MCQ 1511 Mark
The value of $\tan\theta.\tan(90-\theta)$ is equal to :
- A
$\sin^2\theta$
- ✓
$1$
- C
$\cos^2\theta$
- D
$0$
Answer$\tan\theta.\tan(90-\theta)=\tan\theta\times\cot\theta=1$
$\because\tan(90-\theta)=\cot\theta$
View full question & answer→MCQ 1521 Mark
$\sin(45^\circ+\theta)-\cos(45^\circ-\theta)=?$
- A
$2\sin\theta$
- B
$2\cos\theta$
- ✓
$0$
- D
$1$
Answer$\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$
$=\sin(45^\circ+\theta)-\cos[90^\circ-(45^\circ+\theta)]$
$=\sin(45^\circ+\theta)-\sin(45^\circ+\theta)$
$=0$
View full question & answer→MCQ 1531 Mark
$\frac{2\tan^230^\circ\sec^252^\circ\sin^238^\circ}{\text{cosec}^270^\circ-\tan^220^\circ}=?$
- A
$2$
- B
$\frac{1}{2}$
- ✓
$\frac{2}{3}$
- D
$\frac{3}{2}$
AnswerCorrect option: C. $\frac{2}{3}$
$\frac{2\tan^230^\circ\sec^252^\circ\sin^238^\circ}{\text{cosec}^270^\circ-\tan^220^\circ}$
$=\frac{2\times\big(\frac{1}{\sqrt3}\big)^2\times\frac{1}{\cos^252^\circ}\times\sin^2(90^\circ-52^\circ)}{\text{cosec}^2(90^\circ-20^\circ)-\tan^220^\circ}$
$=\frac{2\times\frac{1}{3}\times\frac{1}{\cos^252^\circ}\times\cos^252^\circ}{\sec^220^\circ-\tan^220^\circ}$
$=\frac{\frac{2}{3}}1{}$
$=\frac{2}{3}$
View full question & answer→MCQ 1541 Mark
Choose the correct option. Justify your choice : $\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}= $
- A
$\sec^2 \text{A}$
- B
$-1$
- C
$\cot^2 \text{A}$
- ✓
$\tan^2 \text{A}$
AnswerCorrect option: D. $\tan^2 \text{A}$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}=\frac{\sec^2\text{A}-\tan^2\text{A}+\tan^2\text{A}}{\text{cosec}^2\text{A}-\cot^2\text{A}+\cot^2\text{A}}$
$=\frac{\sec^2\text{A}}{\text{cosec}^2\text{A}}=\frac{\frac{1}{\cos^2\text{A}}}{\frac{1}{\sin^2\text{A}}} $
$=\frac{\sin^2\text{A}}{\cos^2\text{A}}=\tan^2\text{A} $
View full question & answer→MCQ 1551 Mark
If $\text{x}=\text{a}\sec\theta$ and ${y}=\text{b}\tan\theta,$ then $b^2x^2- a^2y^2=$
- A
$ab$
- B
$a^2- b^2$
- C
$a^2+ b^2$
- ✓
$a^2b^2$
AnswerCorrect option: D. $a^2b^2$
Given, $\text{x}=\text{a}\sec\theta,\text{y}=\text{b}\tan\theta$
So,
$\text{b}^2\text{y}^2-\text{a}^2\text{y}^2$
$=\text{b}^2(\text{a}\sec\theta)^2-\text{a}^2(\text{b}\tan\theta)^2$
$=\text{b}^2\text{a}^2\sec^2\theta-\text{a}^2\text{b}^2\tan^2\theta$
$=\text{b}^2\text{a}^2(\sec^2\theta-\tan^2\theta)$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\text{b}^2\text{x}^2-\text{a}^2\text{y}^2=\text{a}^2\text{b}^2$
Hence, the correct option is $(D).$
View full question & answer→MCQ 1561 Mark
The value of $\sin^229^\circ+\sin^261^\circ$ is :
- ✓
$1$
- B
$0$
- C
$2\sin^2{29}^\circ$
- D
$2\cos^2{61}^\circ$
Answer$\sin^2{29}^\circ+\sin^2{61}^\circ$
$=\sin^2{29}^\circ+\sin^2{(90^\circ-29^\circ})$
$=\sin^2{29}+\cos^2{29}^\circ$
$(\sin^2\theta+\cos^2\theta=1)$
View full question & answer→MCQ 1571 Mark
$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ=?$
- A
$0$
- ✓
$\frac{1}{4}$
- C
$4$
- D
$1$
AnswerCorrect option: B. $\frac{1}{4}$
$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ$
$=\Big(\frac{1}{2}\Big)^2+4(1)^2-(2)^2$
$=\frac{1}{4}+4-4$
$=\frac{1}4{}$
View full question & answer→MCQ 1581 Mark
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$ is equal to :
- ✓
$\sin60^\circ$
- B
$\cos60^\circ$
- C
$\tan60^\circ$
- D
$\sin30^\circ$
AnswerCorrect option: A. $\sin60^\circ$
We have to find the value of the following expression
