frac{ A }{2}$
View full question & answer→MCQ 2262 Marks
$\tan \frac{A}{2}$ is equal to
- A
$\sqrt{\frac{1-\sin A}{1+\sin A}}$
- B
$\sqrt{\frac{1+\sin A }{1-\sin A }}$
- ✓
$\sqrt{\frac{1-\cos A}{1+\cos A}}$
- D
$\sqrt{\frac{1+\cos A}{1-\cos A}}$
AnswerCorrect option: C. $\sqrt{\frac{1-\cos A}{1+\cos A}}$
(C)
$\tan \left(\frac{ A }{2}\right)=\frac{\sin \left(\frac{ A }{2}\right)}{\cos \left(\frac{ A }{2}\right)}=\sqrt{\frac{\frac{1-\cos A }{2}}{\frac{1+\cos A }{2}}}=\sqrt{\frac{1-\cos A }{1+\cos A }}$
View full question & answer→MCQ 2272 Marks
If $\sin \alpha=\frac{-3}{5}$, where $\pi<\alpha<\frac{3 \pi}{2}$, then $\cos \left(\frac{\alpha}{2}\right)=$
- ✓
$\frac{-1}{\sqrt{10}}$
- B
$\frac{1}{\sqrt{10}}$
- C
$\frac{3}{\sqrt{10}}$
- D
$\frac{-3}{\sqrt{10}}$
AnswerCorrect option: A. $\frac{-1}{\sqrt{10}}$
(A)
$\cos \left(\frac{\alpha}{2}\right)=-\sqrt{\frac{1+\cos \alpha}{2}}$
$\ldots\left[\because \pi<\alpha<\frac{3 \pi}{2} \Rightarrow \frac{\pi}{2}<\frac{\alpha}{2}<\frac{3 \pi}{4}\right]$
Now, $\cos \alpha=-\sqrt{1-\sin ^2 \alpha}$\[\ldots\left[\because \pi<\alpha<\frac{3 \pi}{2}\right]\]
$=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$
$\therefore \quad \cos \frac{\alpha}{2}=-\sqrt{\frac{1-\frac{4}{5}}{2}}=-\frac{1}{\sqrt{10}}$
View full question & answer→MCQ 2282 Marks
If $\tan A=\frac{1-\cos B}{\sin B}$, then $\tan 2 A$ is equal to
- ✓
$\tan B$
- B
$\tan ^2 B$
- C
$\tan ^2 B+2 \tan B$
- D
$\tan B+2 \tan B$
AnswerCorrect option: A. $\tan B$
(A)
$\tan A =\frac{1-\cos B }{\sin B }=\frac{2 \sin ^2\left(\frac{B}{2}\right)}{2 \sin \left(\frac{B}{2}\right) \cos \left(\frac{ B }{2}\right)}=\tan \frac{ B }{2}$
$\Rightarrow \tan 2 A=\tan B$
View full question & answer→MCQ 2292 Marks
If $\tan \frac{A}{2}=\frac{3}{2}$, then $\frac{1+\cos A}{1-\cos A}=$
- A
- B
- C
$\frac{9}{4}$
- ✓
$\frac{4}{9}$
AnswerCorrect option: D. $\frac{4}{9}$
(D)
Given that, $\tan \frac{ A }{2}=\frac{3}{2}$
$\therefore \frac{1+\cos A }{1-\cos A }=\cot ^2 \frac{A}{2}$
$\frac{1+\cos \theta}{1-\cos \theta}=\cot ^2 \frac{\theta}{2}$, where $\theta \neq 2 n \pi$
$=\left(\frac{2}{3}\right)^2=\frac{4}{9}$
View full question & answer→MCQ 2302 Marks
If $a \tan \theta= b$, then $a \cos 2 \theta+ b \sin 2 \theta=$
Answer(A)
$a \cos 2 \theta+b \sin 2 \theta$
$=a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$
$=a\left(\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}\right)+b\left(\frac{\frac{2 b}{a}}{1+\frac{b^2}{a^2}}\right) \quad \ldots\left[\because \tan \theta=\frac{b}{a}(\right.$ given $\left.)\right]$
$=a\left(\frac{a^2-b^2}{a^2+b^2}\right)+b\left(\frac{2 b a}{a^2+b^2}\right)$
$=\frac{1}{\left(a^2+b^2\right)}\left(a^3-a b^2+2 a b^2\right)=\frac{a\left(a^2+b^2\right)}{a^2+b^2}$
= a
View full question & answer→MCQ 2312 Marks
$(\sec 2 A+1) \sec ^2 A=$
Answer(D)
$(\sec 2 A+1) \sec ^2 A$
$=\left(\frac{1+\tan ^2 A}{1-\tan ^2 A}+1\right)\left(1+\tan ^2 A\right)$
$=\frac{2\left(1+\tan ^2 A\right)}{1-\tan ^2 A}=2 \sec 2 A$
View full question & answer→MCQ 2322 Marks
If $\tan \theta= t$, then $\tan 2 \theta+\sec 2 \theta=$
- ✓
$\frac{1+t}{1-t}$
