Question

Show 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

L.H.S.

${\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$

Put $x=\cos 2\theta$

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

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

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

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

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

On dividing numerator and denominator by $\cos \theta$

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

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

$\Rightarrow {\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }\right)$

$\Rightarrow {\tan }^{-1}\tan (\frac{\pi }{4}-\theta )$

$\Rightarrow \frac{\pi }{4}-\theta$

$\Rightarrow \frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$

Hence, this is the answer.