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt3}}{1+\Big(\frac{1}{\sqrt3}\Big)^2}$
$=\frac{\frac{2}{\sqrt3}}{1+\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt3}}{\frac{4}{3}}$
$\begin{bmatrix}\text{Since}\tan60^\circ=\frac{\sqrt3}{2}\\\text{Since}\tan30^\circ=\frac{1}{\sqrt3}\end{bmatrix}$
$=\frac{\sqrt3}{2}$
$=\sin60^\circ$
Hence the correct option is $(a)$
View full question & answer→MCQ 1591 Mark
$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)=$
Answer$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)$
$=1+\cot\theta-\text{cosec }\theta+\tan\theta+\cot\theta\tan\theta$
$=-\tan\theta\text{ cosec }\theta+\sec\theta+\sec\theta\cot\theta-\sec\theta\text{ cosec }\theta$
$=1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}+\frac{\sin\theta}{\cos\theta}+1-\frac{\sin\theta}{\cos\theta}\times\frac{1}{\sin\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\cos\theta}\times\frac{\cos\theta}{\sin\theta}-\frac{1}{\cos\theta}\times\frac{1}{\sin\theta}$
$=2+\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}-\frac{1}{\sin\theta}-\frac{1}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{1}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}=2$
View full question & answer→MCQ 1601 Mark
If $4\tan\theta = 3$ then $\frac{4\sin\theta-\cos\theta}{{4\sin\theta+\cos\theta}}$ is equal to :
- A
$\frac{3}{4}$
- ✓
$\frac{1}{2}$
- C
$\frac{1}{3}$
- D
$\frac{2}{3}$
AnswerCorrect option: B. $\frac{1}{2}$
Given : $4\tan\theta = 3$
Dividing all terms of $\frac{4\sin\theta-\cos\theta}{{4\sin\theta+\cos\theta}}$ by $\cos\theta$
$=\frac {4\tan\theta-1}{4\tan\theta+1} = \frac{3-1}{3+1} $
$= \frac{2}{4} = \frac{1}{2}$
View full question & answer→MCQ 1611 Mark
If $\cos\text{A}+\cos^2\text{A}=1,$ then $\sin^2\text{A}+\sin^4\text{A}=$
AnswerGiven,
$\cos\text{A}+\cos^2\text{A}=1$
$\Rightarrow\ 1-\cos^2\text{A}=\cos\text{A}$
So $, \sin^2\text{A}+\sin^4\text{A}$
$=\sin^2\text{A}+\sin^2\text{A}\sin^2\text{A}$
$=\sin^2\text{A}+(1-\cos^2\text{A})(1-\cos^2\text{A})$
$=\sin^2\text{A}+\cos\text{A}\cos\text{A}$
$=\sin^2\text{A}+\cos^2\text{A}$
$=1$
Hence, the correct option is $(c).$
View full question & answer→MCQ 1621 Mark
The value of $\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$ is :
AnswerHere we have to find : $\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$
Now,
$\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$
$=\tan(90^\circ-80^\circ)\tan(90^\circ-75^\circ)\tan75^\circ\tan80^\circ$
$=\cot80^\circ\cot75^\circ\tan75^\circ\tan80^\circ$
$=(\cot80^\circ\tan80^\circ)(\cot75^\circ\tan75^\circ)$
$=1\times1$
$[$Since $\cot\theta\tan\theta=1]$
$=1$
Hence the correct option is $(c)$
View full question & answer→MCQ 1631 Mark
If $\sin\theta-\cos\theta=0,$ then the value of $\sin^4\theta+\cos^4\theta$ is :
- A
$1$
- B
$\frac{3}{4}$
- ✓
$\frac{1}{2}$
- D
$\frac{1}{4}$
AnswerCorrect option: C. $\frac{1}{2}$
$\sin\theta-\cos\theta=0$
$\Rightarrow\ \sin\theta=\cos\theta$
$\Rightarrow\ \frac{\sin\theta}{\cos\theta}=1$
$\Rightarrow\ \tan\theta=1$
$\Rightarrow\ \theta=45^\circ$
Now, put the value of $\theta$ in the given equation
$\sin^4\theta+\cos^4\theta$
$=\sin^4{45}^\circ+\cos^4{45}^\circ$
$=\Big(\frac{1}{\sqrt{2}}\Big)^4+\Big(\frac{1}{\sqrt{2}}\Big)^4$
$=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$
$=\frac{1}{2}$
View full question & answer→MCQ 1641 Mark