- B
$\frac{1-t}{1+t}$
- C
$\frac{2 t}{1-t}$
- D
$\frac{2 t}{1+t}$
AnswerCorrect option: A. $\frac{1+t}{1-t}$
(A)
$\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}, \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$
$\therefore \tan 2 \theta+\sec 2 \theta=\frac{2 t}{1-t^2}+\frac{1+t^2}{1-t^2}$
$\ldots[\because \tan \theta= t ($ given $)]$
$=\frac{(1+t)^2}{(1-t)(1+t)}=\frac{1+t}{1-t}$
View full question & answer→MCQ 2332 Marks
$2 \sin A \cos ^3 A-2 \sin ^3 A \cos A =$
- A
$\sin 4 A$
- ✓
$\frac{1}{2} \sin 4 A$
- C
$\frac{1}{4} \sin 4 A$
- D
$\frac{1}{8} \sin 4 A$
AnswerCorrect option: B. $\frac{1}{2} \sin 4 A$
(B)
$2 \sin A \cos ^3 A-2 \sin ^3 A \cos A$
$=2 \sin A \cos A \left(\cos ^2 A-\sin ^2 A\right)$
$=2 \sin A \cos A \cos 2 A$
$-\sin 2 A \cos 2 A$
$=\frac{1}{2} \sin 4 A$
View full question & answer→MCQ 2342 Marks
The value of $\frac{\cot x-\tan x}{\cot 2 x}$ is
Answer(B)
$\frac{\cot x-\tan x}{\cot 2 x}=\frac{\cos ^2 x-\sin ^2 x}{\sin x \cos x} \times \frac{\sin 2 x}{\cos 2 x}$
$-\frac{2 \cos 2 x}{\sin 2 x} \times \frac{\sin 2 x}{\cos 2 x}-2$
View full question & answer→MCQ 2352 Marks
$\sqrt{2+\sqrt{2+2 \cos 4 \theta}}=$
- A
$\cos \theta$
- B
$\sin \theta$
- ✓
$2 \cos \theta$
- D
$2 \sin \theta$
AnswerCorrect option: C. $2 \cos \theta$
(C)
$2+2 \cos 4 \theta=2(1+\cos 4 \theta)$
$=4 \cos ^2 2 \theta$ …(i)
$\therefore \sqrt{2+\sqrt{2+2 \cos 4 \theta}}=\sqrt{2+2 \cos 2 \theta} \ldots[$ From (i) $]$
$=\sqrt{2(1+\cos 2 \theta)}$
$=\sqrt{4 \cos ^2 \theta}=2 \cos \theta$
View full question & answer→MCQ 2362 Marks
$\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}=$
AnswerCorrect option: C. $\tan \theta$
(C)
$\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}=\frac{\sin \theta+2 \sin \theta \cos \theta}{2 \cos ^2 \theta+\cos \theta}$
$=\frac{\sin \theta(1+2 \cos \theta)}{\cos \theta(1+2 \cos \theta)}$
$=\tan \theta$
View full question & answer→MCQ 2372 Marks
$\operatorname{cosec} A-2 \cot 2 A \cos A=$
Answer(A)
$\operatorname{cosec} A -2 \cot 2 A \cos A$
$=\frac{1}{\sin A}-\frac{2 \cos A \cos 2 A}{\sin 2 A}$
$=\frac{1}{\sin A}-\frac{2 \cos A \cos 2 A}{2 \sin A \cos A}$
$=\frac{1-\cos 2 A}{\sin A }=\frac{2 \sin ^2 A}{\sin A }=2 \sin A$
View full question & answer→MCQ 2382 Marks
$1-2 \sin ^2\left(\frac{\pi}{4}+\theta\right)=$
- A
$\cos 2 \theta$
- B
$-\cos 2 \theta$
- C
$\sin 2 \theta$
- ✓
$-\sin 2 \theta$
AnswerCorrect option: D. $-\sin 2 \theta$
(D)
$1-2 \sin ^2\left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{2}+2 \theta\right)$
$\ldots\left[\because \cos 2 \theta=1-2 \sin ^2 \theta\right]$
$=-\sin 2 \theta$
View full question & answer→MCQ 2392 Marks
$1+\cos ^2 2 A$ is equal to
AnswerCorrect option: C. $2\left(\cos ^4 A+\sin ^4 A\right)$
(C)
$1+\cos ^2 2 A=\left(\cos ^2 A+\sin ^2 A\right)^2$$+\left(\cos ^2 A-\sin ^2 A\right)^2$
$=2\left(\cos ^4 A+\sin ^4 A\right)$
View full question & answer→MCQ 2402 Marks
$\sin 4 \theta$ can be written as
- ✓
$4 \sin \theta\left(1-2 \sin ^2 \theta\right) \sqrt{1-\sin ^2 \theta}$
- B
$2 \sin \theta \cos \theta \sin ^2 \theta$
- C
$4 \sin \theta-6 \sin ^3 \theta$
