Question

Prove: ${\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x,-\frac{1}{\sqrt{2}}\le x\le 1$
[Hint: $putx=\cos 2\theta ]$

Answer 1

Put $x=\cos 2\theta$
Then, we have,
L.H.S. $={\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

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

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

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

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

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

$=\frac{\pi }{4}-\theta =\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x=$R.H.S.