The value of $\left(\sin 45^{\circ}+\cos 45^{\circ}\right)$ is :
- A
$1$
- B
$\frac{1}{\sqrt{2}}$
- C
$\frac{\sqrt{3}}{2}$
- ✓
$\sqrt{2}$
AnswerCorrect option: D. $\sqrt{2}$
$\sin 45^{\circ}+\cos 45^{\circ}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$
$=\frac{2}{\sqrt{2}}$
$=\sqrt{2}$
View full question & answer→MCQ 1651 Mark
Choose the correct answer from the given four options. If $\cos(\alpha+\beta)=0,$ then $\sin(\alpha-\beta)$ can be reduced to :
- A
$\cos\beta$
- ✓
$\cos2\beta$
- C
$\sin\alpha$
- D
$\sin2\alpha$
AnswerCorrect option: B. $\cos2\beta$
Given $, \cos(\alpha+\beta)=0=\cos90^\circ$
$[\because\cos90^\circ=0]$
$\Rightarrow\ \alpha+\beta=90^\circ$
$\Rightarrow\ \alpha=90^\circ-\beta\ \ ...(\text{i})$
Now $, \sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ [$put the value from Eq. $(i)]$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
$[\because\sin(90^\circ-\theta)=\cos\theta]$
Hence, $\sin(\alpha-\beta)$ can be reduced to $\cos2\beta.$
View full question & answer→MCQ 1661 Mark
$\sin2\text{A}=2\sin\text{A}$ is true when $A =$
- ✓
$0^\circ$
- B
$30^\circ$
- C
$45^\circ$
- D
$60^\circ$
AnswerCorrect option: A. $0^\circ$
We are given $\sin2\text{A}=2\sin\text{A}.\cos\text{A}$
So,
$\Rightarrow\sin2\text{A}=2\sin\text{A}$
$\Rightarrow2\sin\text{A}.\cos\text{A}=2\sin\text{A}$
$\Rightarrow\cos\text{A}=1$
$\Rightarrow\cos\text{A}=\cos0^\circ$
As $\text{A}=0^\circ$
Hence the correct option is $(a)$
View full question & answer→MCQ 1671 Mark
If $\tan\theta=\frac{1}{\sqrt7}$ then $\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=? $
- A
$\frac{-2}{3}$
- B
$\frac{-3}{4}$
- C
$\frac{2}{3}$
- ✓
$\frac{3}{4}$
AnswerCorrect option: D. $\frac{3}{4}$
$\tan\theta=\frac{1}{\sqrt7}$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\sqrt7$
Now, $\sec^2\theta=(1+\tan^2\theta)$
$1+\Big(\frac{1}{\sqrt7}\Big)^2=1+\frac{1}{7}=\frac{8}{7}$
And, $\text{cosec}^2\theta=(1+\cot^2\theta)$
$=1+\big(\sqrt7\big)^2=1+7=8$
$\therefore\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}=\frac{48}{64}=\frac{3}{4}$
View full question & answer→MCQ 1681 Mark
Choose the correct option and justify your choice : $\frac{1-\tan^245^\circ}{1+\tan^245^\circ} =$
- A
$\tan 90^\circ$
- B
$1$
- C
$\sin 45^\circ$
- ✓
$0$
Answer$=\frac{1-\tan^245^\circ}{1+\tan^245^\circ}$
$=\frac{1-1}{1+1}=0$
View full question & answer→MCQ 1691 Mark
If $\cos\text{A}+\cos^2\text{A}=1$ then $\sin^2\text{A}+\sin^4\text{A}=?$
Answer$\cos\text{A}+\cos^2\text{A}=1$
$\Rightarrow\cos\text{A}=1-\cos^2\text{A}$
$\Rightarrow\cos\text{A}=\sin^2\text{A}$
Now, $\sin^2\text{A}+\sin^4\text{A}$
$=\cos\text{A}+\cos^2\text{A}=1$
View full question & answer→MCQ 1701 Mark
If $A$ and $B$ are acute angles such that $\sin\text{A}=\cos\text{B}$ then $(\text{A}+\text{B})=?$
- A
$45^\circ$
- B
$60^\circ$
- ✓
$90^\circ$
- D
$180^\circ$
AnswerCorrect option: C. $90^\circ$
$\sin\text{A}=\cos\text{B}$
$\Rightarrow\sin\text{A}=\sin(90^\circ-\text{B})$
$\Rightarrow\text{A}=90^\circ-\text{B}$
$\Rightarrow\text{A}+\text{B}=90^\circ$
View full question & answer→MCQ 1711 Mark
If $5 \tan\alpha = 4,$ then the value of. $\frac{5\sin\alpha-3\cos\alpha}{5\sin\alpha+2\cos\alpha}$ is :
- A
$\frac{1}{4}$
- ✓
$\frac{1}{6}$
- C
$\frac{1}{5}$
- D
$\frac{4}{5}$
AnswerCorrect option: B. $\frac{1}{6}$
Given : $5 \tan\alpha = 4,$