- D
$4 \sin \theta+6 \sin ^2 \theta$
AnswerCorrect option: A. $4 \sin \theta\left(1-2 \sin ^2 \theta\right) \sqrt{1-\sin ^2 \theta}$
(A)
$\sin 4 \theta=2 \sin 2 \theta \cos 2 \theta$
$=2.2 \sin \theta \cos \theta\left(1-2 \sin ^2 \theta\right)$
$=4 \sin \theta\left(1-2 \sin ^2 \theta\right) \sqrt{1-\sin ^2 \theta}$
View full question & answer→MCQ 2412 Marks
$\cos 2 \theta$ is not equal to
AnswerCorrect option: C. $\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}$
(C)
$\cos 2 \theta=2 \cos ^2 \theta-1$
$=1-2 \sin ^2 \theta$
$=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$
View full question & answer→MCQ 2422 Marks
The largest value of $\sin \theta \cos \theta$ is
- A
- ✓
$\frac{1}{2}$
- C
$\frac{1}{\sqrt{2}}$
- D
$\frac{\sqrt{3}}{2}$
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$\sin \theta \cos \theta=\frac{1}{2}(\sin 2 \theta)$
Since $-1 \leq \sin 2 \theta \leq 1$
$\Rightarrow-\frac{1}{2} \leq \frac{1}{2}(\sin 2 \theta) \leq \frac{1}{2}$
Largest value is $\frac{1}{2}$.
View full question & answer→MCQ 2432 Marks
If $\sin x+\cos x=\frac{1}{5}$, then $\tan 2 x$ is
- A
$\frac{25}{17}$
- B
$\frac{7}{25}$
- C
$\frac{25}{7}$
- ✓
$\frac{24}{7}$
AnswerCorrect option: D. $\frac{24}{7}$
(D)
$\sin x+\cos x=\frac{1}{5}$
$\Rightarrow \sin ^2 x+\cos ^2 x+2 \sin x \cos x=\frac{1}{25}$
$\Rightarrow \sin 2 x=\frac{-24}{25} \Rightarrow \cos 2 x=\frac{-7}{25}$
$\Rightarrow \tan 2 x=\frac{24}{7}$
View full question & answer→MCQ 2442 Marks
If $\sin A +\cos A =1$, then $\sin 2 A$ is equal to
Answer(C)
Given, $\sin A +\cos A =1$
Squaring on both sides, we get
$(\sin A+\cos A)^2=1$
$\Rightarrow 1+\sin 2 A=1$
$\Rightarrow \sin 2 A=0$
View full question & answer→MCQ 2452 Marks
The value of
$\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}+\sin ^2 \frac{5 \pi}{8}+\sin ^2 \frac{7 \pi}{8}$ is
- A
- ✓
- C
$\frac{3}{8}$
- D
$\frac{1}{8}$
Answer(B)
$\sin ^2 17.5^{\circ}+\sin ^2 72.5^{\circ}$
$\sin \frac{5 \pi}{8}=\sin \left(\pi-\frac{3 \pi}{8}\right)=\sin \frac{3 \pi}{8}$
$\therefore \sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}+\sin ^2 \frac{5 \pi}{8}+\sin ^2 \frac{7 \pi}{8}$
$=2\left[\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}\right]$
$=2\left[\sin ^2 \frac{\pi}{8}+\cos ^2 \frac{\pi}{8}\right]$
$\ldots\left[\because \sin \frac{3 \pi}{8}=\sin \left(\frac{\pi}{2}-\frac{\pi}{8}\right)=\cos \frac{\pi}{8}\right]$
= 2
View full question & answer→MCQ 2462 Marks
$\sin ^2 17.5^{\circ}+\sin ^2 72.5^{\circ}$ is equal to
- A
$\cos ^2 90^{\circ}$
- ✓
$\tan ^2 45^{\circ}$
- C
$\cos ^2 30^{\circ}$
- D
$\sin ^2 45^{\circ}$
AnswerCorrect option: B. $\tan ^2 45^{\circ}$
(B)
$\sin ^2 17.5^{\circ}+\sin ^2 72.5^{\circ}$
$=\sin ^2 17.5^{\circ}+\left[\sin \left(90^{\circ}-17.5^{\circ}\right)\right]^2$
$=\sin ^2 17.5^{\circ}+\cos ^2 17.5^{\circ}$
$=1=\tan ^2 45^{\circ}$
View full question & answer→MCQ 2472 Marks
If ABCD is a cyclic quadrilateral, then the value of cos A - cos B + cos C - cos D =
- ✓
$0$
- B
- C
$2(\cos B-\cos D)$
- D
$2(\cos A-\cos C)$
Answer(A)
Since ABCD is a cyclic quadrilateral.
