$\begin{align}  & \text{Given }2{{\sin }^{-1}}x={{\tan }^{-1\grave{\ }}}y \\ & \text{Let }{{\sin }^{-1}}x=\theta  \\ & \sin \theta =x,\text{ }\cos \theta =\sqrt{1-{{x}^{2}}},\text{ }\tan \theta =\frac{x}{\sqrt{1-{{x}^{2}}}} \\ & \theta ={{\sin }^{-1}}x={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}={{\tan }^{-1}}\left( \frac{x}{\sqrt{1-{{x}^{2}}}} \right) \\ & \text{Also let }{{\tan }^{-1}}y=\beta ,\text{ }\tan \beta =y \\ & \sin \beta =\frac{y}{\sqrt{1+{{y}^{2}}}},\text{ cos}\beta \text{=}\frac{1}{\sqrt{1+{{y}^{2}}}} \\ & \beta ={{\sin }^{-1}}\left( \frac{y}{\sqrt{1+{{y}^{2}}}} \right)={{\cos }^{-1}}\left( \frac{1}{\sqrt{1+{{y}^{2}}}} \right)={{\tan }^{-1}}y \\ & \text{Apply sine of function to both sides of the given equation} \\ & \text{sin(2si}{{\text{n}}^{-1}}x)=\sin ({{\tan }^{-1}}y) \\ & 2\sin ({{\sin }^{-1}}x)\cos ({{\sin }^{-1}}x)=\sin \left( {{\sin }^{-1}}\frac{y}{\sqrt{1+{{y}^{2}}}} \right) \\ & 2x\sin \cos ({{\cos }^{-1}}\sqrt{1-{{x}^{2}}})=\frac{y}{\sqrt{1+{{y}^{2}}}} \\ & 2x\sqrt{1-{{x}^{2}}}=\frac{y}{\sqrt{1+{{y}^{2}}}} \\ & \text{Square both sides} \\ & 4{{x}^{2}}(1-{{x}^{2}})=\frac{{{y}^{2}}}{1+{{y}^{2}}} \\ & 4{{x}^{2}}(1-{{x}^{2}})(1+{{y}^{2}})={{y}^{2}} \\ & 4{{x}^{2}}(1-{{x}^{2}})+4{{x}^{2}}(1-{{x}^{2}}){{y}^{2}}={{y}^{2}} \\ & 4{{x}^{2}}(1-{{x}^{2}})={{y}^{2}}[4{{x}^{2}}(1-{{x}^{2}})-1] \\ & 4{{x}^{2}}(1-{{x}^{2}})={{y}^{2}}[4{{x}^{2}}-4{{x}^{4}}-1] \\ & 4{{x}^{2}}(1-{{x}^{2}})={{y}^{2}}{{(1-2{{x}^{2}})}^{2}} \\ & \text{Find the square root of both sides} \\ & 2x\sqrt{1-{{x}^{2}}}=y(1-2{{x}^{2}}) \\ & y(1-2{{x}^{2}})=2x\sqrt{1-{{x}^{2}}} \\\end{align}$