Dividing all terms of $\frac{5\sin\alpha-3\cos\alpha}{5\sin\alpha+2\cos\alpha}$ by $\cos\alpha$
$=\frac {5\tan\alpha-3}{5\tan\alpha+2} = \frac{4-3}{4+2} = \frac{1}{6}$
View full question & answer→MCQ 1721 Mark
$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}=?$
- A
$\frac32$
- B
$\frac23$
- C
$2$
- ✓
$3$
Answer$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}$
$=\frac{2\sin^263^\circ+2\sin^227+1}{3\cos^217^\circ+3\cos^273^\circ-2}$
$=\frac{2\sin^263^\circ+2\cos^263^\circ+1}{3\cos^217^\circ+3\sin^217^\circ-2}$
$=\frac{2\times1+1}{3\times1-2}$
$=\frac{2+1}{3-2}$
$=3$
View full question & answer→MCQ 1731 Mark
If $A + B = 90^\circ ,$ then $\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin\text{A}\sec\text{B}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$ is equal to :
- A
$\cot^2\text{A}$
- ✓
$\cot^2\text{B}$
- C
$-\tan^2\text{A}$
- D
$-\cot^2\text{A}$
AnswerCorrect option: B. $\cot^2\text{B}$
We have :
$\text{A+B}=90^\circ$
$\Rightarrow\text{B}=90^\circ-\text{A}$
We have to find the value of the following expression
$\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin{\text{A}\sec\text{B}}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$
So,
$\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin{\text{A}\sec\text{B}}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$
$=\frac{\tan\text{A}\tan(90^\circ-\text{A})+\tan\text{A}\cot(90^\circ-\text{A})}{\sin{\text{A}\sec(90^\circ-\text{A})}}-\frac{\sin^2(90^\circ-\text{A})}{\cos^2\text{A}}$
$=\frac{\tan\text{A}\cot{\text{A}}+\tan\text{A}\tan\text{A}}{\sin\text{A}\text{cosecA}}-\frac{\cos^2\text{A}}{\cos^2\text{A}}$
$=1+\tan^2\text{A}-1$
$=\tan^2\text{A}$
$=\tan^2(90^\circ-\text{B})$
$=\cot^2\text{B}$
Hence the correct option is $(b)$
View full question & answer→MCQ 1741 Mark
If $\text{2x}=\sec\text{A}$ and $\frac{2}{\text{x}}=\tan\text{A}$ then $2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=?$
- ✓
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
$\frac{1}{8}$
- D
$\frac{1}{16}$
AnswerCorrect option: A. $\frac{1}{2}$
We know that
$\sec^2\text{A}-\tan^2\text{A}=1$
$\Rightarrow(\text{2x})^2-\Big(\frac{2}{\text{x}}\Big)^2=1$
$\Rightarrow\text{4x}^2-\frac{4}{\text{x}^2}=1$
$\Rightarrow4\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=1$
$\Rightarrow\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}4{}$
$\Rightarrow2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}2{}$
View full question & answer→MCQ 1751 Mark
$2\left(\cos ^4 60^{\circ}+\sin ^4 30^{\circ}\right)-\left(\tan ^2 60^{\circ}+\cot ^2 45^{\circ}\right)$ $+3 \sec ^2 30^{\circ}$ is equal to
- A
$1 / 2$
- B
- C
$3 \times 4$
- D
$1 / 4$
Answer$\begin{aligned} & \text {(d): } 2\left(\cos ^4 60^{\circ}+\sin ^4 30^{\circ}\right)-\left(\tan ^2 60^{\circ}+\cot ^2 45^{\circ}\right) \\ & +3 \sec ^2 30^{\circ} \\ & =2\left[\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^4\right]-\left[(\sqrt{3})^2+(1)^2\right]+3\left(\frac{2}{\sqrt{3}}\right)^2 \\ & =2\left[\frac{1}{16}+\frac{1}{16}\right]-[3+1]+3 \times \frac{4}{3}=2\left[\frac{2}{16}\right]-4+4=\frac{4}{16}=\frac{1}{4} \\ & \end{aligned}$
View full question & answer→MCQ 1761 Mark
If $4 \cos \theta=11 \sin \theta$, then the value of $\frac{11 \cos \theta-7 \sin \theta}{11 \cos \theta+7 \sin \theta}$ is
- ✓
$93 / 149$
- B
$94 / 149$
- C
$91 / 149$
- D
$97 / 149$
AnswerCorrect option: A. $93 / 149$
