Inverse Trigonometric Functions JEE Main PYQs (2020–2026)

Solved Inverse Trigonometric Functions PYQs

3 recent questions with answers. The full set of 79 is free to practise.

1. JEE Main 2026MediumBasic Properties and Identity Conversions
   Q1

   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 :

   • ABoth Statement I and Statement II are true
   • BBoth Statement I and Statement II are false
   • CStatement I is true but Statement II is false
   • DStatement I is false but
     
     Statement II
     
     is true

   Show answer & solution

   Correct answer: (A)

   $$\frac{1}{2}<\frac{6}{11}<\frac{1}{\sqrt{2}}$$
   
   $\Rightarrow {\sin }^{-1}\left(\frac{1}{2}\right)<{\sin }^{-1}\left(\frac{6}{11}\right)<{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$ ….(i)
   
   $$\Rightarrow 3{\sin }^{-1}\left(\frac{1}{2}\right)<3{\sin }^{-1}\left(\frac{6}{11}\right)$$
   
   $$<3{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)\Rightarrow \frac{\pi }{2}<\alpha <\frac{3\pi }{4}$$
   
   $$\Rightarrow cos\alpha <0\dots \text{(ii)}$$
   
   $$\frac{4}{9}<\frac{1}{2}<\frac{1}{\sqrt{2}}\Rightarrow {\cos }^{-1}\left(\frac{4}{9}\right)>\frac{\pi }{3}>\frac{\pi }{4}$$
   
   $$\Rightarrow 3{\cos }^{-1}\left(\frac{4}{9}\right)>\pi >\frac{3\pi }{4}\Rightarrow \beta >\pi \dots \dots (ii)$$
   
   Inequalities $(i)$ and $(i\mathbf{i})$ shows the mathematical logic for $\alpha +\beta >\frac{\pi }{2}+\pi$
   
   $\Rightarrow cos(\alpha +\beta )>0\dots \dots$ (iii)
   
   From equations (i) and (iii) we can conclude :
   
   $$\begin{array}{ccc}\text{ Statement }& \text{ I }& \text{ II }\\ \text{ True }& \text{ Yes }& \text{ Yes }\\ & & \end{array}$$
2. JEE Main 2026EasyEquations and Inequalities involving Inverse Trigonometric Functions
   Q2

   The number of solutions of ${\tan }^{-1}4x+{\tan }^{-1}6x=\frac{\pi }{6}$, where $-\frac{1}{2\sqrt{6}}<x<\frac{1}{2\sqrt{6}}$, is equal to :

   • A2
   • B0
   • C3
   • D1

   Show answer & solution

   Correct answer: (D)

   $$\begin{array}{rl}& {\tan }^{-1}4x+{\tan }^{-1}6x=\frac{\pi }{6}\\ & \begin{array}{rl}& {\tan }^{-1}\left(\frac{10x}{1-24x^2}\right)=\frac{\pi }{6}\\ & 24x^2+10\sqrt{3}x-1=0\\ & x=\frac{-5\sqrt{3}\pm 3\sqrt{14}}{24}\end{array}\end{array}$$
3. JEE Main 2026MediumDomain, Range and Principal Value Branches
   Q3

   Let the maximum value of ${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2$ for $x\in \left[-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\right]$ be $\frac{m}{n}{\pi }^2$ , where $gcd(m,n)=1$ . Then $m+n$ is equal to $\_\_\_\_$ 。

   Numerical answer type

   Show answer & solution

   Correct answer: 65

   $$f(x)={\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2,x\in \left[-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\right]$$
   
   For $x\in [-1,1]$ ,
   
   $${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}.$$
   
   Let $a={\sin }^{-1}x$ . Then ${\cos }^{-1}x=\frac{\pi }{2}-a$ , and
   
   $$f=a^2+{\left(\frac{\pi }{2}-a\right)}^2=2a^2-\pi a+\frac{{\pi }^2}{4}.$$
   
   This is a quadratic in $a$ opening upward, so its maximum on the allowed interval occurs at an endpoint of $a$ .
   
   Now,
   
   $$x=-\frac{\sqrt{3}}{2}\Rightarrow a=-\frac{\pi }{3},x=\frac{1}{\sqrt{2}}\Rightarrow a=\frac{\pi }{4}.$$
   
   Evaluate $f$ :
   
   At $a=\frac{\pi }{4}$ : this is the vertex, giving the minimum.
   
   At $a=-\frac{\pi }{3}$ :
   
   $$f={\left(-\frac{\pi }{3}\right)}^2+{\left(\frac{5\pi }{6}\right)}^2$$
   
   $$=\frac{{\pi }^2}{9}+\frac{25{\pi }^2}{36}=\frac{29{\pi }^2}{36}.$$
   
   Hence the maximum value is
   
   $$\frac{m}{n}{\pi }^2=\frac{29}{36}{\pi }^2,$$
   
   with $gcd(29,36)=1$ .
   
   $$\boxed{m+n=65}$$

Topic-wise question count

Where JEE Main has asked from Inverse Trigonometric Functions, most frequent first.

• Equations and Inequalities involving Inverse Trigonometric Functions24
• Domain, Range and Principal Value Branches20
• Sum and Difference of Angles (tan^-1 x +/- tan^-1 y)16
• Basic Properties and Identity Conversions10
• Summation of Finite and Infinite Inverse Trigonometric Series4
• Multiple Angle Formulas (2 tan^-1 x, 3 sin^-1 x, etc.)4

Year-wise Inverse Trigonometric Functions questions in JEE Main

Number of questions from this chapter in each year's paper(s).