Prove that $\;$ $\sin^{-1} \left(2 x \sqrt{1 - x^2}\right) = 2 \sin^{-1}x$
Let $\;$ $x = \sin \theta$ $\implies$ $\theta = \sin^{-1} \left(x\right)$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
\text{Now, } \sin^{-1} \left(2 x \sqrt{1 - x^2}\right) & = \sin^{-1} \left(2 \sin \theta \sqrt{1 - \sin^2 \theta}\right) \\\\
& = \sin^{-1} \left(2 \sin \theta \sqrt{\cos^2 \theta}\right) \\\\
& = \sin^{-1} \left(2 \sin \theta \cos \theta\right) \\\\
& = \sin^{-1} \left(\sin 2 \theta\right) \\\\
& = 2 \theta \\\\
& = 2 \sin^{-1} \left(x\right) \;\;\; \left[\text{in vew of equation } (1)\right] \\\\
& = \text{RHS} \;\;\; \text{Hence proved}
\end{aligned}$