$\therefore \quad A+C=180^{\circ}$
$\Rightarrow A=180^{\circ}-C$
$\Rightarrow \cos A =\cos \left(180^{\circ}- C \right)=-\cos C$
$\Rightarrow \cos A+\cos C=0$ ….(i)
Also, $B + D =180^{\circ}$
$\Rightarrow \cos B+\cos D=0$ ....(ii)
Subtracting (ii) from (i), we get
cos A - cos B + cos C - cos D = 0
View full question & answer→MCQ 2482 Marks
If ABCD is a cyclic quadrilateral, then $\cos A +\cos B$ is equal to
- A
$0$
- B
$\cos C+\cos D$
- ✓
$-(\cos C+\cos D)$
- D
$\cos C-\cos D$
AnswerCorrect option: C. $-(\cos C+\cos D)$
(C)
Since $A+C=180^{\circ}$ and $B+D=180^{\circ}$
$\therefore \cos A+\cos B=\cos \left(180^{\circ}-C\right)+\cos \left(180^{\circ}-D\right)$
$=-(\cos C +\cos D )$
$\ldots\left[\because \cos \left(180^{\circ}-\theta\right)=-\cos \theta\right]$
View full question & answer→MCQ 2492 Marks
If $\tan A =\frac{1}{2}, \tan B=\frac{1}{3}$, then $\cos 2 A=$
Answer(B)
$\tan (A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}=1$
$\therefore \quad A + B =45^{\circ}$
$\Rightarrow 2 A=90^{\circ}-2 B$
$\Rightarrow \cos 2 A=\sin 2 B$
$\ldots\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
View full question & answer→MCQ 2502 Marks
$\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots+\cos 180^{\circ}=$
Answer(C)
$ (\cos 1^{\circ}+\cos 179^{\circ})+ (\cos 2^{\circ}+\cos 178^{\circ})$
$ +\ldots+\left(\cos 89^{\circ}+\cos 91^{\circ}\right) | \left(\cos 90^{\circ} | \cos 180^{\circ}\right)$
= -1 $\quad\ldots\left[\because \cos \left(180^{\circ}-\theta\right)=-\cos \theta\right]$
View full question & answer→MCQ 2512 Marks
The value of $\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 359^{\circ}$ is equal to
Answer(C)
$\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 359^{\circ}$
$\begin{array}{r}=\left(\sin 1^{\circ}+\sin 359^{\circ}\right)+\left(\sin 2^{\circ}+\sin 358^{\circ}\right)+\ldots \\ +\left(\sin 179^{\circ}+\sin 181^{\circ}\right)+\sin 180^{\circ}\end{array}$
$\begin{array}{l}=\left(\sin 1^{\circ}-\sin 1^{\circ}\right)+\left(\sin 2^{\circ}-\sin 2^{\circ}\right)+\ldots \\ +\left(\sin 179^{\circ}-\sin 179^{\circ}\right)+\sin 180^{\circ}\end{array}$
$\ldots\left[\because \sin \left(360^{\circ}-\theta\right)=-\sin \theta\right]$
= 0
View full question & answer→MCQ 2522 Marks
$\sin \left(\frac{\pi}{10}\right) \sin \left(\frac{3 \pi}{10}\right)=$
- A
$\frac{1}{2}$
- B
$\frac{-1}{2}$
- ✓
$\frac{1}{4}$
- D
AnswerCorrect option: C. $\frac{1}{4}$
(C)
$\sin \frac{\pi}{10} \sin \frac{3 \pi}{10}=\sin 18^{\circ} \cdot \sin 54^{\circ}$
$=\sin 18^{\circ} \cdot \cos 36^{\circ}$
$\ldots\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
$=\frac{\sqrt{5}-1}{4} \cdot \frac{\sqrt{5}+1}{4}=\frac{1}{4}$
View full question & answer→MCQ 2532 Marks
The value of $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}=$
Answer(B)
$\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}$
$=\frac{\cot \left(90^{\circ}-36^{\circ}\right)}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot \left(90^{\circ}-20^{\circ}\right)}$
