MCQ 511 Mark
If $15 \tan ^2 \theta+4 \sec ^2 \theta=23$, then the value of $(\sec \theta+\operatorname{cosec} \theta)^2-\sin ^2 \theta$ is
- A
- ✓
$\frac{15}{2}$
- C
$\frac{9}{2}$
- D
$\frac{11}{2}$
AnswerCorrect option: B. $\frac{15}{2}$
(B) $\frac{15}{2}$
We have,
$15 \tan ^2 \theta+4 \sec ^2 \theta=23$
$\Rightarrow \quad 15 \tan ^2 \theta+4\left(1+\tan ^2 \theta\right)$$=23 \Rightarrow 19 \tan ^2 \theta=19$
$\Rightarrow \tan ^2 \theta=1 \Rightarrow \tan \theta=1 \Rightarrow \theta=45^{\circ}$
$\therefore \quad(\sec \theta+\operatorname{cosec} \theta)^2-\sin ^2 \theta$
$=\left(\sec 45^{\circ}+\operatorname{cosec} 45^{\circ}\right)^2-\sin ^2 45^{\circ}$
$=(\sqrt{2}+\sqrt{2})^2-\left(\frac{1}{\sqrt{2}}\right)^2$
$=8-\frac{1}{2}=\frac{15}{2}$
View full question & answer→MCQ 521 Mark
If $\sin \theta+\cos \theta=\sqrt{2}$, then $\tan \theta+\cot \theta=$
Answer(B)2
$
\begin{array}{l}
\text { We have, } \sin \theta+\cos \theta=\sqrt{2} \\
\Rightarrow \quad(\sin \theta+\cos \theta)^2=2 \\
\Rightarrow \quad \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta=2 \Rightarrow 1+2 \sin \theta \cos \theta=2 \Rightarrow 2 \sin \theta \cos \theta=1 \Rightarrow \sin \theta \cos \theta=\frac{1}{2} \\
\therefore \quad \tan \theta+\cot \theta=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}=\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta}=2
\end{array}
$
View full question & answer→MCQ 531 Mark
$\frac{\tan ^2 \theta}{1+\tan ^2 \theta}+\frac{\cot ^2 \theta}{1+\cot ^2 \theta}=$
- ✓
- B
$2 \tan ^2 \theta$
- C
$2 \cot ^2 \theta$
- D
$2 \sec ^2 \theta$
Answer(A) 1
$\frac{\tan ^2 \theta}{1+\tan ^2 \theta}+\frac{\cot ^2 \theta}{1+\cot ^2 \theta}$
$=\frac{\tan ^2 \theta}{\sec ^2 \theta}+\frac{\cot ^2 \theta}{\operatorname{cosec}^2 \theta}$
$=\tan ^2 \theta \times \frac{1}{\sec ^2 \theta}+\cot ^2 \theta \times \frac{1}{\operatorname{cosec}^2 \theta}$
$=\frac{\sin ^2 \theta}{\cos ^2 \theta} \times \cos ^2 \theta+\frac{\cos ^2 \theta}{\sin ^2 \theta} \times \sin ^2 \theta$
$=\sin ^2 \theta+\cos ^2 \theta=1$
View full question & answer→MCQ 541 Mark
$\tan ^4 \theta+\tan ^2 \theta=$
- A
$\sec ^2 \theta-2 \sec ^4 \theta$
- B
$2 \sec ^2 \theta-2 \sec ^4 \theta$
- C
$\sec ^2 \theta-\sec ^4 \theta$
- ✓
$\sec ^4 \theta-\sec ^2 \theta$
AnswerCorrect option: D. $\sec ^4 \theta-\sec ^2 \theta$
(D) $\sec ^4 \theta-\sec ^2 \theta \tan ^4 \theta+\tan ^2 \theta$
$=\tan ^2 \theta\left(\tan ^2 \theta+1\right)$
$=\tan ^2 \theta \sec ^2 \theta=\left(\sec ^2 \theta-1\right) \sec ^2 \theta$
$=\sec ^4 \theta-\sec ^2 \theta$.