(a) : Given $4 \cos \theta=11 \sin \theta$
$\Rightarrow \cos \theta=\frac{11}{4} \sin \theta$
$\begin{aligned} & \text { Now } \frac{11 \cos \theta-7 \sin \theta}{11 \cos \theta+7 \sin \theta}=\frac{11 \times \frac{11}{4} \sin \theta-7 \sin \theta}{11 \times \frac{11}{4} \sin \theta+7 \sin \theta} \\ & =\frac{\sin \theta\left(\frac{121}{4}-7\right)}{\sin \theta\left(\frac{121}{4}+7\right)}=\frac{121-28}{121+28}=\frac{93}{149}\end{aligned}$
View full question & answer→MCQ 1771 Mark
If $\theta$ is scute and $\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3$, then $\theta$ is equal to is equal to
- ✓
$60^{\circ}$
- B
$30^{\circ}$
- C
$90^{\circ}$
- D
AnswerCorrect option: A. $60^{\circ}$
We have $\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3$
$\Rightarrow \cos ^2 \theta=3\left(\cot ^2 \theta-\cos ^2 \theta\right)$
$\Rightarrow \cos ^2 \theta=3 \cot ^2 \theta-3 \cos ^2 \theta$
$\Rightarrow 4 \cos ^2 \theta=3 \cot ^2 \theta-4 \cos ^2 \theta=\frac{3 \cos ^2 \theta}{\sin ^2 \theta}$
$\Rightarrow \sin ^2 \theta=\frac{3}{4} $
$\Rightarrow \sin \theta=\frac{\sqrt{3}}{2} $
$\Rightarrow \theta=60^{\circ}$
View full question & answer→MCQ 1781 Mark
If $\cos \theta=\frac{\sqrt{3}}{2}$ and $\theta$ is acute, then $(3 \tan \theta$ $\left.-\tan ^2 \theta\right)$ is egual to
- ✓
$\frac{8}{3 \sqrt{3}}$
- B
$\frac{1}{2 \sqrt{3}}$
- C
$\frac{\sqrt{3}}{6}$
- D
AnswerCorrect option: A. $\frac{8}{3 \sqrt{3}}$
(a) ; We have $\cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=30^{\circ}\left(; \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)$Now, $3 \tan \theta-\tan ^3 \theta$
$
=3 \tan 30^{\circ}-\tan ^3 30^{\circ}
$
$
=3 \times \frac{1}{\sqrt{3}}-\left(\frac{1}{\sqrt{3}}\right)^3=\frac{3}{\sqrt{3}}-\frac{1}{3 \sqrt{3}}=\frac{9-1}{3 \sqrt{3}}=\frac{8}{3 \sqrt{3}}
$
View full question & answer→MCQ 1791 Mark
$\tan ^2 \theta \sin ^2 \theta$ is equal to
- ✓
$\tan ^2 \theta-\sin ^2 \theta$
- B
$\tan ^2 \theta+\sin ^2 \theta$
- C
$\frac{\tan ^2 \theta}{\sin ^2 \theta}$
- D
$\sin ^2 \theta \cot ^2 \theta$
AnswerCorrect option: A. $\tan ^2 \theta-\sin ^2 \theta$
$\tan ^2 \theta - \sin ^2 \theta$
$=\frac{\sin ^2 \theta \sin ^2 \theta}{\cos ^2 \theta}$
$=\sin ^2 \theta\left(\frac{1-\cos ^2 \theta}{\cos ^2 \theta}\right)$
$=\frac{\sin ^2 \theta}{\cos ^2 \theta}-\sin ^2 \theta$
$=\tan ^2 \theta-\sin ^2 \theta$
View full question & answer→MCQ 1801 Mark
$\left(\cos ^4 x-\sin ^4 x\right)$ is equal to
- A
$2 \sin ^2 x-1$
- B
$1-2 \cos ^2 x$
- C
$\sin ^2 x-\cos ^2 x$
- ✓
$2 \cos ^2 x-1$
AnswerCorrect option: D. $2 \cos ^2 x-1$
$: \cos ^4 x-\sin ^4 x$
$=\left(\cos ^2 x+\sin ^2 x\right)\left(\cos ^2 x=\sin ^2 x\right)$
$=1\left(\cos ^2 x-\left(1-\cos ^2 x\right)\right\}$
$=\cos ^2 x-1+\cos ^2 x$
$=2 \cos ^2 x-1$
View full question & answer→MCQ 1811 Mark
If $\cos \left(40^{\circ}+x\right)=\cos 60^{\circ}$, then $x$ is equal to
- ✓
$20^{\circ}$
- B
$30^{\circ}$
- C
$CO ^{\circ}$
- D
$0^{\circ}$
AnswerCorrect option: A. $20^{\circ}$
(a): We have, $\cos \left(40^{\circ}+x\right)=\cos 60^{\circ}$
$
\Rightarrow 40^{\circ}+x=60^{\circ} \Rightarrow x=20^{\circ}
$
View full question & answer→MCQ 1821 Mark
If $0^{\circ} \leq \theta \leq 45^{\circ}$ such that $\cos ^2 \theta-\sin ^2 \theta=\alpha$ and $\cos \theta-\sin \theta=b$, then find the value of $\left(\frac{a}{b}\right)^2+b^2$.
