question_answer

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

Answer:

LHS\[={{\tan }^{-\,1}}\left( \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)\] On putting\[{{x}^{2}}=\cos 2\theta ,\]we get \[{{\tan }^{-\,1}}\left( \frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }} \right)\] \[={{\tan }^{-\,1}}\left( \frac{\sqrt{2{{\cos }^{2}}\theta }+\sqrt{2{{\sin }^{2}}\theta }}{\sqrt{2{{\cos }^{2}}\theta }-\sqrt{2{{\sin }^{2}}\theta }} \right)\] \[={{\tan }^{-\,1}}\left( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } \right)\] \[={{\tan }^{-\,1}}\left( \frac{1+\tan \theta }{1-tan\theta } \right)\] [dividing numerator and denominator by \[\cos \theta \]] \[={{\tan }^{-\,1}}\left\{ \tan \left( \frac{\pi }{4}+\theta  \right) \right\}=\frac{\pi }{4}+\theta \] \[=\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}=RHS\] Hence proved.