Solution:

To Prove: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$
Formula Used: $\sin 2 A=2 \times \sin A \times \cos A$
Proof:
$\text { LHS }=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \cdots(1)$
Let $x=\sin A \ldots$ (2)
Substituting (2) in (1),
$\begin{array}{l}
\text { LHS }=\sin ^{-1}\left(2 \sin A \sqrt{1-\sin ^{2} A}\right) \\
=\sin ^{-1}(2 \times \sin A \times \cos A) \\
=\sin ^{-1}(\sin 2 A) \\
=2 A
\end{array}$
From $(2), A=\sin ^{-1} x$
$\begin{array}{l}
2 A=2 \sin ^{-1} x \\
=\text { RHS }
\end{array}$
Therefore, LHS = RHS
Hence proved.