PYQ Vault
EbooksTutor

Inverse Trigonometric Functions

JEE Mains Maths. Try each one first, then open its answer.

Top 10 JEE Mains PYQs to practise first

The questions and kinds of question asked most often come first. Each one says how often and in which papers.

  1. Question 1

    This type asked 10 times: 2021 Paper 1, 2021 Paper 5, 2021 Paper 7, 2021 Paper 19, 27 Jun 2022, 28 Jun 2022, 30 Jan 2023, 24 Jan 2025, 4 Apr 2025, 5 Apr 2026 S1. This one: 5 Apr 2026 S1 See all 10 of this type

    74
    If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\frac{\pi}{4} + \sum_{p = 1}^{11} \tan^{- 1}\left( \frac{2^{p - 1}}{1 + 2^{2p - 1}} \right) = \alpha, then tan⁡α\tan\alpha is equal to ____\_\_\_\_ .

  2. Question 2

    This type asked 6 times: 2021 Paper 4, 2021 Paper 12, 29 Jun 2022, 26 Jul 2022, 28 Jan 2025, 28 Jan 2026 S2. This one: 28 Jan 2026 S2 See all 6 of this type

    56
    Considering the principal values of inverse trigonometric functions, the value of the expression tan⁡(2sin⁡−1(213)−2cos⁡−1(310))\tan\left( 2\sin^{- 1}\left( \frac{2}{\sqrt{13}} \right)- 2\cos^{- 1}\left( \frac{3}{\sqrt{10}} \right) \right) is equal to :

    Tap an option to check your answer.

  3. Question 3

    This type asked 6 times: 2021 Paper 19, 2021 Paper 23, 29 Jun 2022, 28 Jul 2022, 10 Apr 2023, 28 Jan 2025. This one: 28 Jan 2025 See all 6 of this type

    140
    Let [x]\lbrack x\rbrack denote the greatest integer less than or equal to x . Then domain of f(x)=sec⁡−1(2[x]+1)f(x) =\sec^{- 1}(2\lbrack x\rbrack + 1) is :

    Tap an option to check your answer.

  4. Question 4

    Asked 4 times: 2021 Paper 1*, 27 Jun 2022*, 28 Jun 2022*, 30 Jan 2023 (* = numbers or wording changed that year) See all 4 times it was asked

    153
    Let a1=1,a2,a3,a4,….a_{1}= 1,a_{2},a_{3},a_{4},\ldots.. be consecutive natural numbers. Then tan⁡−1(11+a1a2)+tan⁡−1(11+a2a3)+⋯⋅+tan⁡−1(11+a2021a2022)\tan^{- 1}\left( \frac{1}{1 +a_{1}a_{2}} \right)+\tan^{- 1}\left( \frac{1}{1 +a_{2}a_{3}} \right)+ \cdots \cdot +\tan^{- 1}\left( \frac{1}{1 +a_{2021}a_{2022}} \right) is equal to

    Tap an option to check your answer.

  5. Question 5

    This type asked 4 times: 2021 Paper 18, 27 Jul 2022, 24 Jan 2025, 6 Apr 2026 S2. This one: 6 Apr 2026 S2 See all 4 of this type

    63
    If sin⁡(tan⁡−1(x2))=cot⁡(sin⁡−11−x2),x∈(0,1)\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 :

    Tap an option to check your answer.

  6. Question 6

    This type asked 4 times: 28 Jul 2022, 9 Apr 2024, 29 Jan 2025, 28 Jan 2026 S1. This one: 28 Jan 2026 S1 See all 4 of this type

    73
    If k=tan⁡(π4+12cos⁡−1(23))+tan⁡(12sin⁡−1(23))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(kx−1)=sin⁡−1x−cos⁡−1x\sin^{- 1}(kx - 1) = \sin^{- 1}x - \cos^{- 1}x is......... .

  7. Question 7

    This type asked 4 times: 2021 Paper 24, 27 Jun 2022, 31 Jan 2023, 31 Jan 2024. This one: 31 Jan 2024 See all 4 of this type

    166
    If a=sin⁡−1(sin⁡(5))a =\sin^{- 1}(\sin(5)) and b=cos⁡−1(cos⁡(5))b =\cos^{- 1}(\cos(5)), then a2+b2a^{2}+b^{2} is equal to

    Tap an option to check your answer.

  8. Question 8

    This type asked 3 times: 14 Jun 2022, 22 Jan 2025, 21 Jan 2026 S2. This one: 21 Jan 2026 S2 See all 3 of this type

    73
    Let the maximum value of (sin⁡−1x)2+(cos⁡−1x)2\left( \sin^{- 1}x \right)^{2} + \left( \cos^{- 1}x \right)^{2} for x∈[−32,12]x \in \left\lbrack - \frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}} \right\rbrack be mnπ2\frac{m}{n}\pi^{2}, where gcd(m,n)=1gcd(m,n) = 1. Then m+nm + n is equal to ____\_\_\_\_ .

  9. Question 9

    This type asked 3 times: 2021 Paper 20, 2021 Paper 25, 26 Jul 2022. This one: 26 Jul 2022 See all 3 of this type

    166
    If 0<x<120 < x <\frac{1}{\sqrt{2}} and sin⁡−1xα=cos⁡−1xβ\frac{\sin^{- 1}x}{\alpha}=\frac{\cos^{- 1}x}{\beta}, then a value of sin⁡(2παα+β)\sin\left( \frac{2\pi\alpha}{\alpha+\beta} \right) is

    Tap an option to check your answer.

  10. Question 10

    Asked 2 times: 27 Jan 2024*, 22 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    68
    The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{- 1}4x +\tan^{- 1}6x =\frac{\pi}{6}, where −126<x<126-\frac{1}{2\sqrt{6}}< x <\frac{1}{2\sqrt{6}} is equal to

    Tap an option to check your answer.