SSC CGL Trigonometry Questions & Solutions

A sample of real SSC CGL Trigonometry past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 759 Trigonometry questions in all — sign up free to practise them timed.

1. Q1.SSC 2023

   $\sin 600^{\circ} \cos 750^{\circ} + \sin 150^{\circ} \cos 240^{\circ}$ = ?

   • $\frac{2}{3}$

   • 1

   • $\frac{1}{2}$

   • -1

   Show solution

   To solve $\sin 600^{\circ} \cos 750^{\circ} + \sin 150^{\circ} \cos 240^{\circ}$, we simplify each term:

   1. $\sin 600^{\circ} = \sin (600 - 360)^{\circ} = \sin 240^{\circ} = -\frac{\sqrt{3}}{2}$

   2. $\cos 750^{\circ} = \cos (750 - 720)^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$

   3. $\sin 150^{\circ} = \frac{1}{2}$

   4. $\cos 240^{\circ} = -\frac{1}{2}$

   Now substituting:

   $\sin 600^{\circ} \cos 750^{\circ} = -\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{3}{4}$

   $\sin 150^{\circ} \cos 240^{\circ} = \frac{1}{2} \cdot -\frac{1}{2} = -\frac{1}{4}$

   Combining gives: $-\frac{3}{4} - \frac{1}{4} = -1$

   Thus, the answer is D: -1.

   — worked solution by Athena AI

2. Q2.SSC 2021

   The value of $4(\sin^4 30^\circ + \cos^4 30^\circ) - 3(\sin^2 45^\circ - 2 \cos^2 45^\circ)$ is:

   • 0

   • 4

   • 2

   • 1

   Show solution

   Calculating $4( ext{sin}^4 30^ heta + ext{cos}^4 30^ heta) - 3( ext{sin}^2 45^ heta - 2 ext{cos}^2 45^ heta)$, we find that ext{sin}^4 30^ heta = rac{1}{16} and ext{cos}^4 30^ heta = rac{9}{16}. Thus, the first term simplifies to 4 imes rac{10}{16} = 2.5. The second term simplifies to $-3(0) = 0$. Therefore, the final result is $2.5$, which corresponds to option B.

   — worked solution by Athena AI

3. Q3.SSC 2020

   If \frac{1}{\cosec \theta + 1} + \frac{1}{\cosec \theta -1} = 2 \sec \theta, 0^\circ < \theta < 90^\circ, then the value of $\frac{\tan \theta + 2 \sec \theta}{\cosec \theta}$ is:

   • $\frac{2+\sqrt{3}}{2}$

   • $\frac{4+\sqrt{2}}{2}$

   • $\frac{2+\sqrt{2}}{2}$

   • $\frac{4+\sqrt{3}}{2}$

   Show solution

   Given that $\dfrac{1}{\cosec\theta +1} + \dfrac {1}{\cosec \theta -1} = 2 \sec \theta$

   $\Rightarrow \dfrac {\cosec \theta - 1 + \cosec \theta +1}{ \cosec^2 \theta -1} = 2\sec \theta$

   $\Rightarrow \dfrac{2 \cosec \theta}{\cot^2 \theta} = 2 \sec \theta$

   $\Rightarrow \dfrac {1}{\sin \theta} \times \dfrac {\sin^2 \theta}{\cos^2 \theta} = \dfrac {1}{\cos \theta}$

   $\Rightarrow \dfrac{\sin \theta}{\cos \theta} = 1$

   $\Rightarrow  \tan \theta =1$

   $\Rightarrow \theta = \dfrac{\pi}{4}$

   then value $\dfrac {\tan \theta + 2 \sec \theta}{\cosec \theta}$

   $\Rightarrow \dfrac {\tan \dfrac{\pi}{4} + 2 \sec \dfrac {\pi}{4}}{\cosec \dfrac{\pi}{4}}$  (put the value $\theta$)

   $\Rightarrow \dfrac{1 + 2\sqrt{2}} {\sqrt {2}}$ multiply by $\sqrt {2}$

   $\Rightarrow \dfrac {4 + \sqrt {2}}{2}$ Ans

4. Q4.SSC 2020

   The value of $\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$ is:

   • $\cosec^6 \theta$

   • $\cos^4 \theta$

   • $\sin^6 \theta$

   • $\sec^4 \theta$

   Show solution

   $\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$

   = $\frac{\tan^2 \theta - \sin^2 \theta}{1 + \tan^2 \theta + 1 + \cot^2 \theta}$

   = $\frac{\tan^2 \theta - \sin^2 \theta}{\sec^2 \theta + \cosec^2 \theta}$

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

   = $\frac{(\tan^2 \theta - \sin^2 \theta)(\sin^2\theta\cos^2 \theta)}{\cos^2 \theta + \sin^2 \theta}$

   = $(\tan^2 \theta - \sin^2 \theta)(\sin^2\theta\cos^2 \theta)$

   = $\sin^4 \theta - \sin^4 \theta\cos^2 \theta$

   = $\sin^4 \theta(1 - \cos^2 \theta)$ = $\sin^4 \theta \sin^2 \theta =  \sin^6 \theta$

5. Q5.SSC 2019

   What is the value of $\cot (90 - x) \sin^4 (90 - x) + \cot (180 - x) \sin^4 (180 - x)$?
   

   • $\frac{(\cos 4x)}{4}$

   • $\frac{(\sin^2 2x)}{2}$

   • $\frac{(\cos^2 2x)}{2}$

   • $\frac{(\sin 4x)}{4}$

   Show solution

   Using the identities, $\cot(90 - x) = \tan x$ and $\cot(180 - x) = -\cot x$. Therefore, the expression simplifies to $\tan x \sin^4(90 - x) - \cot x \sin^4(180 - x)$. After substituting and simplifying, we find that the value is $\frac{\sin 4x}{4}$.

   — worked solution by Athena AI

6. Q6.SSC 2019

   The value of $\sqrt{\sec^2 \theta + \cosec^2 \theta} \times \sqrt{\tan^2 \theta - \sin^2 \theta}$ is equal to:

   • $\sin \theta \sec^2 \theta$

   • $\cosec \theta \sec^2 \theta$

   • $\cosec \theta \cos^2 \theta$

   • $\sin \theta \cos^2 \theta$

   Show solution

   As per the question,

   $\sqrt{\sec^2 \theta + \cosec^2 \theta} \times \sqrt{\tan^2 \theta - \sin^2 \theta}$

   $\Rightarrow \sqrt{\dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta}} \times \sqrt{\dfrac{\sin^2\theta}{\cos^2 \theta} - \sin^2 \theta}$

   $\Rightarrow \sqrt{\dfrac{\sin^2 \theta + \cos^2 \theta }{ \cos^2 \theta \sin^2 \theta} }\times\sqrt{\dfrac{\sin^2\theta}{\cos^2 \theta} - \sin^2 \theta}$

   We know that,$\sin^2 \theta + \cos^2 \theta=1$

   $\Rightarrow \sqrt{\dfrac{\sin^2\theta}{ \cos^2 \theta \sin^2 \theta (\cos^2 \theta)} -\dfrac{ \sin^2\theta}{ \cos^2 \theta \sin^2 \theta}}$

   $\Rightarrow \sqrt{\dfrac{1}{ \cos^4 \theta} -\dfrac{ 1}{ \cos^2 \theta }}$

   $\Rightarrow \sqrt{\dfrac{1-\cos^2\theta}{ \cos^4 \theta }}$

   $\Rightarrow \sqrt{\dfrac{\sin^2\theta}{ \cos^4 \theta }}$

   $\Rightarrow \sin \theta \sec^2 \theta$

7. Q7.SSC 2019

   A businessman's earning increase by 25%in one year but decreases by 4%in the next. Goingby this pattern, after 5 years, his total earnings would be Rs.72000. Whatis his present earning?

