Prove that $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$,where $|x| < \frac{1}{\sqrt{3}}$.

(A) Let $x = \tan \theta$. Then $\theta = \tan ^{-1} x$.
Consider the $R.H.S.$:
$\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right) = \tan ^{-1}\left(\frac{3 \tan \theta - \tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)$
Using the trigonometric identity $\tan 3\theta = \frac{3 \tan \theta - \tan ^{3} \theta}{1-3 \tan ^{2} \theta}$,we get:
$= \tan ^{-1}(\tan 3 \theta) = 3 \theta$
Since $3\theta = \theta + 2\theta$,we can write:
$= \tan ^{-1} x + 2 \tan ^{-1} x$
Using the formula $2 \tan ^{-1} x = \tan ^{-1} \left(\frac{2x}{1-x^2}\right)$ for $|x|<1$:
$= \tan ^{-1} x + \tan ^{-1} \left(\frac{2x}{1-x^2}\right) = L.H.S.$
Hence,the identity is proved.