$=1+1 \quad \ldots\left[\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right]$
= 2
View full question & answer→MCQ 2542 Marks
If $8 \theta=\pi$, then $\cos 7 \theta+\cos \theta$ is equal to
Answer(A)
$\cos 7 \theta+\cos \theta=\cos (8 \theta-\theta)+\cos \theta$
$=\cos (\pi-\theta)+\cos \theta$ $\ldots[\because 8 \theta=\pi($ given $)]$
$=-\cos \theta+\cos \theta=0$
View full question & answer→MCQ 2552 Marks
$\sin 15^{\circ}+\cos 105^{\circ}=$
Answer(A)
$\sin 15^{\circ}+\cos 105^{\circ}$
$=\sin 15^{\circ}+\cos \left(90^{\circ}+15^{\circ}\right)$
$=\sin 15^{\circ}-\sin 15^{\circ}$
$\ldots\left[\because \cos \left(90^{\circ}+\theta\right)=-\sin \theta\right]$
= 0
View full question & answer→MCQ 2562 Marks
Which of the following is the correct identity?
- A
$\cot \left(\frac{\pi}{2}+A\right)=\tan A$
- ✓
$\sec \left(\frac{7 \pi}{2}-A\right)=-\operatorname{cosec} A$
- C
$\sin (n \pi+A)=(-1)^{2 n} \sin A$
- D
$\sin (\pi-A)=-\sin A$
AnswerCorrect option: B. $\sec \left(\frac{7 \pi}{2}-A\right)=-\operatorname{cosec} A$
(B)
$\sec \left(\frac{7 \pi}{2}-A\right)=\sec \left(2 \pi+\frac{3 \pi}{2}-A\right)$
$=\sec \left(\frac{3 \pi}{2}- A \right)=-\operatorname{cosec} A$
View full question & answer→MCQ 2572 Marks
$\frac{\cos \left(90^{\circ}+\theta\right) \sec (-\theta) \tan \left(180^{\circ}-\theta\right)}{\sin \left(360^{\circ}+\theta\right) \sec \left(180^{\circ}+\theta\right) \cot \left(90^{\circ}-\theta\right)}=$
Answer(C)
$\frac{\cos \left(90^{\circ}+\theta\right) \sec (-\theta) \tan \left(180^{\circ}-\theta\right)}{\sin \left(360^{\circ}+\theta\right) \sec \left(180^{\circ}+\theta\right) \cot \left(90^{\circ}-\theta\right)}$
$=\frac{(-\sin \theta)(\sec \theta)(-\tan \theta)}{(\sin \theta)(-\sec \theta) \tan \theta}$
= -1
View full question & answer→MCQ 2582 Marks
At $x=\frac{5 \pi}{6}$, the value of $2 \sin 3 x+3 \cos 3 x$ is
Answer(D)
Let $f (x)=2 \sin 3 x+3 \cos 3 x$
$\therefore f\left(\frac{5 \pi}{6}\right)=2 \sin \left(\frac{5 \pi}{2}\right)+3 \cos \left(\frac{5 \pi}{2}\right)$
$=2 \sin \left(2 \pi | \frac{\pi}{2}\right) | 3 \cos \left(2 \pi | \frac{\pi}{2}\right)$
$=2 \sin \frac{\pi}{2}+3 \cos \frac{\pi}{2}=2(1)+3(0)=2$
View full question & answer→MCQ 2592 Marks
The value of
$\cos \left(270^{\circ}+\theta\right) \cos \left(90^{\circ}-\theta\right)-\sin \left(270^{\circ}-\theta\right)cos \ \theta \ is $
Answer(D)
$\cos \left(270^{\circ}+\theta\right) \cos \left(90^{\circ}-\theta\right)-\sin \left(270^{\circ}-\theta\right) \cos \theta$
$=\sin \theta \cdot \sin \theta+\cos \theta \cdot \cos \theta=1$
View full question & answer→MCQ 2602 Marks
$\tan A +\cot \left(180^{\circ}+ A \right)+\cot \left(90^{\circ}+ A \right) +\cot \left(360^{\circ}- A =\right)$
- ✓
$0$
- B
$2 \tan A$
- C
$2 \cot A$
- D
$2(\tan A-\cot A)$
Answer(A)
$\tan A +\cot \left(180^{\circ}+ A \right)+\cot ( \left.90^{\circ}+ A \right) +\cot \left(360^{\circ}- A \right)$
$=\tan A +\cot A -\tan A -\cot A =0$
View full question & answer→MCQ 2612 Marks
$\sin (\pi+\theta) \sin (\pi-\theta) \operatorname{cosec}^2 \theta=$
- A
- ✓
$-1$
- C
