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CUET · MATHS · PYQ PAPER 2023

Let \(\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\). Then \(y\) is :

  1. A \(\frac{3 x-x^3}{1-3 x^2}\)
  2. B \(\frac{3 x+x^3}{1-3 x^2}\)
  3. C \(\frac{3 x-x^3}{1+3 x^2}\)
  4. D \(\frac{3 x+x^3}{1+3 x^2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{3 x-x^3}{1-3 x^2}\)

Step-by-step Solution

Detailed explanation

\( \tan ^{-1} y = \tan ^{-1}\left(\frac{x + \frac{2 x}{1-x^2}}{1 - x\left(\frac{2 x}{1-x^2}\right)}\right) \) \( \tan ^{-1} y = \tan ^{-1}\left(\frac{x(1-x^2) + 2x}{1-x^2 - 2x^2}\right) \) \( \tan ^{-1} y = \tan ^{-1}\left(\frac{x-x^3+2x}{1-3x^2}\right) \)…
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