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