View full question & answer→MCQ 551 Mark
If $\cos A=\frac{3}{5}$, then the value $9+9 \tan ^2 A$ is
Answer(C)25
$
\text {} 9+9 \tan ^2 A=9\left(1+\tan ^2 A\right)=9 \sec ^2 A=\frac{9}{\cos ^2 A}=9 \times\left(\frac{5}{3}\right)^2=25
$
View full question & answer→MCQ 561 Mark
$(1+\tan A-\sec A)(1+\tan A+\sec A)=$
- ✓
$2 \tan A$
- B
$2 \sin A$
- C
$2 \sec A$
- D
$2 \cot A$
AnswerCorrect option: A. $2 \tan A$
(A)$2 \tan A$
$
\begin{aligned}
\text {} & (1+\tan A-\sec A)(1+\tan A+\sec A) \\
= & (1+\tan A)^2-\sec ^2 A=1+\tan ^2 A+2 \tan A-\sec ^2 A=\sec ^2 A+2 \tan A-\sec ^2 A=2 \tan
\end{aligned}
$
View full question & answer→MCQ 571 Mark
The value of $\left(1+\tan ^2 \theta\right)(1+\sin \theta)(1-\sin \theta)(1+\cos \theta)(1-\cos \theta)\left(1+\cot ^2 \theta\right)$ is
Answer(A)1
$\begin{array}{l}\left(1+\tan ^2 \theta\right)(1+\sin \theta)(1-\sin \theta)(1+\cos \theta)(1-\cos \theta)\left(1+\cot ^2 \theta\right) \\ =\sec ^2 \theta\left(1-\sin ^2 \theta\right)\left(1-\cos ^2 \theta\right) \operatorname{cosec}^2 \theta=\frac{1}{\cos ^2 \theta} \times \cos ^2 \theta \times \sin ^2 \theta \times \frac{1}{\sin ^2 \theta}=1\end{array}$
View full question & answer→MCQ 581 Mark
The value of $\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}$ is
Answer(B)1
$\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}=\cos ^2 \theta+\frac{1}{\operatorname{cosec}^2 \theta}=\cos ^2 \theta+\sin ^2 \theta=1$
View full question & answer→MCQ 591 Mark
If $\cos A+\cos ^2 A=1$, then $\sin ^2 A+\sin ^4 A=$
View full question & answer→MCQ 601 Mark
If $a \cos \theta+b \sin \theta=m$ and $a \sin \theta-b \cos \theta=n$, then $a^2+b^2=$
- A
$m^2-n^2$
- B
$m^2 n^2$
- C
$n^2-m^2$
- ✓
$m^2+n^2$
AnswerCorrect option: D. $m^2+n^2$
View full question & answer→MCQ 611 Mark
If $\sin \theta+\sin ^2 \theta=1$, then $\cos ^2 \theta+\cos ^4 \theta=$
View full question & answer→MCQ 621 Mark
If $x=r \sin \theta \cos \phi, y=r \sin \theta \sin \phi$ and $z=r \cos \theta$, then
- ✓
$x^2+y^2+z^2=r^2$
- B
$x^2+y^2-z^2=r^2$
- C
$x^2-y^2+z^2=r^2$
- D
$z^2+y^2-x^2=r^2$
AnswerCorrect option: A. $x^2+y^2+z^2=r^2$
View full question & answer→MCQ 631 Mark
If $a \cot \theta+b \operatorname{cosec} \theta=p$ and $b \cot \theta+a \operatorname{cosec} \theta=q$, then $p^2-q^2=$
- A
$a^2-b^2$
- ✓
$b^2-a^2$
- C
$a^2+b^2$
- D
$b-a$
AnswerCorrect option: B. $b^2-a^2$
View full question & answer→MCQ 641 Mark
If $a \cos \theta+b \sin \theta=4$ and $a \sin \theta-b \cos \theta=3$, then $a^2+b^2=$
View full question & answer→MCQ 651 Mark
$2\left(\sin ^6 \theta+\cos ^6 \theta\right)-3\left(\sin ^4 \theta+\cos ^4 \theta\right)$ is equal to
View full question & answer→MCQ 661 Mark
$\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$ is equal to
View full question & answer→MCQ 671 Mark
If $1+\sin ^2 \alpha=3 \sin \alpha \cos \alpha$, then the values of $\cot \alpha$ are
View full question & answer→MCQ 681 Mark
If $\cos (\alpha+\beta)=0$, then $\sin (\alpha-\beta)$ can be reduced to
- A
$\cos \beta$
- ✓
$\cos 2 \beta$
- C
$\sin \alpha$
- D
$\sin 2 \alpha$
AnswerCorrect option: B. $\cos 2 \beta$
View full question & answer→MCQ 691 Mark
If $a \cos \theta-b \sin \theta=c$, then $a \sin \theta+b \cos \theta=$
AnswerCorrect option: B. $\pm \sqrt{a^2+b^2-c^2}$
View full question & answer→MCQ 701 Mark
If $\sin \theta-\cos \theta=0$, then the value of $\sin ^4 \theta+\cos ^4 \theta$ is
- A
- B
$\frac{3}{4}$
- ✓
$\frac{1}{2}$
- D
$\frac{1}{4}$
AnswerCorrect option: C. $\frac{1}{2}$
View full question & answer→MCQ 711 Mark
If $A$ and $B$ are acute angles such that $\sin (A-B)=0$ and $2 \cos (A+B)-1=0$, then $A=$
- A
$60^{\circ}$
- ✓
$30^{\circ}$
- C
$45^{\circ}$
- D
$15^{\circ}$
AnswerCorrect option: B. $30^{\circ}$
