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  4. If 0 le x<1 , then sin tan -1(1-x2/2x)+ cos -1(1-x2/1+x2) is equal to

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Q. If $ 0\le x<1 $ , then $ \sin \left\{ {{\tan }^{-1}}\frac{1-{{x}^{2}}}{2x}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right\} $ is equal to

Jharkhand CECEJharkhand CECE 2015

Solution:

We know that, $ {{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}x, $
if $ -1 < x < 1 $ and $ {{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x, $
if $ 0\le x\le \infty $
$ \therefore $ $ \sin \left\{ {{\tan }^{-1}}\frac{1-{{x}^{2}}}{2x}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right\} $
$ =\sin \left\{ {{\cot }^{-1}}\frac{2x}{1-{{x}^{2}}}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right\} $
$ =\sin \left\{ \frac{\pi }{2}-{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right\} $
$ =\sin \left\{ \frac{\pi }{2}-2{{\tan }^{-1}}x+2{{\tan }^{-1}}x \right\} $ ,
if $ 0\le x<1 $
$ =\sin \frac{\pi }{2}=1 $