IIT JEE Trigonometry Questions with Solutions PDF – Practice Important Problems

IIT JEE Trigonometry Last 10 Years Most Asked Topics

JEE Main trigonometry questions asked in the last 10 years included questions from different Trigonometry topics. The table below shows the topics that were asked most often in the exam.

Also Check: JEE Main Chapter-Wise Weightage

JEE Main 2026 Trigonometry Most Asked Concepts

JEE trigonometry questions asked in 2026 included questions from different trigonometry topics. The table below shows the topics that were asked most often in the JEE exam.

Also check: How to Prepare for JEE Main 2027?

Types of IIT JEE Trigonometry Questions Asked in Exam

Different types of JEE Main mathematics trigonometry questions were asked from this chapter. Refer to the types of JEE trigonometry questions given below:

1. Questions based on formulas

2. Questions on trigonometric equations

3. Questions on multiple and compound angles

4. Questions on Inverse trigonometric functions, kind of directly

5. Mixed questions involving calculus and a few variants

6. Solving questions by using those identities

7. Coordinate Geometry applications

8. Complex Number applications

Difficulty Level of JEE Trigonometry Questions

JEE Main usually goes with direct formula based or identity-kind of Trigonometry JEE questions, while JEE Advanced leans more towards the conceptual grasp and multi concept applications type questions:

Also Check: JEE Main 2027 Important Formulas

IIT JEE Trigonometry Previous Year Questions

Practising Trigonometry JEE Mains PYQ is one of the best way to prepare for IIT JEE. The questions below will help you get a feel for the exam pattern, the key topics and also the general level of difficulty they are asking in Trigonometry.

Question 1. Let $S=\{\theta \in(-2 \pi, 2 \pi): \cos \theta+1=\sqrt{3} \sin \theta\}$.

Then $\sum_{\theta \in \mathrm{S}} \theta$ is equal to:

(1) $-\frac{2 \pi}{3}$

(2) $-\frac{4 \pi}{3}$

(3) $\frac{2 \pi}{3}$

(4) $\frac{4 \pi}{3}$

Solution: Option (2)

$\begin{aligned} & \cos \theta+1=\sqrt{3} \sin \theta \\ & \frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}+1=\sqrt{3}\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right) \\ & 2=2 \sqrt{3} \tan \frac{\theta}{2} \\ & \Rightarrow \tan \frac{\theta}{2}=\frac{1}{\sqrt{3}} \quad \theta \in(-2 \pi, 2 \pi), \frac{\theta}{2} \in(-\pi, \pi) \\ & \frac{\theta}{2}=-\frac{5 \pi}{6}, \frac{\pi}{6} \\ & \theta=\frac{-5 \pi}{3}, \frac{\pi}{3} \\ & \text { Sum }=-\frac{-5 \pi}{3}+\frac{\pi}{3}=\frac{-4 \pi}{3}\end{aligned}$

Question 2. Let $0<\alpha<1, \beta=\frac{1}{3 \alpha}$ and $\tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta) =\frac{\pi}{4}$. Then $6(\alpha+\beta)$ is equal to:

(1) 6

(2) 7

(3) 8

(4) 9

Solution: Option (2)

$\begin{aligned} & \beta=\frac{1}{3 \alpha}, \tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)=\frac{\pi}{4} \\ & \Rightarrow \tan \left(\tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)\right)=1 \\ & \Rightarrow \frac{1-\alpha+1-\beta}{1-(1-\alpha)(1-\beta)}=1 \\ & \Rightarrow 2-\alpha-\beta=1-(1-\alpha-\beta+\alpha \beta) \\ & \Rightarrow 2-\alpha-\beta=\alpha+\beta-\alpha \beta \\ & \Rightarrow 2(\alpha+\beta)=2+\alpha \beta \\ & \Rightarrow 2(\alpha+\beta)=2+\frac{1}{3}=\frac{7}{3} \\ & \alpha+\beta=\frac{7}{6} \\ & 6(\alpha+\beta)=7\end{aligned}$