AnswerWe have,
$\left(\frac{a}{b}\right)^2+b^2-\left(\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos \theta-\sin \theta}\right)^2+(\cos \theta-\sin \theta)^2$
$=\left(\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta}\right)^2$
$=\left(\cos \theta+\sin ^2 \theta+1-2 \sin \theta \cos \theta \left[\because \cos ^2 \theta+\sin ^2 \theta-2 \sin \theta \cos \theta\right.\right.$
$=\cos ^2 \theta+\sin ^2 \theta+2 \sin \theta \cos \theta+1-2 \sin \theta \cos \theta$
$=1+1-2$
View full question & answer→MCQ 1831 Mark
Find the value of $5+\frac{\left(1+\tan ^2 \theta\right) \sin \theta \cos \theta}{\tan \theta}$.
AnswerWe have, $5+\frac{\left(1+\tan ^2 \theta\right) \sin \theta \cos \theta}{\tan \theta}$
$=5+\frac{\operatorname{scc}^2 \theta \sin \theta \cos \theta}{\frac{\sin \theta}{\cos \theta}}\left[\because 1+\tan ^2 \theta=\sec ^2 \theta, \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=5+\frac{\sec ^2 \theta \sin \theta \cos ^2 \theta}{\sin \theta}\left[\because \sec \theta=\frac{1}{\cos \theta}\right]$
$=5+1$
$=6$
View full question & answer→MCQ 1841 Mark
Find the value of $(1+\cos \theta)(1-\cos \theta) \times\left(1+\cot ^2 \theta\right)$.
Answer$\begin{aligned} & \text {(a) : We have, }(1+\cos \theta)(1-\cos \theta)\left(1+\cot ^2 \theta\right) \\ & =\left(1-\cos ^2 \theta\right)\left(1+\cot ^2 \theta\right)\left[\because(a+b)(a-b)=a^2-b^2\right] \\ & =\sin ^2 \theta\left(1+\frac{\cos ^2 \theta}{\sin ^2 \theta}\right) \quad\left(\because 1-\cos ^2 \theta=\sin ^2 \theta, \cot \theta=\frac{\cos \theta}{\sin \theta}\right) \\ & =\sin ^2 \theta\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta}\right)=1\end{aligned}$
View full question & answer→MCQ 1851 Mark
If $\sin A-\cos A=\frac{\sqrt{3}-1}{2}$ and $\sin A \cos A=\frac{\sqrt{3}}{x}$, then find the value of $x$.
AnswerWe have, $\sin A-\cos A=\frac{\sqrt{3}-1}{2}$
Squaring both sides, we get
$\sin ^2 A+\cos ^2 A-2 \sin A \cos A=\frac{3+1-2 \sqrt{3}}{4}$
$\Rightarrow 1-2 \sin A \cos A=\frac{4-2 \sqrt{3}}{4} \quad\left[\because \sin ^2 A+\cos ^2 A=1\right]$
$\Rightarrow 1-2 \sin A \cos A=1-\frac{\sqrt{3}}{2}$
$\Rightarrow 2 \sin A \cos A=\frac{\sqrt{3}}{2} \rightarrow \sin A \cos A=\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{x}$
Hence, $x=4$.
View full question & answer→MCQ 1861 Mark
In a $\triangle A B C$, right angled at $B$, if $\tan A=\sqrt{3}$, then find the value of $2 \sin A \cot A$.
Answer(a) : We have, $\tan A=\sqrt{3}$
$
\Rightarrow \angle A=60^{\circ} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right]
$
Now, $2 \sin A \cot A=2 \sin 60^{\circ} \cot 60^{\circ}$
$
=2 \times \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}=1
$
View full question & answer→MCQ 1871 Mark
In $\triangle A B C$, right angled at $B, B C=5 cm$, $\angle B A C=30^{\circ}$, find the length of the side $A C$.
- A
$5 cm$
- B
$2 cm$
- C
$7 cm$
- D
$10 cm$
View full question & answer→MCQ 1881 Mark
$\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}+\sin \theta \cos \theta$ is equal to
AnswerWe have, $\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}+\sin \theta \cos \theta$
$=\frac{(\sin \theta+\cos \theta)\left(\sin ^2 \theta+\cos ^2 \theta-\sin \theta \cos \theta\right)}{\sin \theta+\cos \theta}+\sin \theta \cos \theta [\because\left(a^3+b^3\right)]$
$=(a+b)\left(\alpha^2+b^2-a b\right) $
$=1-\sin \theta \cos \theta+\sin \theta \cos \theta-1$
View full question & answer→MCQ 1891 Mark
If \cot $\theta=\frac{15}{8}$, then evaluate $\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}$.
