SSC Questions on Trigonometry PDF

0
371
SSC Questions on Trigonometry PDF
SSC Questions on Trigonometry PDF

SSC Questions on Trigonometry PDF

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

Download SSC Questions on Trigonometry PDF

Take a free SSC-CGL mock test

Get 200 SSC mocks for just Rs. 249. Enroll here

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

Take a free mock test for SSC CGL

Download SSC CGL Previous Papers PDF

18000+ Questions – Free SSC Study Material

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) $\sqrt{2}\tan\theta$

FREE SSC EXAM YOUTUBE VIDEOS

More SSC CGL Important Questions and Answers PDF

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) $\frac{1}{4}$

SSC CGL Previous Papers Download PDF

SSC CGL Free Mock Test

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,

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

Get 200 SSC mocks for just Rs. 249. Enroll here

SSC Free Previous Papers App

We hope this Trigonometry Questions for SSC  Exams will be highly useful for your preparation.

LEAVE A REPLY

Please enter your comment!
Please enter your name here