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

माना \(\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\), जहाँ \(|x|<\frac{1}{\sqrt{3}}\) है, तो \(y\) का एक मान है

  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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