Question

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

Answer 1

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

Put $x=\cos 2\theta ........................\left(1\right)$

Then,

$\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\right\}$

$=\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-(1-2{\sin }^2\theta )}{1+(2{\cos }^2\theta -1)}}\right\}$

$=\sin \left\{2{\tan }^{-1}\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}\right\}$

$=\sin \left\{2{\tan }^{-1}\sqrt{{\tan }^2\theta }\right\}$

$=\sin \left(2{\tan }^{-1}\tan \theta \right)$

$=\sin 2\theta$

By equation (1)

$x=\cos 2\theta$

$2\theta ={\cos }^{-1}x$

Put this value in $\sin 2\theta$

$=\sin \left({\cos }^{-1}x\right)$

$=\sin {\sin }^{-1}\sqrt{1-x^2}$ Since $({\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2})$

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

Hence, the value is $\sqrt{1-x^2}$.