Answer:
Option A
Explanation:
Given,
$\tan ^{-1}y= \tan ^{-1} x+\tan ^{-1}\left(\frac{2x}{1-x^{2}}\right)$
where $|x|<\frac{1}{\sqrt{3}}$
$\Rightarrow$ $\tan ^{-1}y=\tan ^{-1}\left\{\frac{x+\frac{2x}{1-x^{2}}}{1-x\left(\frac{2x}{1-x^{2}}\right)}\right\}$
$[\therefore \tan ^{-1}x+ \tan ^{-1} y= \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$
x>0, y>0,xy<1]
= $\tan ^{-1}\left(\frac{x-x^{3}+2x}{1-x^{2}-2x^{2}}\right)$
$\tan ^{-1}y= \tan ^{-1}\left(\frac{3x-x^{3}}{1-3x^{2}}\right)$
$\Rightarrow$ $y= \frac{3x-x^{3}}{1-3x^{2}}$
Aliter
$|x|<\frac{1}{\sqrt{3}}\Rightarrow -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$
Let $x=\tan\theta$
$\Rightarrow$ $-\frac{\pi}{6}<\theta<\frac{\pi}{6}$
$\therefore$ $\tan^{-1} y =\theta+\tan^{-1}(tan 2\theta)$
= $\theta+ 2\theta=3\theta$
$\Rightarrow$ $y=tan 3\theta$
$\Rightarrow$ $y=\frac{3 \tan\theta-\tan^{3}\theta}{1-3tan^{2}\theta}$
$\Rightarrow$ $y=\frac{3 x-x^{3}}{1-3x^{2}}$
