Trigonometry Questions for SSC MTS 2022

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Question 1: If $\cot \theta = \frac{15}{8}, \theta$ is an acute angle, then find the value of $\frac{(1 – \cos \theta)(2 + 2 \cos \theta)}{(2 – 2 \sin \theta)(1 + \sin \theta)}$.

a) $\frac{225}{64}$

b) $\frac{64}{225}$

c) $\frac{16}{15}$

d) $\frac{8}{15}$

1) Answer (B)

Solution:

$\cot\theta=\frac{15}{8}$

$\tan\theta=\frac{8}{15}$

$\sec\theta=\sqrt{1+\tan^2\theta\ }=\sqrt{1+\left(\frac{8}{15}\right)^2}=\sqrt{\frac{289}{225}\ }=\frac{17}{15}$

$\cos\theta=\frac{15}{17}$

$\sin\theta=\sqrt{1-\cos^2\theta\ }=\sqrt{1-\left(\frac{15}{17}\right)^2}=\sqrt{1-\frac{225}{289}}=\sqrt{\frac{64}{289}}=\frac{8}{17}$

$\frac{(1-\cos\theta)(2+2\cos\theta)}{(2-2\sin\theta)(1+\sin\theta)}=\frac{\left(1-\frac{15}{17}\right)\left[2+2\left(\frac{15}{17}\right)\right]}{\left[2-2\left(\frac{8}{17}\right)\right]\left(1+\frac{8}{17}\right)}$

$=\frac{\left(\frac{2}{17}\right)\left[\frac{64}{17}\right]}{\left[\frac{18}{17}\right]\left(\frac{25}{17}\right)}$

$=\frac{2\times64}{18\times25}$

$=\frac{64}{225}$

Hence, the correct answer is Option B

Question 2: If $2 \cos^2 \theta – 5 \cos \theta + 2 = 0, 0^\circ < \theta < 90^\circ$, then the value of $(\sec \theta + \tan \theta)$ is:

a) $2 + \sqrt{3}$

b) $1 – \sqrt{3}$

c) $1 + \sqrt{3}$

d) $2 – \sqrt{3}$

2) Answer (A)

Solution:

$2\cos^2\theta-5\cos\theta+2=0$

$2\cos^2\theta-4\cos\theta-\cos\theta+2=0$

$2\cos\theta\left(\cos\theta-2\right)-1\left(\cos\theta-2\right)=0$

$\left(\cos\theta-2\right)\left(2\cos\theta-1\right)=0$

$\cos\theta=2$  or $\cos\theta=\frac{1}{2}$

$\cos\theta=2$ is not possible.

So, $\cos\theta=\frac{1}{2}$

$\sec\theta=2$

$\tan\theta=\sqrt{\sec^2\theta-1}=\sqrt{4-1}=\sqrt{3}$

$\therefore$  $\left(\sec\theta+\tan\theta\right)=2+\sqrt{3}$

Hence, the correct answer is Option A

Question 3: Find the value of $\tan35^{\circ}\cot40^{\circ}\tan45^{\circ}\cot50^{\circ}\tan55^{\circ}$.

a) $\frac{1}{\sqrt{2}}$

b) $\frac{1}{2}$

c) -1

d) 1

3) Answer (D)

Solution:

$\tan35^{\circ}\cot40^{\circ}\tan45^{\circ}\cot50^{\circ}\tan55^{\circ}=\tan35^{\circ}\cot40^{\circ}\tan45^{\circ}\cot\left(90-40^{\circ}\right)\tan\left(90-35^{\circ}\right)$

$=\tan35^{\circ}\cot40^{\circ}\left(1\right)\tan40^{\circ}\cot35^{\circ}$

$=1$

Hence, the correct answer is Option D

Question 4: If $\sin(20 + x)^\circ = \cos 60^\circ, 0 \leq (20 + x) \leq 90$, then find the value of $2 \sin^2(3x + 15)^\circ – \cosec^2(2x + 10)^\circ$.

a) 3

b) -3

c) $-\frac{1}{3}$

d) -2

4) Answer (B)

Solution:

$\sin(20+x)^{\circ}=\cos60^{\circ}$

$\sin(20+x)^{\circ}=\sin30^{\circ}$

$20+x=30$

$x=10$

$2\sin^2(3x+15)^{\circ}-\operatorname{cosec}^2(2x+10)^{\circ}=2\sin^245^{\circ}-\operatorname{cosec}^230^{\circ}$

$=2\left(\frac{1}{\sqrt{2}}\right)^2-\left(2\right)^2$

$=2\left(\frac{1}{2}\right)-4$

$=1-4$

$=-3$

Hence, the correct answer is Option B

Question 5: If $\sin^6 \theta + \cos^6 \theta = \frac{1}{3}, 0^\circ < \theta < 90^\circ$, then what is the value of $\sin \theta \cos \theta$?

a) $\frac{\sqrt{2}}{3}$

b) $\frac{\sqrt{6}}{6}$

c) $\frac{\sqrt{2}}{\sqrt{3}}$

d) $\frac{2}{9}$

5) Answer (A)

Solution:

$\sin^6\theta+\cos^6\theta=\frac{1}{3}$

$\left(\sin^2\theta\right)^3+\left(\cos^2\theta\right)^3=\frac{1}{3}$

$\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^4\theta-\sin^2\theta\ \cos^2\theta+\cos^4\theta\ \right)=\frac{1}{3}$

$\left(1\right)\left(\sin^4\theta+\cos^4\theta+2\sin^2\theta\ \cos^2\theta-3\sin^2\theta\ \cos^2\theta\ \right)=\frac{1}{3}$

$\left(\sin^2\theta+\cos^2\theta\right)^2-3\sin^2\theta\ \cos^2\theta\ =\frac{1}{3}$

$1-3\sin^2\theta\ \cos^2\theta\ =\frac{1}{3}$

$3\sin^2\theta\ \cos^2\theta\ =\frac{2}{3}$

$\sin^2\theta\ \cos^2\theta\ =\frac{2}{9}$

$\sin\theta\ \cos\theta=\frac{\sqrt{2}}{3}$

Hence, the correct answer is Option A

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Question 6: If $\sin(\theta+30^{\circ})=\frac{3}{\sqrt{12}}$, then the value of $\theta$ is equal to:

a) $15^\circ$

b) $60^\circ$

c) $30^\circ$

d) $45^\circ$

6) Answer (C)

Solution:

Given,  $\sin(\theta+30^{\circ})=\frac{3}{\sqrt{12}}$

$\Rightarrow$  $\sin(\theta+30^{\circ})=\frac{3}{2\sqrt{3}}$

$\Rightarrow$  $\sin(\theta+30^{\circ})=\frac{\sqrt{3}\times\sqrt{3}}{2\sqrt{3}}$

$\Rightarrow$  $\sin(\theta+30^{\circ})=\frac{\sqrt{3}}{2}$

$\Rightarrow$  $\sin(\theta+30^{\circ})=\sin60^{\circ\ }$

$\Rightarrow$  $\theta+30^{\circ}=60^{\circ\ }$

$\Rightarrow$  $\theta=30^{\circ\ }$

Hence, the correct answer is Option C

Question 7: If $\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}=5$ and $\theta$ is an acute angle, then the value of $\frac{3\cos^2\theta+1}{3\cos^2\theta-1}$ is:

a) 4

b) 3

c) 1

d) 2

7) Answer (A)

Solution:

Given,  $\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}=5$

$\Rightarrow$  $\sec\theta+\tan\theta=5\sec\theta\ -5\tan\theta\ $

$\Rightarrow$  $6\tan\theta=4\sec\theta$

$\Rightarrow$  $6\frac{\sin\theta\ }{\cos\theta\ }=4\frac{1}{\cos\theta\ }$

$\Rightarrow$  $\sin\theta=\frac{2}{3}$

$\cos^2\theta\ =1-\sin^2\theta\ =1-\left(\frac{2}{3}\right)^2=\frac{5}{9}$

$\therefore\ $ $\frac{3\cos^2\theta+1}{3\cos^2\theta-1}=\frac{3\left(\frac{5}{9}\right)+1}{3\left(\frac{5}{9}\right)-1}$

