Question

Prove that: ${\tan }^{-1}\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x,\frac{-1}{\sqrt{2}}\le x\le 1.$

Answer 1

Take LHS,

Put $x=cos2\theta$

$ta{n}^{-1}\left[\frac{\sqrt{1+cot2\theta }-\sqrt{1-cot2\theta }}{\sqrt{1+cot2\theta }+\sqrt{1-cot2\theta }}\right]$

$=ta{n}^{-1}\left[\frac{\sqrt{1+2co{s}^2\theta -1}-\sqrt{1-1+2si{n}^2\theta }}{\sqrt{1+2co{s}^2\theta -1}+\sqrt{1-1+2si{n}^2\theta }}\right]$

$ta{n}^{-1}\left[\frac{cos\theta -sin\theta }{cos\theta +sin\theta }\right]$

$=ta{n}^{-1}\left[\frac{1-tan\theta }{1+tan\theta }\right]$

$=ta{n}^{-1}\left[\frac{tan\left(\frac{\pi }{4}\right)-tan\theta }{1+tan\left(\frac{\pi }{4}\right).tan\theta }\right]$

$=ta{n}^{-1}\left[tan\right(\frac{\pi }{4}-\theta \left)\right]$

$\frac{\pi }{4}-\theta$ $as\left\{\begin{array}{c}x=cos2\theta \\ so,\theta =\frac{co{s}^{-1}}{2}\end{array}\right\}$

$=\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}x$ = RHS

Hence proved.