- A
$\frac{64}{225}$
- B
$\frac{32}{225}$
- ✓
$\frac{225}{64}$
- D
$\frac{225}{32}$
AnswerCorrect option: C. $\frac{225}{64}$
We have, $\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}$
$=\frac{2(1+\sin \theta)(1-\sin \theta)}{2(1+\cos \theta)(1-\cos \theta)}$
$=\frac{\left(1-\sin ^2 \theta\right)}{\left(1-\cos ^2 \theta\right)}$
${\left[\because(e+b)(e-b)=a^2-b^2\right]}$
$=\frac{\cos ^2 \theta}{\sin ^2 \theta} \quad\left[\because \cos ^2 \theta+\sin ^2 \theta-1\right]$
$=\left(\frac{\cos \theta}{\sin \theta}\right)^2-(\cos \theta)^2$
$=\left(\frac{15}{8}\right)^2=\frac{225}{64}\left[\because \cot \theta-\frac{15}{8}\right]$
View full question & answer→MCQ 1901 Mark
Find the value of $x$, if $\cos 2 x=\sin 60^{\circ}-\cos 30^{\circ} =\cos 60^{\circ} \cdot \sin 30^{\circ}$.
- ✓
$30^{\circ}$
- B
$60^{\circ}$
- C
$45^{\circ}$
- D
AnswerCorrect option: A. $30^{\circ}$
Wehave, $\cos 2 x-\sin 60^{\circ}-\cos 30^{\circ}-\cos 60^{\circ} \cdot \sin 30^{\circ}$
$=\cos 2 x=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$
$\Rightarrow \cos 2 x=\frac{3}{4}-\frac{1}{4}=\frac{1}{2} \rightarrow 2 x=600^{\circ}\ \left[\because \cos 60^{\circ}=\frac{1}{2}\right]$
$=x=30^{\circ}$
View full question & answer→MCQ 1911 Mark
If $\theta$ is an acute angle and $\sin \theta=\cos \theta$, then find the value of $2 \tan ^2 \theta-1$.
AnswerWe have, $\sin \theta-\cos \theta$
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{\cos \theta}{\cos \theta}$
$[$Dividing both sides by $\cos \theta]$
$\Rightarrow \tan \theta-1 $
$\Rightarrow 0=45^{\circ} \quad\left[\because \tan 45^{\circ}=1\right]$
$\therefore 2 \tan ^2 \theta-1=2 \tan ^2 45^{\circ}-1=2(1)^2-1=2-1=1$
View full question & answer→MCQ 1921 Mark
If $\tan \theta=\sqrt{3}$, then find the value of $\sin \theta \cos \theta$.
- A
$\frac{\sqrt{3}}{2}$
- B
$\sqrt{3}$
- ✓
$\frac{\sqrt{3}}{4}$
- D
$\frac{1}{\sqrt{3}}$
AnswerCorrect option: C. $\frac{\sqrt{3}}{4}$
( c ) ; We have, $\tan \theta=\sqrt{3}$
$\begin{aligned} & \Rightarrow \theta=60^{\circ} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right] \\ & \therefore \quad \sin \theta \cos \theta=\sin 60^{\circ} \cos 60^{\circ}=\frac{\sqrt{3}}{2} \cdot \frac{1}{2}=\frac{\sqrt{3}}{4}\end{aligned}$
View full question & answer→MCQ 1931 Mark
Find the acute angle $\theta$, satisfying the equation $\sec ^2 \theta+\tan ^2 \theta=3$.
- A
$30^{\circ}$
- ✓
$45^{\circ}$
- C
$60^{\circ}$
- D
AnswerCorrect option: B. $45^{\circ}$
We have, $\sec ^2 \theta+\tan ^2 \theta-3$
$\left.\Rightarrow 1+\tan ^2 \theta+\tan ^2 \theta=3\ [\because \sec ^2 \theta=1+\tan ^2 \theta\right]$
$\Rightarrow 1+2 \tan ^2 \theta=3 $
$\Rightarrow 2 \tan ^2 \theta=2$
$\Rightarrow \tan ^2 \theta=1$
$ \Rightarrow(\tan \theta)^2=1^2$
$\Rightarrow \tan \theta=1 $
$\Rightarrow \theta=45^{\circ} \ \left[\because \tan 45^{\circ}=1\right]$
View full question & answer→MCQ 1941 Mark
If $\tan A=\sqrt{2}-1$, then what is the value of $\frac{\tan A}{1+\tan ^2 A}$ ?
