A) \[1\]
B) \[0\]
C) \[-1\]
D) \[\frac{\pi }{2}\]
Correct Answer: A
Solution :
\[\sin \left\{ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right\}\] Put \[x=\tan \theta ,\] \[=sin\left\{ {{\tan }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{2\,\tan \theta } \right)+{{\cos }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right) \right\}\] \[=\sin \{{{\tan }^{-1}}\,(\cot \,2\theta )+{{\cos }^{-1}}(\cos \,2\theta )\}\] \[=\sin \left\{ {{\tan }^{-1}}\tan \left( \frac{\pi }{2}-2\theta \right)+{{\cos }^{-1}}(\cos 2\theta ) \right\}\] \[=\sin \left\{ \frac{\pi }{2}-2\theta +2\theta \right\}=\sin \left( \frac{\pi }{2} \right)=1\]
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