SSC Questions on Trigonometry PDF

Download SSC Trigonometry Questions with answers PDF based on previous papers very useful for SSC  exams. Top-10 Very Important Questions for SSC Exams.

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Question 1: $5tan\theta = 4$, then the value of $(\frac{5sin\theta – 3cos\theta}{5sin\theta + 3cos\theta})$ is

a) $\frac{1}{7}$

b) $\frac{2}{7}$

c) $\frac{5}{7}$

d) $\frac{2}{5}$

Question 2: The least value of $(4sec^2\theta + 9cosec^2\theta)$ is

a) 1

b) 19

c) 25

d) 7

Question 3: If $x=cosec\theta-sin\theta$ and $y=sec\theta-cos\theta$, then the value of $x²y²(x² + y² + 3)$

a) 0

b) 1

c) 2

d) 3

Question 4: If $ 0 \leq \theta \leq \frac{\pi}{2}$, $2ycos\theta=sin\theta$ and $\frac{x}{2cosec\theta}=y$, then the value of $x^2-4y^2$ is

a) 1

b) 2

c) 3

d) 4

Question 5: If $sin\theta + \sin^2\theta = 1$, then the value of cos¹²$\theta$ + 3cos¹⁰$\theta$ + cos⁶$\theta$ +  3cos⁸$\theta$  – 1 is

a) 0

b) 1

c) -1

d) 2

Question 6: The value of $\frac{1}{cosec\theta – cot\theta} – \frac{1}{sin\theta}$

a) $cot\theta$

b) $cosec\theta$

c) $tan\theta$

d) $1$

Question 7: If $\cos\theta + \sin\theta = \sqrt{2}\cos\theta$, then $\cos\theta – \sin\theta$ is

a) -$\sqrt{2}\cos\theta$

b) -$\sqrt{2}\sin\theta$

c) $\sqrt{2}\sin\theta$

d) $\sqrt{2}\tan\theta$

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Question 8: If $cos^4\theta-sin^4\theta=\frac{2}{3}$, then the value of $1-2sin^2\theta$ is,

a) 0

b) $\frac{2}{3}$

c) $\frac{1}{3}$

d) $\frac{4}{3}$

Question 9: If $tan\theta$ = 3/4 and $\theta$ is acute, then $cosec\theta$ is equal to

a) $\frac{5}{3}$

b) $2$

c) $\frac{1}{2}$

d) $4$

Question 10: The value of $\frac{1}{1 + tan^2\theta}$ + $\frac{1}{1 + cot^2\theta}$ is

a) 1

b) 2

c) $\frac{1}{2}$

d) $\frac{1}{4}$

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Answers & Solutions:

1) Answer (A)

Taking $cos\theta$ outside in numerator and in denominator and making $tan\theta$
hence eq will be  $(\frac{5tan\theta – 3}{5tan\theta + 3})$
As it is given that $5tan\theta$ = 4
after putting values and solving we will get the equation reduced to 1/7.

2) Answer (A)

$4sec^2\theta+9cosec^2\theta$
or $4+4tan^2\theta+9+9cot^2\theta$
or $13+4tan^2\theta+9cot^2\theta$
or $ 13+4tan^2\theta+\frac{9}{tan^2\theta} $
or $  13-12+(2tan\theta+\frac{3}{tan\theta})^2 $    (eq. (1) )
or now above expression to be minimum, equation $(2tan\theta+\frac{3}{tan\theta})^2$ should be minimum.
So applying $A.M.\geq G.M. $
$\frac{(2tan\theta +\frac{3}{tan\theta})}{2} \geq \sqrt{6}$
or ${(2tan\theta+\frac{3}{tan\theta})}=2\sqrt{6}$ ( for value to be minimum)
After putting above value in eq.(1) , we will get least value of expression as 25.

3) Answer (B)

$x=cosec\theta – sin\theta=\frac{cos^2\theta}{sin\theta}=cot\theta cos\theta$
Similarly $y=tan\theta sin\theta$
$xy=sin\theta cos\theta$
$x^2+y^2+3=(sec^2\theta +cosec^2\theta )$
Now putting above values in given equation, and after solving it will be reduced to 1

4) Answer (A)

$2y=tan\theta$
$x=2ycosec\theta$
Hence value of $x^2 – 4y^2 $ = $4y^2(cosec^2\theta – 1)$
or $tan^2\theta cot^2\theta$ = 1

5) Answer (A)

Given equation can be written as $(cos^4\theta + cos^2\theta)^3 -1$
as $sin\theta + sin^2\theta = 1$
or $sin\theta = cos^2\theta$
putting above value in given equation it will be
$(sin^2\theta + sin\theta)^3 -1 = 0$

6) Answer (A)

$\frac{sin\theta}{1-cos\theta} – \frac{1}{sin\theta}$

or $\frac{cos\theta – cos^2\theta}{(1-cos\theta)sin\theta}$ = $cot\theta$

7) Answer (C)

$\sin^2 \theta + \cos^2 \theta = 1$
So, $\sin^2 \theta + \cos^2 \theta + 2\sin\theta * \cos \theta = 2 \cos^2\theta$
Hence, $\cos^2 \theta – \sin^2 \theta = 2 \sin\theta*\cos\theta$
So, $\cos\theta – \sin\theta = \sqrt{2}\sin\theta$

8) Answer (B)

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

9) Answer (A)

$\frac{sin \theta}{cos \theta} = \frac{3}{4}$
So, $\frac{sin^2\theta}{cos^2\theta}=\frac{9}{16}$
Hence, $sin^2 \theta = \frac{9}{9+16}=\frac{9}{25}$
So, $cosec \theta = \frac{5}{3}$

10) Answer (A)

$1 + \tan ^2 \theta = \sec ^2 \theta$
$1 + \cot ^2 \theta = \csc ^2 \theta$
So, the given fraction becomes,

$\frac{1}{\sec ^2 \theta} + \frac{1}{\csc ^2 \theta} = \sin^2\theta + \cos^2 \theta = 1$

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