- ✓
$\frac{\sqrt{2}}{4}$
- B
$\frac{4}{\sqrt{2}}$
- C
$\frac{1}{\sqrt{3}}$
- D
$\frac{\sqrt{3}}{4}$
AnswerCorrect option: A. $\frac{\sqrt{2}}{4}$
We have, $\frac{\tan A}{1+\tan ^2 A}=\frac{\sqrt{2}-1}{1+(\sqrt{2}-1)^2}$
$=\frac{\sqrt{2}-1}{1+2+1-2 \sqrt{2}}=\frac{\sqrt{2}-1}{4-2 \sqrt{2}}=\frac{\sqrt{2}-1}{2 \cdot 2-2 \sqrt{2}}=\frac{\tan A=\sqrt{2}-1]}{2 \sqrt{2} \sqrt{2}-2 \sqrt{2}}$
$=\frac{\sqrt{2}-1}{2 \sqrt{2}(\sqrt{2}-1)}=\frac{1}{2 \sqrt{2}}=\frac{1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}$
View full question & answer→MCQ 1951 Mark
If $4 \tan \theta=3$, then $\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)$ is equal to
- A
$\frac{2}{3}$
- B
$\frac{1}{3}$
- ✓
$\frac{1}{2}$
- D
$\frac{3}{4}$
AnswerCorrect option: C. $\frac{1}{2}$
(c) : Given, $4 \tan \theta=3 \Rightarrow \tan \theta=\frac{3}{4}$
$
\therefore \frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}=\frac{4 \frac{\sin \theta}{\cos \theta}-1}{4 \frac{\sin \theta}{\cos \theta}+1}
$
[Dividing numerator and denominatoe by $\cos \theta$ ]
$
=\frac{4 \tan \theta-1}{4 \tan \theta+1}
$
$
\left[\because \tan \theta-\frac{\sin \theta}{\cos \theta}\right]
$
$=\frac{4\left(\frac{3}{4}\right)-1}{4\left(\frac{3}{4}\right)+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}$
View full question & answer→MCQ 1961 Mark
If $\sin A+\sin ^2 A=1$, then the value of the expression $\left(\cos ^2 A+\cos ^4 A\right)$ is
Answer(a) : Given $\sin A+\sin ^2 A=1$
$\Rightarrow \sin A=1-\sin ^2 A=\cos ^2 A \quad\left[\because \cos ^2 A+\sin ^2 A=1\right]$
On equaring both sides, we get
$
\begin{aligned}
& \sin ^2 A=\cos ^4 A \\
\Rightarrow \quad & 1-\cos ^2 A=\cos ^4 A \Rightarrow \cos ^2 A+\cos ^4 A-1
\end{aligned}
$
View full question & answer→MCQ 1971 Mark
Given that $\sin \theta=\frac{a}{b}$, then $\cos \theta$ is equal to
- A
$\frac{b}{\sqrt{b^2-a^2}}$
- B
$\frac{b}{a}$
- ✓
$\frac{\sqrt{b^2-a^2}}{b}$
- D
$\frac{a}{\sqrt{b^2-a^2}}$
AnswerCorrect option: C. $\frac{\sqrt{b^2-a^2}}{b}$
Given $\sin \theta=\frac{a}{b}$
$\because \cos \theta=\sqrt{1-\sin ^2 \theta} \ \left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$\Rightarrow \cos \theta=\sqrt{1-\left(\frac{a}{b}\right)^2}$
$=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{b^2-a^2}}{b}$
View full question & answer→MCQ 1981 Mark
If $\sin A=\frac{1}{2}$, then the value of $\cot A$ is
- ✓
$\sqrt{3}$
- B
$\frac{1}{\sqrt{3}}$
- C
$\frac{\sqrt{3}}{2}$
- D
AnswerCorrect option: A. $\sqrt{3}$
(a) : Given, $\sin A=\frac{1}{2}$
$
\because \quad \cos A=\sqrt{1-\sin ^2 A} \quad\left[\because \sin ^2 A+\cos ^2 A=1\right]
$
$
\Rightarrow \cos A=\sqrt{1-\left(\frac{1}{2}\right)^2}=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}
$
Now, $\cot A=\frac{\cos A}{\sin A}=\frac{\sqrt{3} / 2}{1 / 2}=\sqrt{3}$
View full question & answer→MCQ 1991 Mark
If $\cos A=\frac{4}{5}$, then the value of $\tan A$ is
- A
$\frac{3}{5}$
- ✓
$\frac{3}{4}$
- C
$\frac{4}{3}$
- D
$\frac{5}{3}$
AnswerCorrect option: B. $\frac{3}{4}$
Civen, $\cos A=4 / 5$
$\because \sin A=\sqrt{1-\cos ^2 A} \ \left[\because \sin ^2 A+\cos ^2 A=1\right]$
$\Rightarrow \sin A=\sqrt{1-\left(\frac{4}{5}\right)^2}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
Now, $\tan A=\frac{\sin A}{\cos A}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
View full question & answer→MCQ 2001 Mark
Evaluate $: \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}$
- A
$2\sin\theta$
- B
$2 \cos \theta$
- ✓
2 cosec $\theta$
- D
$2 \sec \theta$
AnswerCorrect option: C. 2 cosec $\theta$
(c) : We have, $\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}$
$
=\frac{(\sec \theta-1)+(\sec \theta+1)}{\sqrt{\sec ^2 \theta-1}}=\frac{2 \sec \theta}{\sqrt{\tan ^2 \theta}} \quad\left[\because \sec ^2 \theta-1=\tan ^2 \theta\right]
$
$
=\frac{2 \sec \theta}{\tan \theta}=\frac{2}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}=\frac{2}{\sin \theta}=2 \operatorname{cosec} \theta
$
View full question & answer→