Question 3. If $\sin \left(\tan ^{-1}(x \sqrt{2})\right)=\cot \left(\sin ^{-1} \sqrt{1-x^2}\right), x \in(0,1)$, then the value of $x$ is :

(1) $\frac{1}{2}$

(2) $\frac{1}{3}$

(3) $\frac{2}{3}$

(4) $\frac{5}{8}$

Solution: Option (1)

$\begin{aligned} & \frac{x \sqrt{2}}{\sqrt{2 x^2+1}}=\frac{1}{\sqrt{1-x^2}} \\ & 2\left(1-x^2\right)=\left(2 x^2+1\right) \\ & 2-2 x^2=2 x^2+1 \\ & x^2=\frac{1}{4} \\ & x=\frac{1}{2}\end{aligned}$

Question 4. Let $\alpha=3 \sin ^{-1}\left(\frac{6}{11}\right)$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$, where inverse trigonometric functions take only the principal values.

Given below are two statements:

Statement I : $\cos (\alpha+\beta)>0$

Statement II : $\cos (\alpha)<0$

In the light of the above statements, choose the correct answer from the options given below:

(1) Both Statement I and Statement II are true

(2) Both Statement I and Statement II are false

(3) Statement I is true but Statement II is false

(4) Statement I is false but Statement II is true

Solution: Option (1)

$\begin{aligned} & \frac{1}{2}<\frac{6}{11}<\frac{1}{\sqrt{2}} \\ & \sin ^{-1}\left(\frac{1}{2}\right)<\sin ^{-1}\left(\frac{6}{11}\right)<\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ & \frac{\pi}{6}<3 \sin ^{-1}\left(\frac{6}{11}\right)<\frac{\pi}{4} \\ & \frac{\pi}{2}<\alpha<\frac{3 \pi}{4} \quad \therefore \cos \alpha<0 \\ & 0<\frac{4}{9}<\frac{1}{2} \\ & \frac{\pi}{3}<\cos ^{-1}\left(\frac{4}{9}\right)<\frac{\pi}{2} \\ & \pi<3 \cos ^{-1}\left(\frac{4}{9}\right)<\frac{3 \pi}{2} \\ & \pi<\beta<\frac{3 \pi}{2} \\ & \text { Now } \frac{3 \pi}{2}<\alpha+\beta<\frac{9 \pi}{4} \\ & \therefore \cos (\alpha+\beta)>0\end{aligned}$

Question 5: If $S=\left\{\theta \in[-\pi, \pi]: \cos \theta \cos \frac{5 \theta}{2}=\cos 7 \theta \cos \frac{7 \theta}{2}\right\}$, then $\mathrm{n}(\mathrm{S})$ is equal to $\_\_\_\_$

(1) 19

(2) 17

(3) 15

(4) 14

Solution: Option (1)

$\begin{aligned} & \cos \theta \cos \frac{5 \theta}{2}=\cos 7 \theta \cos \frac{7 \theta}{2} \\ & \cos \frac{7 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{21 \theta}{2}+\cos \frac{7 \theta}{2} \\ & \cos \frac{21 \theta}{2}-\cos \frac{3 \theta}{2}=0 \\ & -2 \sin 6 \theta \sin \frac{9 \theta}{2}=0 \\ & \sin 6 \theta=0 \\ & \theta=0, \pm \frac{\pi}{6}, \pm \frac{2 \pi}{6}, \ldots \pm \frac{5 \pi}{6}, \pm \pi \quad(13 \text { solutions }) \\ & \sin \frac{9 \theta}{2}=0 \\ & \theta= \pm \frac{2 \pi}{9}, \pm \frac{4 \pi}{9}, \pm \frac{8 \pi}{9} \quad(6 \text { more solutions }) \\ & \text { Total }=19 \text { solutions }\end{aligned}$