$=\frac{\frac{5}{3}+1}{\frac{5}{3}-1}$

$=\frac{8}{2}$

$=4$

Hence, the correct answer is Option A

Question 8: If sin(A – B) = $\frac{1}{2}$ and cos(A + B) = $\frac{1}{2}$ where A > B > 0$^\circ$ and A + B is an acute angle, then the value of A is:

a) $30^\circ$

b) $60^\circ$

c) $45^\circ$

d) $15^\circ$

8) Answer (C)

Solution:

Given,  sin(A – B) = $\frac{1}{2}$

$\Rightarrow$  sin(A – B) = sin 30$^{\circ\ }$

$\Rightarrow$  A – B = 30$^{\circ\ }$ ……..(1)

cos(A + B) = $\frac{1}{2}$ and A + B is an acute angle

$\Rightarrow$  cos(A + B) = cos 60$^{\circ\ }$

$\Rightarrow$  A + B = 60$^{\circ\ }$ ……..(2)

Adding equations (1) and (2),

2A = 90$^{\circ\ }$

$\Rightarrow$  A = 45$^{\circ\ }$

Hence, the correct answer is Option C

Question 9: If $2x=\sin\theta$ and $\frac{2}{x}=\cos\theta$, then the value of $4\left(x^2+\frac{1}{x^2}\right)$ is:

a) 1

b) 0

c) 2

d) 4

9) Answer (A)

Solution:

Given, $2x=\sin\theta$ and $\frac{2}{x}=\cos\theta$

From the trigonometric identities,

$\sin^2\theta\ +\cos^2\theta\ =1$

$\Rightarrow$ $\left(2x\right)^2+\left(\frac{2}{x}\right)^2=1$

$\Rightarrow$  $4x^2+\frac{4}{x^2}=1$

$\Rightarrow$  $4\left(x^2+\frac{1}{x^2}\right)=1$

Hence, the correct answer is Option A

Question 10: If X = $\tan 40^\circ$, then the value of $2 \tan 50^\circ$ will be:

a) $\frac{2}{X}$

b) 2X

c) $\frac{1}{X}$

d) $\frac{1}{2X}$

10) Answer (A)

Solution:

Given, X = $\tan 40^\circ$

$\therefore\ 2\tan50^{\circ}=2\tan\left(90-40^{\circ}\right)$

$=2\cot40^{\circ}$

$=\frac{2}{\tan40^{\circ\ }}$

$=\frac{2}{X\ }$

Hence, the correct answer is Option A

Question 11: If $2\cot\theta=3$, then $\frac{\sqrt{13}\cos\theta-3\tan\theta}{3\tan\theta+\sqrt{13}\sin\theta}$ is:

a) $\frac{3}{4}$

b) $\frac{1}{4}$

c) $\frac{2}{3}$

d) $\frac{1}{5}$

11) Answer (B)

Solution:

Given, $2\cot\theta=3$

$\Rightarrow$  $\cot\theta=\frac{3}{2}$

$\tan\theta=\frac{2}{3}$

$\operatorname{cosec}\theta\ =\sqrt{1+\cot^2\theta\ }=\sqrt{1+\left(\frac{3}{2}\right)2}=\sqrt{1+\frac{9}{4}}=\sqrt{\frac{13}{4}}=\frac{\sqrt{13}}{2}$

$\sin\theta\ =\frac{2}{\sqrt{13}}$

$\cos\theta\ =\sqrt{1-\sin^2\theta\ }=\sqrt{1-\frac{4}{13}}=\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^2}=\sqrt{\frac{9}{13}}=\frac{3}{\sqrt{13}}$

$\therefore\ $ $\frac{\sqrt{13}\cos\theta-3\tan\theta}{3\tan\theta+\sqrt{13}\sin\theta}=\frac{\sqrt{13}\left(\frac{3}{\sqrt{13}}\right)-3\left(\frac{2}{3}\right)}{3\left(\frac{2}{3}\right)+\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)}$

$=\frac{3-2}{2+2}$

$=\frac{1}{4}$

Hence, the correct answer is Option B

Question 12: If  $x\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta$ and  $x\sin\theta=y\cos\theta$, then the value of $x^2 + y^2$ is:

a) 4

b) 0

c) 2

d) 1

12) Answer (D)

