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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

Suppose \({\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)\) , where \(\left| x \right| < \frac{1}{{\sqrt 3 }}\), then one of the value of \( y \) is 

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}\) \(|x|<\frac{1}{\sqrt{3}}\) \(\Rightarrow \quad \tan ^{-1} \frac{2 x}{1-x^{2}}=2 \tan ^{-1} x\) \(\Rightarrow \quad \tan ^{-1} y=\tan ^{-1} x+2 \tan ^{-1} x\) \( = 3{\tan ^{ - 1}}x\)…
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