Question 6: The sum of all the integral values of $p$ such that the equation $3 \sin ^2 x+12 \cos x-3=p, x \in R$, has at least one solution, is :

(1) -54

(2) -60

(3) -75

(4) -84

Solution: Option (3)

$\begin{aligned} & P=12 \cos x-3 \cos ^2 x \\ & P=-3\left(\cos ^2 x-4 \cos x\right) \\ & P=-3\left((\cos x-2)^2-2\right) \\ & \text { put } \cos x=-1 \Rightarrow P=-15 \\ & \text { put } \cos x=1 \Rightarrow P=9 \\ & -15 \leq P \leq 9 \\ & \text { Sum of all integers }=-(10+11+\ldots+15)=-75\end{aligned}$

Question 7: Let $\tan A, \tan B$, where $A, B \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, be the roots of the quadratic equation $x^2-2 x-5=0$.

Then $20 \sin ^2\left(\frac{A+B}{2}\right)$ is equal to :

(1) $10+\sqrt{10}$

(2) $10-2 \sqrt{10}$

(3) $10-3 \sqrt{10}$

(4) $10-\sqrt{10}$

Solution: Option (3)

$\begin{aligned} & \mathrm{x}^2-2 \mathrm{x}-5=0 \\ & \tan \mathrm{~A}+\tan \mathrm{B}=2 ; \tan \mathrm{Atan} \mathrm{B}=-5 \\ & \therefore \tan (\mathrm{~A}+\mathrm{B})=\frac{2}{1-(-5)}=\frac{1}{3} \\ & \Rightarrow \cos (\mathrm{~A}+\mathrm{B})=\frac{3}{\sqrt{10}} \\ & \therefore 20\left(\sin ^2\left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right)\right)=\frac{10}{2}(1-\cos (\mathrm{A}+\mathrm{B})) \\ & =10\left(1-\frac{3}{\sqrt{10}}\right)=(10-3 \sqrt{10})\end{aligned}$

Question 8: $\max _{0 \leq x \leq \pi}\left(16 \sin \left(\frac{x}{2}\right) \cos ^3\left(\frac{x}{2}\right)\right)$ is equal to:

(1) $\frac{3 \sqrt{3}}{2}$

(2) $3 \sqrt{3}$

(3) $4 \sqrt{3}$

(4) $6 \sqrt{3}$

Solution: Option (2)

$\begin{aligned} & \mathrm{E}=16 \sin \frac{\mathrm{x}}{2} \cos ^3 \frac{\mathrm{x}}{2} \\ & \mathrm{E}=4 \sin \mathrm{x}[1+\cos \mathrm{x}] \\ & \frac{\mathrm{dE}}{\mathrm{dx}}=4[\cos \mathrm{x}+\cos 2 \mathrm{x}] \\ & =8 \cos \frac{3 \mathrm{x}}{2} \cos \frac{\mathrm{x}}{2}=0 \\ & \Rightarrow \cos \frac{3 \mathrm{x}}{2}=0 \text { or } \cos \frac{\mathrm{x}}{2}=0 \\ & \Rightarrow \mathrm{x}=\left\{\frac{\pi}{3}, \pi\right\} \text { are critical points of the function } \\ & \mathrm{E}(0)=0 \\ & \mathrm{E}(\pi)=0 \\ & \mathrm{E}\left(\frac{\pi}{3}\right)=3 \sqrt{3} \\ & \therefore \text { maximum value of } \mathrm{E}=3 \sqrt{3}\end{aligned}$

Question 9: If $\mathrm{A}=\frac{\sin 3^{\circ}}{\cos 9^{\circ}}+\frac{\sin 9^{\circ}}{\cos 27^{\circ}}+\frac{\sin 27^{\circ}}{\cos 81^{\circ}}$ and $\mathrm{B}=\tan 81^{\circ}- \tan 3^{\circ}$, then $\frac{\mathrm{B}}{\mathrm{A}}$ is equal to________.

