Trigonometry is one of the most important topics in the JEE Main and JEE Advanced Mathematics syllabus. In IIT JEE trigonometry questions based on trigonometric identities, equations, inverse trigonometric functions, and the properties of triangles are asked either directly or just hidden inside other chapters such as Calculus, Coordinate Geometry, Complex Numbers, Vectors, and 3D Geometry. So if you solve JEE Main trigonometry questions properly, it not only helps candidates score strong marks from this chapter, but it also improves their grasp of many other mathematical ideas, in a cleaner way.

In this article, students can go through the important IIT JEE trigonometry questions, topic wise weightage, JEE previous years question trends, the difficulty level, and a few JEE preparation tips too.

Practice JEE Mains Questions free PDF - JEE Main & Advanced Trigonometry Previous Year Questions

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.

Concept Name	No of Questions
Trigonometric Identities	46
Height and Distance	44
Double Angle Formula and Reduction Formula	39
Trigonometric Equations	34
Trigonometric Functions of Acute Angles	19
Complementary Angles	18
Basic relation b/w sides and angle of triangle and Sine Rule	14
General Solution of some Standard Equations (Part 2)	14
Graph of Trigonometric Function (Part 1)	14
Trigonometric Ratios of some Special Angles	13
Domain and range of Inverse Trigonometric Function (Part 1)	12
Inverse Trigonometric Function	12
Piecewise function	11
Trigonometric Ratio for Compound Angles (Part 1)	11
Sum of n-term of a GP	10
Sum-to-Product and Product-to-Sum Formulas	10
Trigonometric Ratio for Compound Angles (Part 2)	10
Algebra of Limits	9
Cosine Rule	8
Sum of angles in terms of arctan	8
Triple Angle Formula	8
Domain and range of Inverse Trigonometric Function (Part 2)	7
General Solution of some Standard Equations (Part 1)	7
Conversion of one ITF to other	6
Relating f-1(x) with f-1( -x)	6
Sum and difference of angles in terms of arctan (Part 2)	6
Trigonometric Ratio for Compound Angles (Some more Result)	6
Half Angle Formula	5
Maximum and Minimum value of Trigonometric Function	5
Total Questions	412

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.

Concept Name	JEE Main 2026 January Session	JEE Main 2026 April Session
Half-Angle Formula	1	0
Sum-to-Product and Product-to-Sum Formulas	1	0
Trigonometric Equations	1	2
General Solution of some Standard Equations (Part 1)	2	1
Inverse Trigonometric Function	2	4
Maximum and Minimum value of Trigonometric Function	2	2
Trigonometric Identities	2	1
Trigonometric Ratio for Compound Angles (Part 1)	2	2
Domain and range of Inverse Trigonometric Function (Part 1)	3	0
Principal Value of function f-1 (f (x))	0	1
Total Questions	16	13

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:

Exam	Difficulty Level	No. of Questions Asked
JEE Main Trigonometry Questions	Moderate	25-30
JEE Advanced Trigonometry Questions	Moderate to Difficult	5-6

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.

Class 11 Maths NCERT

RD Sharma Objective Maths

Cengage Maths

Arihant Skills in Maths

Problems Plus in IIT Maths

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