A) \[1\]
B) \[-1\]
C) \[0\]
D) None of the above
Correct Answer: A
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 \[0\le x<1\] \[=\sin \frac{\pi }{2}=1\]
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