SSC Questions on Trigonometry PDF

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SSC Questions on Trigonometry PDF
SSC Questions on Trigonometry PDF

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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Download SSC Questions on Trigonometry PDF

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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