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JEE Advanced 2023 · Paper 2

JEE Advanced 2023 Mathematics question on Inverse Trigonometric Functions

For any y∈Ry\in\mathbb{R}, let cot⁡−1(y)∈(0,π)\cot^{-1}(y)\in(0,\pi) and tan⁡−1(y)∈(−π2,π2)\tan^{-1}(y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right). Then the sum of all the solutions of the equation tan⁡−1 ⁣(6y9−y2)+cot⁡−1 ⁣(9−y26y)=2π3\tan^{-1}\!\left(\dfrac{6y}{9-y^2}\right)+\cot^{-1}\!\left(\dfrac{9-y^2}{6y}\right)=\dfrac{2\pi}{3} for 0<∣y∣<30<|y|<3, is equal to
  1. A.
    23−32\sqrt{3}-3
  2. B.
    3−233-2\sqrt{3}
  3. C.
    43−64\sqrt{3}-6
  4. D.
    6−436-4\sqrt{3}
Show answer

Correct option: C

[JEE Advanced 2023 · Paper 2]

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