View full question & answer→MCQ 721 Mark
$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$ is equal
View full question & answer→MCQ 731 Mark
The value of $(1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)$ is
View full question & answer→MCQ 741 Mark
$\cos ^4 A-\sin ^4 A$ is equal to
- A
$2 \cos ^2 A+1$
- ✓
$2 \cos ^2 A-1$
- C
$2 \sin ^2 A-1$
- D
$2 \sin ^2 A+1$
AnswerCorrect option: B. $2 \cos ^2 A-1$
View full question & answer→MCQ 751 Mark
$\sec ^4 A-\sec ^2 A$ is equal to
- A
$\tan ^2 A-\tan ^4 A$
- B
$\tan ^4 A-\tan ^2 A$
- ✓
$\tan ^4 A+\tan ^2 A$
- D
$\tan ^2 A+\tan ^4 A$
AnswerCorrect option: C. $\tan ^4 A+\tan ^2 A$
View full question & answer→MCQ 761 Mark
If $\sec \theta+\tan \theta=x$, then $\tan \theta=$
- A
$\frac{x^2+1}{x}$
- B
$\frac{x^2-1}{x}$
- C
$\frac{x^2+1}{2 x}$
- ✓
$\frac{x^2-1}{2 x}$
AnswerCorrect option: D. $\frac{x^2-1}{2 x}$
View full question & answer→MCQ 771 Mark
If $\sec \theta+\tan \theta=x$, then $\sec \theta=$
- A
$\frac{x^2+1}{x}$
- ✓
$\frac{x^2+1}{2 x}$
- C
$\frac{x^2-1}{2 x}$
- D
$\frac{x^2-1}{x}$
AnswerCorrect option: B. $\frac{x^2+1}{2 x}$
View full question & answer→MCQ 781 Mark
If $\triangle A B C$ is right angled at $C$, then the value of $\cos (A+B)$ is
- ✓
$0$
- B
- C
$\frac{1}{2}$
- D
$\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 791 Mark
If $2 \sin ^2 \beta-\cos ^2 \beta=2$, then $\beta$ is equal to
- A
$0^{\circ}$
- ✓
$90^{\circ}$
- C
$45^{\circ}$
- D
$30^{\circ}$
AnswerCorrect option: B. $90^{\circ}$
View full question & answer→MCQ 801 Mark
If $x=a \sec \theta \cos \phi, y=b \sec \theta \sin \phi$ and $z=c \tan \theta$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}=$
- A
$\frac{z^2}{c^2}$
- B
$1-\frac{z^2}{c^2}$
- C
$\frac{z^2}{c^2}-1$
- ✓
$1+\frac{z^2}{c^2}$
AnswerCorrect option: D. $1+\frac{z^2}{c^2}$
View full question & answer→MCQ 811 Mark
If $x=a \sec \theta$ and $y=b \tan \theta$, then $b^2 x^2-a^2 y^2=$
- A
$a b$
- B
$a^2-b^2$
- C
$a^2+b^2$
- ✓
$a^2 b^2$
AnswerCorrect option: D. $a^2 b^2$
View full question & answer→MCQ 821 Mark
If $x=a \cos \theta$ and $y=b \sin \theta$, then $b^2 x^2+a^2 y^2=$
- ✓
$a^2 b^2$
- B
$a b$
- C
$a^4 b^4$
- D
$a^2+b^2$
AnswerCorrect option: A. $a^2 b^2$
View full question & answer→MCQ 831 Mark
$\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$ is equal to
AnswerCorrect option: C. $2 \operatorname{cosec} \theta$
View full question & answer→MCQ 841 Mark
$\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$ is equal to
AnswerCorrect option: C. $\sin \theta+\cos \theta$
View full question & answer→MCQ 851 Mark
$\frac{\sin \theta}{1+\cos \theta}$ is equal to
- A
$\frac{1+\cos \theta}{\sin \theta}$
- B
$\frac{1-\cos \theta}{\cos \theta}$
- ✓
$\frac{1-\cos \theta}{\sin \theta}$
- D
$\frac{1-\sin \theta}{\cos \theta}$
AnswerCorrect option: C. $\frac{1-\cos \theta}{\sin \theta}$
View full question & answer→MCQ 861 Mark
The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is
- A
$\cot \theta-\operatorname{cosec} \theta$
- ✓
$\operatorname{cosec} \theta+\cot \theta$
- C
$\operatorname{cosec}^2 \theta+\cot ^2 \theta$
- D
$(\cot \theta+\operatorname{cosec} \theta)^2$
AnswerCorrect option: B. $\operatorname{cosec} \theta+\cot \theta$
View full question & answer→MCQ 871 Mark
$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ is equal to
- ✓
$\sec \theta+\tan \theta$
- B
$\sec \theta-\tan \theta$
- C
$\sec ^2 \theta+\tan ^2 \theta$
- D
$\sec ^2 \theta-\tan ^2 \theta$
AnswerCorrect option: A. $\sec \theta+\tan \theta$
View full question & answer→MCQ 881 Mark
If $\tan \alpha+\cot \alpha=2$, then $\tan ^{2020} \alpha+\cot ^{2020} \alpha=$
View full question & answer→MCQ 891 Mark
If $x=2 \sin ^2 \theta$ and $y=2 \cos ^2 \theta+1$, then $x+y$ is equal to
View full question & answer→