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JEE Main Mathematics question on Inverse Trigonometric Functions

If tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2)\tan^{-1} y = \tan^{-1} x + \tan^{-1}\left(\dfrac{2x}{1-x^2}\right), where ∣x∣<13|x| < \dfrac{1}{\sqrt{3}}.

Then, the value of yy is
  1. A.
    3x−x31−3x2\dfrac{3x - x^3}{1 - 3x^2}
  2. B.
    3x+x21−3x2\dfrac{3x + x^2}{1 - 3x^2}
  3. C.
    3x−x31+3x2\dfrac{3x - x^3}{1 + 3x^2}
  4. D.
    3x+x31+3x2\dfrac{3x + x^3}{1 + 3x^2}
Show answer and solution

Correct option: A

Solution

Since |x| < 1/√3, we have 2tan⁻¹x is valid. Note that tan⁻¹(2x/(1-x²)) = 2tan⁻¹x (the double angle formula for arctangent, valid when |x| < 1).

So the right side becomes tan⁻¹x + 2tan⁻¹x = 3tan⁻¹x.

Thus y = tan(3tan⁻¹x). Using the triple angle formula tan(3θ) = (3tanθ - tan³θ)/(1 - 3tan²θ) with θ = tan⁻¹x gives y = (3x - x³)/(1 - 3x²).

Previous-year question, JEE Main.

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