<s>[INST] If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \frac{2x}{1-x^2} \right) $ where $|x| < \frac {1}{\sqrt {3}}$, then what is $y$? If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right)$ where $|x| < \dfrac {1}{\sqrt {3}}$ then find the value of $y$.
. . .
Let $\tan^{-1} y= A$
$$y=\tan A$$
$$\tan^{-1} x=B$$
$$x=\tan B$$
$$
\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right) =C
$$
$$\dfrac {2x}{1-x^2}=\tan C$$.
 [/INST] We use the identity $$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}.$$
Letting $a=\tan^{-1}(x)$ and $b=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, we have $$y=\tan(a+b)=\frac{x+\frac{2x}{1-x^2}}{1-\frac{2x^2}{1-x^2}}=\frac{3x-x^3}{1-3x^2}$$
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