Question

Prove that : $${\tan }^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2;1<x<1$$.

Answer 1

Put $$x^2=\cos 2\theta$$, we have 

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

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

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

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

$$={\tan }^{-1}[\tan (\frac{\pi }{4}+\theta )]$$

$$=\frac{\pi }{4}+\theta [\because -1<x<1\Rightarrow 0<x^2<1\Rightarrow 0<2\theta <\frac{\pi }{2}\Rightarrow 0<\theta <\frac{\pi }{4}]$$

$$=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2[\because x^2=\cos 2\theta \Rightarrow 2\theta ={\cos }^{-1}{x}^2]$$

Hence proved.