$\sin \theta$
- D
$-\sin \theta$
Answer(B)
$\sin (\pi+\theta) \sin (\pi-\theta) \operatorname{cosec}^2 \theta$
$=-\sin \theta \sin \theta \frac{1}{\sin ^2 \theta}$
= -1
View full question & answer→MCQ 2622 Marks
The value of $\tan \left(-945^{\circ}\right)$ is
Answer(A)
$\tan \left(-945^{\circ}\right)=\tan \left[-\left(945^{\circ}\right)\right]$
$=-\tan \left[\left(2 \times 360^{\circ}+225^{\circ}\right)\right]$
$=-\tan \left(225^{\circ}\right)$
$=-\tan 45^{\circ}$
$\ldots\left[\because \tan \left(180^{\circ}+\theta\right)=\tan \theta\right]$
$=-1$
View full question & answer→MCQ 2632 Marks
$\sin 765^{\circ}$ is equal to
- A
- B
$0$
- C
$\frac{\sqrt{3}}{2}$
- ✓
$\frac{1}{\sqrt{2}}$
AnswerCorrect option: D. $\frac{1}{\sqrt{2}}$
(D)
$\sin 765^{\circ}=\sin (720+45)^{\circ}$
$=\sin \left(2 \times 360^{\circ}+45^{\circ}\right)$
$=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 2642 Marks
The value of $\sin \frac{31}{3} \pi$ is
- ✓
$\frac{\sqrt{3}}{2}$
- B
$\frac{1}{\sqrt{2}}$
- C
$\frac{-\sqrt{3}}{2}$
- D
$\frac{-1}{\sqrt{2}}$
AnswerCorrect option: A. $\frac{\sqrt{3}}{2}$
(A)
$\sin \left(\frac{31}{3} \pi\right)=\sin \left(10 \pi+\frac{\pi}{3}\right)$
$=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 2652 Marks
$\frac{\cot ^2 15^{\circ}-1}{\cot ^2 15^{\circ}+1}=$
- A
$\frac{1}{2}$
- ✓
$\frac{\sqrt{3}}{2}$
- C
$\frac{3 \sqrt{3}}{4}$
- D
$\sqrt{3}$
AnswerCorrect option: B. $\frac{\sqrt{3}}{2}$
(B)
$\frac{\cot ^2 15^{\circ}-1}{\cot ^2 15^{\circ}+1}=\frac{\frac{\cos ^2 15^{\circ}}{\sin ^2 15^{\circ}}-1}{\frac{\cos ^2 15^{\circ}}{\sin ^2 15^{\circ}}+1}$
$=\frac{\cos ^2 15^{\circ}-\sin ^2 15^{\circ}}{\cos ^2 15^{\circ}+\sin ^2 15^{\circ}}=\cos \left(30^{\circ}\right)$
$\left[\because \cos ^2 A-\sin ^2 B=\cos (A+B) \cos (A-B)\right]$
$=\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 2662 Marks
$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)=$
AnswerCorrect option: A. $\frac{1}{2} \cos 2 \theta$
(A)
$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)$
$=\cos \left(\frac{\pi}{6}+\theta+\frac{\pi}{6}-\theta\right) \cos \left(\frac{\pi}{6}+\theta-\frac{\pi}{6}+\theta\right)$
$\ldots\left[\because \cos ^2 A-\sin ^2 B=\cos (A+B) \cos (A-B)\right]$
$=\cos \frac{2 \pi}{6} \cos 2 \theta=\frac{1}{2} \cos 2 \theta$
View full question & answer→MCQ 2672 Marks
The value of $\cos ^2 45^{\circ}-\sin ^2 15^{\circ}$ is
AnswerCorrect option: B. $\frac{\sqrt{3}}{4}$
(B)
$\cos ^2 45^{\circ}-\sin ^2 15^{\circ}$
$=\cos (45+15)^{\circ} \cdot \cos (45-15)^{\circ}$
$\ldots\left[\because \cos ^2 A-\sin ^2 B=\cos (A+B) \cos (A-B)\right]$
$=\cos 60^{\circ} \cos 30^{\circ}$
$=\frac{\sqrt{3}}{4}$
View full question & answer→MCQ 2682 Marks
If $\tan A -\tan B =x$ and $\cot B -\cot A =y$, then $\cot (A-B)=$
AnswerCorrect option: D. $\frac{1}{x}+\frac{1}{y}$
(D)
$\cot (A-B)=\frac{1}{\tan (A-B)}$
$=\frac{1+\tan A \tan B}{\tan A-\tan B}$
$=\frac{1}{\tan A-\tan B}+\frac{\tan A \tan B}{\tan A-\tan B}$
$=\frac{1}{\tan A-\tan B}+\frac{1}{\cot B-\cot A}$
$=\frac{1}{x}+\frac{1}{y}$
