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 :
- A \(\frac{3 x-x^3}{1-3 x^2}\)
- B \(\frac{3 x+x^3}{1-3 x^2}\)
- C \(\frac{3 x-x^3}{1+3 x^2}\)
- D \(\frac{3 x+x^3}{1+3 x^2}\)
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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