Question

Let $\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$, where $|x|<\frac{1}{\sqrt{3}}$. Then a value of $y$ is

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

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

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

(4) $\frac{3 x+x^3}{1+3 x^2}$

Answer 1

SOLUTION : $\frac{-1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}} ; x=\tan \theta \quad ; \quad \frac{-\pi}{6}<\theta<\frac{\pi}{6}$

$\tan ^{-1} y=\theta+\tan ^{-1} \tan 2 \theta=\theta+2 \theta=3 \theta \\$

$y=\tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta} ; y=\frac{3 x-x^3}{1-3 x^2} $

Above Correct Answer is Option (1)