Prove that: $\tan (2\tan^{-1} (x))=2\tan (\tan^{-1} (x)+\tan^{-1} (x^3))$

Question

Prove that: $\tan (2\tan^{-1} (x))=2\tan (\tan^{-1} (x)+\tan^{-1} (x^3))$

My Attempt' Let $\tan^{-1} (x)=A$ $$x=\tan (A)$$ Now, L.H.S.$=\tan (2\tan^{-1} (x))$ $$=\tan (2A)$$ $$=\dfrac {2\tan (A)}{1-\tan^2 (A)}$$ $$=\dfrac {2x}{1-x^2}$$

Answer 1

It is good, indeed from

$$\tan (A+B) = \frac{\tan A+\tan B}{1- \tan A \tan B} $$

for RHS we obtain

$$2\tan (\tan^{-1} (x)+\tan^{-1} (x^3))=2\frac{x+x^3}{1- x^4}=\frac{2x(1+x^2)}{(1-x^2)(1+x^2)}=\frac{2x}{1- x^2}=LHS$$

Answer 2

$$\begin{align}2\tan (\tan^{-1} (x)+\tan^{-1} (x^3)) =&2\tan \left(\tan^{-1} \dfrac{x + x^3}{1 - x^4}\right) \\= \ & 2\left( \dfrac{x + x^3}{1 - x^4}\right) = 2\left( \dfrac{x }{1 - x^2}\right) \\= \ & \dfrac{2\tan(\tan^{-1} x)}{1 - \tan^2 (\tan^{-1} x)} = \tan(2 \tan^{-1}x) \end{align}$$