View full question & answer→MCQ 2692 Marks
A positive acute angle is divided into two parts whose tangents are $\frac{1}{2}$ and $\frac{1}{3}$. Then the angle is
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{5}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: A. $\frac{\pi}{4}$
(A)
Let $\theta=\alpha+\beta$, where $\tan \alpha=\frac{1}{2}, \tan \beta=\frac{1}{3}$
$\therefore \tan \theta=\tan (\alpha+\beta)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}=1 \Rightarrow \theta=\frac{\pi}{4}$
View full question & answer→MCQ 2702 Marks
If $\tan A=\frac{a}{a+1}$ and $\tan B=\frac{1}{2 a+1}$, then the value of $A+B$ is
- A
$0$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
(D)
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$
$=\frac{\frac{a}{a+1}+\frac{1}{2 a+1}}{1-\frac{a}{a+1} \cdot \frac{1}{2 a+1}}$
$=\frac{2 a^2+a+a+1}{2 a^2+2 a+a+1-a}$
$=\frac{2 a^2+2 a+1}{2 a^2+2 a+1}$
$=1=\tan \frac{\pi}{4}$
$\therefore A+B=\frac{\pi}{4}$
View full question & answer→MCQ 2712 Marks
If $\tan (A+B)=p, \tan (A-B)=q$, then the value of $\tan 2 A$ in terms of $p$ and $q$ is
- A
$\frac{p+q}{p-q}$
- B
$\frac{p-q}{1+p q}$
- ✓
$\frac{p+q}{1-p q}$
- D
$\frac{1+p q}{1-p}$
AnswerCorrect option: C. $\frac{p+q}{1-p q}$
(C)
$2 A=( A + B )+( A - B )$
$\Rightarrow \tan 2 A=\frac{\tan ( A + B )+\tan ( A - B )}{1-\tan ( A + B ) \tan ( A - B )}$
$=\frac{p+q}{1-p q}$
View full question & answer→MCQ 2722 Marks
$\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}$ is equal to
- A
$\tan 53^{\circ}$
- ✓
$\tan 37^{\circ}$
- C
$\tan 82^{\circ}$
- D
$\tan 62^{\circ}$
AnswerCorrect option: B. $\tan 37^{\circ}$
(B)
$\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}=\frac{1-\tan 8^{\circ}}{1+\tan 8^{\circ}}$
$=\tan \left(45^{\circ}-8^{\circ}\right)=\tan 37^{\circ}$
View full question & answer→MCQ 2732 Marks
$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=$
- ✓
$\tan 54^{\circ}$
- B
$\tan 36^{\circ}$
- C
$\tan 18^{\circ}$
- D
$\tan 9^{\circ}$
AnswerCorrect option: A. $\tan 54^{\circ}$
(A)
$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\frac{1+\tan 9^{\circ}}{1-\tan 9^{\circ}}$
$=\tan \left(45^{\circ}+9^{\circ}\right)$ $\ldots\left[\because \tan 45^{\circ}=1\right]$
$=\tan 54^{\circ}$
View full question & answer→MCQ 2742 Marks
$\tan 15^{\circ}=$
- A
$\frac{1}{3}$
- B
$\sqrt{3}-2$
- ✓
$2-\sqrt{3}$
- D
$2+\sqrt{3}$
AnswerCorrect option: C. $2-\sqrt{3}$
(C)
$\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$\ldots\left[\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right]$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=2-\sqrt{3}$
View full question & answer→MCQ 2752 Marks
Given that $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$, then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$ is equal to
- A
$\frac{1}{2}$
- ✓
$\frac{1}{3}$
- C
$\frac{1}{4}$
- D
$\frac{1}{8}$
AnswerCorrect option: B. $\frac{1}{3}$
(B)
$\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$
$\Rightarrow \cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}$
