\[\sin \left\{ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right\}\]is equal to [Kurukshetra CEE 2001]

\[\sin \left\{ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{2x} \right)+{{\

1.	\[\sin \left\{ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right\}\]is equal to [Kurukshetra CEE 2001]
A.	0
B.	1
C.	\[\sqrt{2}\]
D.	\[\frac{1}{\sqrt{2}}\]
Answer» C. \[\sqrt{2}\]

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\[\sin \left\{ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right\}\]is equal to [Kurukshetra CEE 2001]

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