Solution:

Given, $x\sin\theta=y\cos\theta$

$\Rightarrow$  $y=\frac{x\sin\theta}{\cos\theta\ }$

$x\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta$

$\Rightarrow$  $x\sin^3\theta+\frac{x\sin\theta\ }{\cos\theta\ }\cos^3\theta=\sin\theta\cos\theta$

$\Rightarrow$  $x\sin^3\theta+x\sin\theta\ \cos^2\theta=\sin\theta\cos\theta$

$\Rightarrow$  $x\sin\theta\ \left(\sin^2\theta+\cos^2\theta\right)=\sin\theta\cos\theta$

$\Rightarrow$  $x\sin\theta\ =\sin\theta\cos\theta$

$\Rightarrow$  $x=\cos\theta$

$\therefore\ $ $y=\frac{x\sin\theta}{\cos\theta\ }=\frac{\cos\theta\ \sin\theta}{\cos\theta\ }$

$\Rightarrow$  $y=\sin\theta\ $

$\therefore\ $ $x^2+y^2=\cos^2\theta\ +\sin^2\theta=1\ $

Hence, the correct answer is Option D

Question 13: If $\sqrt{3}\cos\theta=\sin\theta$, then the value of $\frac{4\sin^2\theta-5\cos\theta}{3\cos\theta+1}$ is:

a) $\frac{1}{4}$

b) $\frac{1}{5}$

c) 5

d) $\frac{2}{5}$

13) Answer (B)

Solution:

Given,  $\sqrt{3}\cos\theta=\sin\theta$

$\Rightarrow$  $\sqrt{3}=\frac{\sin\theta}{\cos\theta\ }$

$\Rightarrow$  $\tan\theta\ =\sqrt{3}$

$\Rightarrow$  $\tan\theta\ =\tan60^{\circ\ }$

$\Rightarrow$  $\theta =60^{\circ\ }$

$\therefore\ $ $\frac{4\sin^2\theta-5\cos\theta}{3\cos\theta+1}=\frac{4\sin^260^{\circ\ }-5\cos60^{\circ\ }}{3\cos60^{\circ}+1}$

$=\frac{4\left(\frac{\sqrt{3}}{2}\right)^2-5\left(\frac{1}{2}\right)}{3\left(\frac{1}{2}\right)+1}$

$=\frac{4\left(\frac{3}{4}\right)-\frac{5}{2}}{\frac{3}{2}+1}$

$=\frac{3-\frac{5}{2}}{\frac{5}{2}}$

$=\frac{\frac{1}{2}}{\frac{5}{2}}$

$=\frac{1}{5}$

Hence, the correct answer is Option B

Question 14: If $\cos3\theta=\sin(\theta-34^{\circ})$, then the value of $\theta$ as an acute angle is:

a) $56^\circ$

b) $17^\circ$

c) $31^\circ$

d) $34^\circ$

14) Answer (C)

Solution:

Given,  $\cos3\theta=\sin(\theta-34^{\circ})$

$\Rightarrow$  $\cos3\theta=\cos\left[90-(\theta-34^{\circ})\right]$

$\Rightarrow$  $\cos3\theta=\cos\left[90-\theta+34^{\circ}\right]$

$\Rightarrow$  $\cos3\theta=\cos\left[124^{^{\circ}}-\theta\right]$

$\Rightarrow$  $3\theta=124^{^{\circ}}-\theta$

$\Rightarrow$  $4\theta=124^{^{\circ}}$

$\Rightarrow$  $\theta=31^{^{\circ}}$

Hence, the correct answer is Option C

Question 15: The value of $\sin60^{\circ}\cos30^{\circ}-\cos60^{\circ}\sin30^{\circ}$ is:

a) 1

b) $\frac{1}{2}$

c) $\frac{\sqrt 3}{2}$

d) $\frac{1}{\sqrt 2}$

15) Answer (B)

Solution:

$\sin60^{\circ}\cos30^{\circ}-\cos60^{\circ}\sin30^{\circ}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac{1}{2}\times\frac{1}{2}$

$=\frac{3}{4}-\frac{1}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Hence, the correct answer is Option B

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