$=2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}-2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}$
$\Rightarrow 3 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}=\cos \frac{\alpha}{2} \cos \frac{\beta}{2}$
$\Rightarrow \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1}{3}$
View full question & answer→MCQ 2762 Marks
The value of $\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$ is
- A
$\sqrt{2} \sin ^2 x$
- B
$\sqrt{2} \sin x$
- C
$\sqrt{2} \cos ^2 x$
- ✓
$\sqrt{2} \cos x$
AnswerCorrect option: D. $\sqrt{2} \cos x$
(D)
$\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$
$=\cos \frac{\pi}{4} \cos x$$-\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cos x+\sin \frac{\pi}{4} \sin x$
$=2 \cos \frac{\pi}{4} \cos x$
$=\frac{2}{\sqrt{2}} \cos x$
$=\sqrt{2} \cos x$
View full question & answer→MCQ 2772 Marks
The value of $\cos 15^{\circ}-\sin 15^{\circ}$ is equal to
- ✓
$\frac{1}{\sqrt{2}}$
- B
$\frac{1}{2}$
- C
$-\frac{1}{\sqrt{2}}$
- D
$0$
AnswerCorrect option: A. $\frac{1}{\sqrt{2}}$
(A)
$\cos 15^{\circ}-\sin 15^{\circ}=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos 15^{\circ}-\frac{1}{\sqrt{2}} \sin 15^{\circ}\right)$
$=\sqrt{2} \cos \left(45^{\circ}+15^{\circ}\right)$
$=\sqrt{2} \cos 60^{\circ}$
$=\sqrt{2} \cdot \frac{1}{2}$
$=\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 2782 Marks
$\frac{1}{4}\left(\sqrt{3} \cos 23^{\circ}-\sin 23^{\circ}\right)=$
AnswerCorrect option: D. $\frac{1}{2} \cos 53^{\circ}$
(D)
$\frac{1}{4}\left(\sqrt{3} \cos 23^{\circ}-\sin 23^{\circ}\right)$
$-\frac{1}{2}\left(\cos 30^{\circ} \cos 23^{\circ}-\sin 30^{\circ} \sin 23^{\circ}\right)$$\ldots\left[\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\frac{1}{2}\right]$
$=\frac{1}{2} \cos \left(30^{\circ}+23^{\circ}\right)=\frac{1}{2} \cos 53^{\circ}$
View full question & answer→MCQ 2792 Marks
If $\cos (A+B)=\alpha \cos A \cos B+\beta \sin A \sin B$, then $(\alpha, \beta)$ is
- A
$(-1,-1)$
- B
$(-1,1)$
- ✓
$(1,-1)$
- D
$(1,1)$
AnswerCorrect option: C. $(1,-1)$
(C)
$\cos ( A + B )=\alpha \cos A \cos B +\beta \sin A \sin B$
Dut, $\cos (A+B)-\cos A \cos B-\sin A \sin B$
$\Rightarrow \alpha=1, \beta=-1$
View full question & answer→MCQ 2802 Marks
$\cos 38^{\circ} \cos 8^{\circ}+\sin 38^{\circ} \sin 8^{\circ}$ is equal to
- ✓
$\cos 30^{\circ}$
- B
$\cos 60^{\circ}$
- C
$\cos 45^{\circ}$
- D
$\cos 38^{\circ}$
AnswerCorrect option: A. $\cos 30^{\circ}$
(A)
$\cos 38^{\circ} \cos 8^{\circ}+\sin 38^{\circ} \sin 8^{\circ}$
$=\cos \left(38^{\circ}-8^{\circ}\right)=\cos 30^{\circ}$
View full question & answer→MCQ 2812 Marks
The value of sin $105^{\circ}$ is
- ✓
$\frac{\sqrt{3}+1}{2 \sqrt{2}}$
- B
$\frac{\sqrt{3}-1}{2 \sqrt{2}}$
- C
$-\frac{\sqrt{3}+1}{2 \sqrt{2}}$
- D
$\frac{1-\sqrt{3}}{2 \sqrt{2}}$
AnswerCorrect option: A. $\frac{\sqrt{3}+1}{2 \sqrt{2}}$
(A)
$\sin 105^{\circ}=\sin \left(60^{\circ}+45^{\circ}\right)$
$=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}$
$=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
View full question & answer→