Question

Show that ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2{\sin }^{-1}x$.

Answer 1

We have
$$\begin{gathered} \mathrm{LHS}={\sin }^{-1}\left(2x\sqrt{1-x^2}\right) \\ \mathrm{Putting}x=\sin a,\mathrm{we}\mathrm{get} \\ ={\sin }^{-1}\left(2\sin a\sqrt{1-{\sin }^{2}a}\right) \\ ={\sin }^{-1}\left(2\sin a\cos a\right) \\ ={\sin }^{-1}\left(\sin 2a\right) \\ =2a \\ =2{\sin }^{-1}x\left(\because x=\sin a\right) \end{gathered}$$