Inverse Trigonometric Functions JEE Main previous year questions with solutions

5 solved JEE Main questions on Inverse Trigonometric Functions, free to read — no sign-in needed. The full chapter has 91 questions; sign in to attempt the remaining 86 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Converting one Inverse T function to other Inverse T
   If $\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1-x^2})$, $x \in (0,1)$, then the value of $x$ is :
   1. A.$\dfrac{1}{2}$
   2. B.$\dfrac{1}{3}$
   3. C.$\dfrac{2}{3}$
   4. D.$\dfrac{5}{8}$
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   Answer: (A)

   Let $\alpha = \tan^{-1}(x\sqrt{2}) \Rightarrow \tan\alpha = x\sqrt{2}$ From the right-angled triangle, $\sin\alpha = \dfrac{x\sqrt{2}}{\sqrt{1+2x^2}}$ Let $\beta = \sin^{-1}\sqrt{1-x^2} \Rightarrow \sin\beta = \sqrt{1-x^2}$ From the right-angled triangle, $\cos\beta = \sqrt{1 - (1-x^2)} = x$ $\cot\beta = \dfrac{\cos\beta}{\sin\beta} = \dfrac{x}{\sqrt{1-x^2}}$ Given $\sin\alpha = \cot\beta$, substituting the values: $\dfrac{x\sqrt{2}}{\sqrt{1+2x^2}} = \dfrac{x}{\sqrt{1-x^2}}$ Since $x \in (0,1)$, $x \neq 0$. Dividing by $x$ and squaring both sides: $\dfrac{2}{1+2x^2} = \dfrac{1}{1-x^2}$ $2(1-x^2) = 1+2x^2$ $2 - 2x^2 = 1 + 2x^2$ $4x^2 = 1$ $x^2 = \dfrac{1}{4}$ Since $x \in (0,1)$, $x = \dfrac{1}{2}$ Answer: $\dfrac{1}{2}$
2. Q2JEE Main 2025 (04 Apr, Shift 2)Infinite Series
   <p>The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+$ $\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots .$ is :-</p>
   1. A.<p>$\frac{\pi}{2}+\tan ^{-1}\left(\frac{1}{2}\right)$</p>
   2. B.$\frac{\pi}{2}-\cot ^{-1}\left(\frac{1}{2}\right)$
   3. C.$\frac{\pi}{2}+\cot ^{-1}\left(\frac{1}{2}\right)$
   4. D.<p>$\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)$</p>
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   Answer: (D)

   <p>$\begin{aligned} &amp; \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{4}{4 \mathrm{n}^2+3}\right) \\ &amp; \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\left(\mathrm{n}+\frac{1}{2}\right)-\left(\mathrm{n}-\frac{1}{2}\right)}{1+\left(\mathrm{n}+\frac{1}{2}\right)\left(\mathrm{n}-\frac{1}{2}\right)}\right) \\ &amp; \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\mathrm{n}-\frac{1}{2}\right) \\ &amp; \mathrm{T}_1+\mathrm{T}_2+\ldots+\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ &amp; \mathrm{S}_{\infty}=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$ option (4)</p>
3. Q3JEE Main 2024 (29 Jan, Shift 2)Multiple Angles
   Let $x=\dfrac{m}{n}(m,n$ are co-prime natural numbers) be a solution of the equation $\cos \left(2{\sin }^{-1}x\right)=\dfrac{1}{9}$ and let $\alpha ,\beta (\alpha \gt \beta )$ be the roots of the equation $m{x}^{2}-nx-m+n=0$. Then the point $(\alpha ,\beta )$ lies on the line
   1. A.$3x+2y=2$
   2. B.$5x-8y=-9$
   3. C.$3x-2y=-2$
   4. D.$5x+8y=9$
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   Answer: (D)

