Inverse Trigonometric Functions JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Inverse Trigonometric Functions, free to read — no sign-in needed. The full chapter has 14 questions; sign in to attempt the remaining 11 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   Considering only the principal values of the inverse trigonometric functions, the value of $\cot^{-1}(\cot(-11)) + 10 \sin\left(2 \cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)\right) + 10 \sin(2 \tan^{-1}(2))$ is
   1. A.$3\pi + 7$
   2. B.$7$
   3. C.$4\pi + 7$
   4. D.$3\pi - 5$
   Show answer & solution

   Answer: (C)

   For the first term, $\cot^{-1}(\cot(-11))$: The principal value branch of $\cot^{-1}(x)$ is $(0, \pi)$. Since the cotangent function is periodic with period $\pi$, $\cot(-11) = \cot(-11 + 4\pi)$. Using $\pi \approx 3.14$, we have $4\pi \approx 12.56$, so $-11 + 4\pi \approx 1.56$, which lies in the interval $(0, \pi)$. Thus, $\cot^{-1}(\cot(-11)) = 4\pi - 11$. For the second term, $10 \sin\left(2 \cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)\right)$: Since $\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$, we get: $10 \sin\left(2 \times \dfrac{\pi}{4}\right) = 10 \sin\left(\dfrac{\pi}{2}\right) = 10(1) = 10$. For the third term, $10 \sin(2 \tan^{-1}(2))$: Let $\theta = \tan^{-1}(2) \Rightarrow \tan(\theta) = 2$. Using the multiple angle formula $\sin(2\theta) = \dfrac{2\tan(\theta)}{1 + \tan^2(\theta)}$, we obtain: $\sin(2\theta) = \dfrac{2(2)}{1 + 2^2} = \dfrac{4}{5}$. Therefore, $10 \sin(2 \tan^{-1}(2)) = 10 \times \dfrac{4}{5} = 8$. Adding the three evaluated terms together: $(4\pi - 11) + 10 + 8 = 4\pi + 7$. Answer: $4\pi + 7$
2. Q2JEE Advanced Adv 2024 (Paper 2)
   Considering only the principal values of the inverse trigonometric functions, the value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$ is
   1. A.$\frac{7}{24}$
   2. B.$\frac{-7}{24}$
   3. C.$\frac{-5}{24}$
   4. D.$\frac{5}{24}$
   Show answer & solution

   Answer: (B)

   Let $E = \tan\left(\sin^{-1}\left(\frac{3}{5}\right) - 2\cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$. Evaluate the inverse trigonometric expressions: Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$, so $\sin\alpha = \frac{3}{5}$ and $\tan\alpha = \frac{3}{4}$. Let $\beta = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)$, so $\cos\beta = \frac{2}{\sqrt{5}}$ and $\tan\beta = \frac{1}{2}$. Compute the double angle: Using $\tan(2\beta) = \frac{2\tan\beta}{1 - \tan^2\beta}$, we obtain $\tan(2\beta) = \frac{2\cdot\frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{4}{3}$. Apply the tangent subtraction formula: $E = \tan(\alpha - 2\beta) = \frac{\tan\alpha - \tan(2\beta)}{1 + \tan\alpha\tan(2\beta)} = \frac{\frac{3}{4} - \frac{4}{3}}{1 + \frac{3}{4}\cdot\frac{4}{3}} = \frac{-\frac{7}{12}}{2} = -\frac{7}{24}$. Final result: $\boxed{-\frac{7}{24}}$
3. Q3JEE Advanced Adv 2021 (Paper 1)
   For any positive integer $n$, let ${S}_{n}:(0,\infty )\rightarrow R$ be defined by ${S}_{n}\left(x\right)=\sum _{k=1}^{n}{\cot }^{-1}\left(\dfrac{1+k(k+1){x}^{2}}{x}\right)$ where for any $x\in R,{\cot }^{-1}(x)\in (0,\pi )$ and ${\tan }^{-1}(x)\in \left(-\dfrac{\pi }{2},\dfrac{\pi }{2}\right)$. Then which of the following statements is (are) TRUE ?
   1. A.${S}_{10}\left(x\right)=\dfrac{\pi }{2}-{\tan }^{-1}\left(\dfrac{1+11{x}^{2}}{10x}\right)$, for all $x\gt 0$
   2. B.$\lim _{n\rightarrow \infty }\cot \left({S}_{n}(x)\right)=x$, for all $x\gt 0$
   3. C.The equation ${S}_{3}(x)=\dfrac{\pi }{4}$ has a root in $(0,\infty )$
   4. D.$\tan \left({S}_{n}(x)\right)\leq \dfrac{1}{2}$, for all $n\geq 1$ and $x\gt 0$
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   Answer: A,B

