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 $\tan^{-1} y= A$
$$y=\tan A$$
$$\tan^{-1} x=B$$
$$x=\tan B$$
$$
\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right) =C
$$
$$\dfrac {2x}{1-x^2}=\tan C$$.
 A: We know that $$tan2A= \frac{2tanA}{1-tan^2A}$$
Now, substitute $tanA=x\implies tan^{-1}x=A$
Plugging this into first equation, $tan2(tan^{-1}x)= \frac{2x}{1-x^2}$
Again taking $tan^{-1}$ on both sides we get
$$2tan^{-1}x = tan^{-1}\frac{2x}{1-x^2}$$
Plugging this into the question,
$$tan^{-1}y= tan^{-1}x + 2tan^{-1}x = 3tan^{-1}x$$
We know that $$tan3A=tan(2A+A)= \frac{tanA+tan2A}{1-tanAtan2A}$$
This can be reduced to $$tan3A=\frac{3tanA-tan^3A}{1-3tan^2A}$$
By a similiar logic as before, $$3tan^{-1}x= tan^{-1}\frac{3x-x^3}{1-3x^2}$$
Now the question reduces to
$$tan^{-1}y=tan^{-1}\frac{3x-x^2}{1-3x^2}$$
So, we conclude that $$y=\frac{3x-x^3}{1-3x^2}$$
NOTE: Since it is given that $|x|\lt\frac{1}{\sqrt3}$ we can safely apply the above formulae without adding or subtracting $\pi$,because x is well within the domain.
A: 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 $$y=\tan(a+b)=\frac{x+\frac{2x}{1-x^2}}{1-\frac{2x^2}{1-x^2}}=\frac{3x-x^3}{1-3x^2}$$
