Question

Prove that $\sin \left[2{tan}^{-1}\left\{\sqrt{\frac{1-x}{1+x}}\right\}\right]=\sqrt{{1-x}^2}$.

Answer 1

$\sin \left[2{\tan }^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]=\sqrt{1-x^2}$

$\Rightarrow {\sin }^{-1}\left[{\tan }^{-1}\left(\frac{2\sqrt{\frac{1-x}{1+x}}}{1-\left(\frac{1-x}{1+x}\right)}\right)\right]\Rightarrow {\sin }^{-1}\left[{\tan }^{-1}\left(\frac{2\sqrt{\frac{1-x}{1+x}}}{1-\left(\frac{1+x-1+x}{(1+x)}\right)}\right)\right]$

$\Rightarrow \sin \left({\tan }^{-1}\left(2\sqrt{\frac{1-x}{1+x}}\frac{(1+x)}{x}\right)\right)$

$\Rightarrow \sin \left({\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right)$

$\Rightarrow \sin \left({\sin }^{-1}\frac{\sqrt{1-x^2}}{1}\right)$

$\Rightarrow \sqrt{1-x^2}$