   ${S}_{n}\left(x\right)=\sum _{k=1}^{n}{\cot }^{-1}\left(\dfrac{1+k(k+1){x}^{2}}{x}\right)$ $=\sum _{k=1}^{n}{\cot }^{-1}\left(\dfrac{1+kx(k+1)x}{(k+1)x-kx}\right)$ $=\sum _{k=1}^{n}{\tan }^{-1}\left(\dfrac{(k+1)x-kx}{1+kx(k+1)x}\right)$ $=\sum _{k=1}^{n}\left[{\tan }^{-1}(k+1)x-{\tan }^{-1}kx\right]$ $=\left[{\tan }^{-1}(n+1)x-{\tan }^{-1}nx\right]+\ldots +\left[{\tan }^{-1}3x-{\tan }^{-1}2x\right]+\left[{\tan }^{-1}2x-{\tan }^{-1}x\right]$ $={\tan }^{-1}(n+1)x-{\tan }^{-1}x$ $={\tan }^{-1}\left(\dfrac{(n+1)x-x}{1+(n+1){x}^{2}}\right)$ $={\tan }^{-1}\left(\dfrac{nx}{1+(n+1){x}^{2}}\right)$ 1. ${S}_{10}\left(x\right)={\tan }^{-1}\left(\dfrac{10x}{1+11{x}^{2}}\right)$ $=\dfrac{\pi }{2}-{\cot }^{-1}\left(\dfrac{10x}{1+11{x}^{2}}\right)$ $=\dfrac{\pi }{2}-{\tan }^{-1}\left(\dfrac{1+11{x}^{2}}{10x}\right)$ 2. $\lim _{n\rightarrow \infty }\cot \left({S}_{n}(x)\right)=\lim _{n\rightarrow \infty }\cot \left({\tan }^{-1}\left(\dfrac{nx}{1+(n+1){x}^{2}}\right)\right)$ $=\lim _{n\rightarrow \infty }\cot \left({\cot }^{-1}\left(\dfrac{1+(n+1){x}^{2}}{nx}\right)\right)$ $=\lim _{n\rightarrow \infty }\left(\dfrac{1+(n+1){x}^{2}}{nx}\right)$ $=\dfrac{{x}^{2}}{x}=x$ 3. ${S}_{3}\left(x\right)={\tan }^{-1}\left(\dfrac{3x}{1+4{x}^{2}}\right)=\dfrac{\pi }{4}$ $\Rightarrow \left(\dfrac{3x}{1+4{x}^{2}}\right)=\tan \dfrac{\pi }{4}$ $\Rightarrow \dfrac{3x}{1+4{x}^{2}}=1$ $\Rightarrow 1+4{x}^{2}=3x$ $\Rightarrow 4{x}^{2}-3x+1=0$ $D\lt 0$ Hence, no real roots. 4. $\tan \left({S}_{n}(x)\right)=\tan \left({\tan }^{-1}\left(\dfrac{nx}{1+(n+1){x}^{2}}\right)\right)=\left(\dfrac{nx}{1+(n+1){x}^{2}}\right)$ $\left(\dfrac{nx}{1+(n+1){x}^{2}}\right)\leq \dfrac{1}{2}\Rightarrow 2nx\leq 1+(n+1){x}^{2}$ $\Rightarrow 2nx\leq (n+1){x}^{2}+1$ $\Rightarrow (n+1){x}^{2}-2nx+1\geq 0\forall n\geq 1;x\gt 0$ Let, $y=(n+1){x}^{2}-2nx+1$ $D=4{n}^{2}-4(n+1)$ and $n\in N$ $D\lt 0$ for $n=1$ Hence, no solution if $n=1$.