   Given: $\cos \left(2{\sin }^{-1}x\right)=\dfrac{1}{9}$ Let, ${\sin }^{-1}x=\theta$ $\Rightarrow \cos \left(2\theta \right)=\dfrac{1}{9}$ $\Rightarrow 1-2{\sin }^{2}\theta =\dfrac{1}{9}$ $\Rightarrow \dfrac{8}{9}=2{\sin }^{2}\theta$ $\Rightarrow \sin \theta =\pm \dfrac{2}{3}$ $\Rightarrow x=\pm \dfrac{2}{3}$ It is given that $m$ and $n$ are co-prime natural numbers, $\Rightarrow m=2,n=3$ Also, $m{x}^{2}-nx-m+n=0$. $\Rightarrow 2{x}^{2}-3x-2+3=0$ $\Rightarrow 2{x}^{2}-3x+1=0$ $\Rightarrow 2{x}^{2}-2x-x+1=0$ $\Rightarrow 2x\left(x-1\right)-\left(x-1\right)=0$ $\Rightarrow \left(2x-1\right)\left(x-1\right)=0$ $\Rightarrow x=1,\dfrac{1}{2}$ $\Rightarrow \alpha =1,\beta =\dfrac{1}{2}$ Out of the given options, $\left(1,\dfrac{1}{2}\right)$ lines on $5x+8y=9$.
4. Q4JEE Main 2023 (13 Apr, Shift 1)Sum and difference of angles
   <p>If $S=\left\{x\in ℝ:{\sin }^{-1}\left(\dfrac{x+1}{\sqrt{{x}^{2}+2x+2}}\right)-{\sin }^{-1}\left(\dfrac{x}{\sqrt{{x}^{2}+1}}\right)=\dfrac{\pi }{4}\right\}$ then $\sum _{x\in S}\left(\sin \left(\left({x}^{2}+x+5\right)\dfrac{\pi }{2}\right)-\cos \left(\left({x}^{2}+x+5\right)\pi \right)\right)$ is equal to ______.</p>
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   Answer: 4

   <p>Given, ${\sin }^{-1}\left(\dfrac{x+1}{\sqrt{{x}^{2}+2x+2}}\right)-{\sin }^{-1}\left(\dfrac{x}{\sqrt{{x}^{2}+1}}\right)=\dfrac{\pi }{4}$ $\Rightarrow {\sin }^{-1}\left(\dfrac{x+1}{\sqrt{{x}^{2}+x+2}}\right)=\dfrac{\pi }{4}+{\sin }^{-1}\dfrac{x}{\sqrt{{x}^{2}+1}}$ $\Rightarrow \left(\dfrac{x+1}{\sqrt{{x}^{2}+x+2}}\right)=\sin \left(\dfrac{\pi }{4}+{\sin }^{-1}\dfrac{x}{\sqrt{{x}^{2}+1}}\right)$ $\Rightarrow \left(\dfrac{x+1}{\sqrt{{x}^{2}+x+2}}\right)=\dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{{x}^{2}+1}}+\dfrac{1}{\sqrt{2}}\times \dfrac{x}{\sqrt{{x}^{2}+1}}$ $\Rightarrow \dfrac{x+1}{\sqrt{{x}^{2}+x+2}}=\dfrac{x+1}{\sqrt{2}\sqrt{{x}^{2}+1}}$ $\Rightarrow \left(x+1\right)\left(\sqrt{2}\sqrt{{x}^{2}+1}-\sqrt{{x}^{2}+x+2}\right)=0$ $\Rightarrow x=-1$ or $\sqrt{{x}^{2}+x+2}=\sqrt{2}\cdot \sqrt{{x}^{2}+1}$ Now solving, $\sqrt{{x}^{2}+x+2}=\sqrt{2}\cdot \sqrt{{x}^{2}+1}$ we get, ${x}^{2}+x+2=2\left({x}^{2}+1\right)$ $\Rightarrow {x}^{2}-x=0$ $\Rightarrow x=0,x=1$ {rejected as $x=1$ will not satisfy the given equation} Hence, $S=\left\{0,1\right\}$ Now solving, $\sum _{n\in S}\left(\sin \left({x}^{2}+x+5\right)\dfrac{\pi }{2}-\cos \left({x}^{2}+x+5\right)\pi \right)$ $=\left(\sin \dfrac{5\pi }{2}-\cos 5\pi +\sin \dfrac{5\pi }{2}-\cos 5\pi \right)$ $=1-\left(-1\right)+1-\left(-1\right)=4$</p>
5. Q5JEE Main 2022 (25 Jun, Shift 2)Principal Value and Basics
   The value of ${\tan }^{-1}\left[\dfrac{\cos \left(\dfrac{15\pi }{4}\right)-1}{\sin \left(\dfrac{\pi }{4}\right)}\right]$ is equal to
   1. A.$-\dfrac{\pi }{4}$
   2. B.$-\dfrac{\pi }{8}$
   3. C.$-\dfrac{5\pi }{12}$
   4. D.$-\dfrac{4\pi }{9}$
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   Answer: (B)

   Given, ${\tan }^{-1}\left(\dfrac{\cos \left(\dfrac{15\pi }{4}\right)-1}{\sin \dfrac{\pi }{4}}\right)$ $={\tan }^{-1}\left(\dfrac{\cos \left(4\pi -\dfrac{\pi }{4}\right)-1}{\sin \dfrac{\pi }{4}}\right)$ $={\tan }^{-1}\left(\dfrac{\dfrac{1}{\sqrt{2}}-1}{\dfrac{1}{\sqrt{2}}}\right)$ $={\tan }^{-1}\left(1-\sqrt{2}\right)=-{\tan }^{-1}\left(\sqrt{2}-1\right)$ $=-\dfrac{\pi }{8}$