(1) 2

(2) 3

(3) 1

(4) 4

Solution: Option (1)

$\begin{aligned} & \text { Consider } \mathrm{E}=\frac{\sin \theta}{\cos 3 \theta} \Rightarrow \mathrm{E}=\frac{2 \sin \theta \cos \theta}{2 \cos 3 \theta \cos \theta} \\ & \Rightarrow \mathrm{E}=\frac{\sin 2 \theta}{2 \cos 3 \theta \cos \theta} \Rightarrow \mathrm{E}=\frac{\sin (30-\theta)}{2 \cos 3 \theta \cos \theta} \\ & \Rightarrow \mathrm{E}=\frac{1}{2}[\tan 3 \theta-\tan \theta] \\ & \mathrm{A}=\frac{1}{2}\left[\tan 9^{\circ}-\tan 3^{\circ}+\tan 27^{\circ}-\tan 9^{\circ}+\tan 81^{\circ}-\tan 27^{\circ}\right] \\ & \therefore \mathrm{A}=\frac{1}{2}\left[\tan 81^{\circ}-\tan 3^{\circ}\right] \\ & \therefore \frac{\mathrm{B}}{\mathrm{A}}=2\end{aligned}$

Question 10: Let $P=\left\{\theta \in[0,4 \pi]: \tan ^2 \theta \neq 1\right\}$ and $S=\left\{a \in Z: 2\left(\cos ^8 \theta-\sin ^8 \theta\right) \sec 2 \theta=a^2, \theta \in P\right\}$. Then $\mathrm{n}(\mathrm{S})$ is :

(1) 0

(2) 1

(3) 2

(4) 4

Solution: Option (1)

$\begin{aligned} & 2\left(\cos ^8 \theta-\sin ^8 \theta\right) \sec 2 \theta=\mathrm{a}^2 \\ & 2\left(\cos ^4 \theta+\sin ^4 \theta\right)\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^2 \theta-\sin ^2 \theta\right) \sec 2 \theta=\mathrm{a}^2 \\ & 2\left(\cos ^4 \theta+\sin ^4 \theta\right)=\mathrm{a}^2 \\ & 2\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)=\mathrm{a}^2 \\ & 2\left(1-\frac{\sin ^2 2 \theta}{2}\right)=\mathrm{a}^2, \mathrm{a} \in \mathrm{Z} \\ & \mathrm{a}^2=2-\sin ^2 2 \theta \in[1,2] \\ & \mathrm{a}^2=1 \text { at } \sin ^2 2 \theta=1 \\ & 2 \theta=(2 \mathrm{n}+1) \frac{\pi}{2} \\ & \theta=(2 \mathrm{n}+1) \frac{\pi}{4} \notin \mathrm{P} \\ & \mathrm{n}(\mathrm{S})=0\end{aligned}$

Question 11. Let $S=\{x \in[-\pi, \pi]: \sin x(\sin x+\cos x)=a$, $\mathrm{a} \in \mathbf{Z}\}$. Then $\mathrm{n}(\mathrm{S})$ is equal to :

(1) 3

(2) 6

(3) 7

(4) 9

Solution: Option (4)

$\sin x(\sin x+\cos x) \in\left[\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$

2 integer will be there $\Rightarrow \mathrm{a}=0,1$
If $\mathrm{a}=0 \quad \sin \mathrm{x}(\sin \mathrm{x}+\cos \mathrm{x})=0$
$\Rightarrow \sin x=0 \quad$ or $\quad \sin x+\cos x=0$

3 solutions $x=-\frac{\pi}{4}, \frac{3 \pi}{4} \quad 2$ solutions

Total 9 solution

Question 12: If $\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right)=K$, then the value of $\sin \left(\frac{10 \mathrm{~K} \pi}{3}\right)$ is :

(1) $\frac{\sqrt{3}+1}{2 \sqrt{2}}$

(2) $\frac{\sqrt{3}-1}{\sqrt{2}}$

(3) $\frac{\sqrt{3}}{2}$

(4) $\frac{1}{2}$

Solution: Option (1)