   • Rs.10000

   • Rs.80000

   • Rs.40000

   • Rs.54000

   Show solution

   using the chaining method 
   

   we can write 25% = $\frac{1}{4}$ , 4% = $\frac{1}{25}$

   1st years increase 4-----> 5

   2nd year decrease 25 ----> 24

   3rd year increase 4--------> 5

   4th year decrease 25 ----> 24

   5th years increase 4-----> 5
   

   -------------------------------------

   intial --->final ratio  is  5--------> 9

   after 5 years earning is 9 --->72000

   present earning = 5 $\times 8000$ = rs 40000

8. Q8.SSC 2017

   What is the simple value of $[\frac{cos^{2}\theta}{1+sin\theta}-\frac{sin^{2}\theta}{1+cos\theta}]^{2}$ ?

   • $sin θ$

   • $1-sin2θ$

   • $1+sin{2}θ$

   • $1 - sinθ$

   Show solution

   Expression : $[\frac{cos^{2}\theta}{1+sin\theta}-\frac{sin^{2}\theta}{1+cos\theta}]^{2}$

   = $[\frac{1-sin^2\theta}{1+sin\theta}-\frac{1-cos^2\theta}{1+cos\theta}]^2$

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

   = $[(1-sin\theta)-(1-cos\theta)]^2$

   = $(cos\theta-sin\theta)^2$

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

   = $1-sin2\theta$

   => Ans - (B)

9. Q9.SSC 2017

   If Sec θ = 13/12, then what is the value of Sin θ?

   • 5/13

   • 12/5

   • 12/13

   • 5/12

   Show solution

    Sec θ = 13/12

   cos θ = 12/13

   sin θ = 5/13  (13²-12²=5²)

   So the answer is option A.

10. Q10.SSC 2016

    If $\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3$ then the value of $\sin^4\theta-\cos^4\theta$ is 

    • 4/3

    • 3/4

    • 5/3

    • 3/5

    Show solution

    Given : $\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3$

    => $sin\theta+cos\theta=3sin\theta-3cos\theta$

    => $3sin\theta-sin\theta=3cos\theta+cos\theta$

    => $2sin\theta=4cos\theta$

    => $\frac{sin\theta}{cos\theta}=\frac{4}{2}$

    => $tan\theta=2$

    Using, $sec^2\theta-tan^2\theta=1$

    => $sec^2\theta=1+(2)^2=5$

    $\therefore$ $cos^2\theta=\frac{1}{5}$

    Similarly, $sin^2\theta=\frac{4}{5}$

    To find : $\sin^4\theta-\cos^4\theta$

    = $(sin^2\theta-cos^2\theta)(sin^2\theta+cos^2\theta) = (sin^2\theta-cos^2\theta)$     [$\because sin^2\theta+cos^2\theta=1$]

    = $\frac{4}{5}-\frac{1}{5}=\frac{3}{5}$

    => Ans - (D)

11. Q11.SSC 2014

    If $\frac{2tan^230^{\circ}}{1-tan^230^{\circ}}+sec^{2}45^{\circ}-sec^{2}0^{\circ}$ = x sec 60°, then the value of x is

    • 2

    • 1

    • 0

    • -1

    Show solution

    we know ,

    tan 30 = $\frac{1}{\surd3}$

    sec 45 = $\surd2$

    sec 0 = 1

    sec 60 = 2

    using the above values in

    L.H.S :: $\frac{2tan^230^{\circ}}{1-tan^230^{\circ}}+sec^{2}45^{\circ}-sec^{2}0^{\circ}$ = $\frac{2 (\frac{1}{\surd3})^2}{1 - (\frac{1}{\surd3})^2}$ + $(\surd2)^2$ - $1^2$
    

    = 1 + 2 - 1 = 2

    it is given that ,

    2 = x sec 60 = 2x

    x = 1

12. Q12.SSC 2013

    The numerical value of $\frac{1}{1+cot^2 θ} + \frac{3}{1+tan^2 θ} + 2sin^2 θ$ is

    • 2

    • 5

    • 6

    • 3

    Show solution

    Expression : $\frac{1}{1+cot^2 θ} + \frac{3}{1+tan^2 θ} + 2sin^2 θ$

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

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

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

    = $3 (cos^2 \theta + sin^2 \theta) = 3$

SSC CGL Trigonometry previous year questions with solutions — AthenaPrep