Prove that $\;$ $2 \tan^{-1} \left(x\right) = \sin^{-1} \left(\dfrac{2 x}{1 + x^2}\right)$
Let $\;$ $\sin^{-1} \left(\dfrac{2 x}{1 + x^2}\right) = \theta$ $\;\;\; \cdots \; (1)$
Then, $\;$ $\sin \theta = \dfrac{2x}{1 + x^2}$
$\begin{aligned}
\therefore \; \cos \theta & = \sqrt{1 - \sin^2 \theta} \\\\
& = \sqrt{1 - \dfrac{4x^2}{\left(1 + x^2\right)^2}} \\\\
& = \sqrt{\dfrac{1 + x^4 - 2 x^2}{\left(1 + x^2\right)^2}} \\\\
& = \sqrt{\left(\dfrac{1 - x^2}{1 + x^2}\right)^2} = \dfrac{1 - x^2}{1 + x^2} \\\\
\therefore \; \tan \theta & = \dfrac{\sin \theta}{\cos \theta} \\\\
& = \dfrac{\dfrac{2x}{1 + x^2}}{\dfrac{1 - x^2}{1 + x^2}} = \dfrac{2x}{1 - x^2}
\end{aligned}$
$\implies$ $\theta = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) = 2 \tan^{-1} \left(x\right)$ $\;\;\; \cdots \; (2)$
$\therefore \;$ We have from equations $(1)$ and $(2)$,
$2 \tan^{-1} \left(x\right) = \sin^{-1} \left(\dfrac{2 x}{1 + x^2}\right)$ $\;\;\;$ Hence proved