$\begin{aligned} & \mathrm{K}=\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} \\ & =\frac{1}{4} \sin 30^{\circ}=\frac{1}{8} \\ & \sin 10 \mathrm{~K} \frac{\pi}{3}=\sin \left(10 \times \frac{1}{8} \cdot \frac{\pi}{3}\right)=\sin \frac{5 \pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\end{aligned}$

Question 13: Considering the principal values of inverse trigonometric functions, the value of the expression $\tan \left(2 \sin ^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos ^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to:

(1) $-\frac{33}{56}$

(2) $\frac{16}{63}$

(3) $\frac{33}{56}$

(4) $-\frac{16}{63}$

Solution: Option (3)

$\operatorname{Sin}^{-1} \frac{2}{\sqrt{13}}=\theta, \operatorname{Cos}^{-1} \frac{3}{\sqrt{10}}=\phi, \tan (2 \theta-2 \phi)=\frac{33}{56}$

Question 14: If $k=\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{2}{3}\right)\right)+\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin ^{-1}(k x-1)=\sin ^{-1} x-\cos ^{-1} x$ is $\_\_\_\_$

(1) 1

(2) 2

(3) 0

(4) 3

Solution: Option (1)

$K=\tan \theta+\cot \theta=\frac{1}{\sin \theta \cos \theta}=\frac{2}{\sin 2 \theta}$

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

$K=\frac{2}{\frac{2}{3}}=3$

$\begin{aligned} & \sin ^{-1}(3 x-1)=\sin ^{-1} x-\cos ^{-1} x \\ & \sin ^{-1}(3 x-1)=\frac{\pi}{2}-2 \cos ^{-1} x \\ & 3 x-1=\sin ^{-1}\left(\frac{\pi}{2}-2 \cos ^{-1} x\right) \\ & 3 x-1=2 x^2-1 \Rightarrow x=0, \frac{3}{2} \\ & x=\frac{3}{2}(\text { Rejected }) \\ & x=0 \text { only } \\ & \text { No of solution }=1\end{aligned}$

Question 15: The number of elements in the set $\left\{x \in\left[0,180^{\circ}\right]: \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)\right\}$ is $\_\_\_\_$

(1) 4

(2) 3

(3) 2

(4) 1

Solution: Option (1)

$\begin{aligned}
& \tan (x+100)=\tan \left(x+50^0\right) \tan x \tan (x-50) \\
& \Rightarrow \frac{\tan (x+100)}{\tan x}=\tan (x+50) \tan (x-50) \\
& \Rightarrow \frac{\sin (x+100) \cos x}{\cos (x+100) \sin x}=\frac{\sin \left(x+50^0\right) \sin \left(x-50^0\right)}{\cos \left(x+50^0\right) \cos \left(x-50^0\right)}
\end{aligned}$

Using Componendo and Dividendo, we get

$\begin{aligned}
& 2 \sin (2 x+100) \cos 2 x+\sin 200=0 \\
& \Rightarrow \sin (4 x+100)+\sin 100^0+\sin (200)=0 \\
& \Rightarrow \sin (4 x+100)=-2 \sin 150 \sin 50 \\
& \sin (4 x+100)=-\sin (50)=\sin \left(-50^0\right)
\end{aligned}$

$\begin{aligned}
& 4 x+100=n \pi+(-1)^n(-50) \\
& x=\frac{n \pi+(-1)^{n+1} 50-100^0}{4}, n \in I \\
& x=\frac{130}{4}, \frac{210}{4}, \frac{490}{4}, \frac{570}{4} \text { in }[0,180]
\end{aligned}$

Number of solutions $=4$

Best Books for JEE Main Mathematics Trigonometry Questions

While preparing for IIT JEE, it is very important to follow the best books because they help properly cover all the concepts. Refer to the books given below for solving IIT problems on trigonometry.