\[{{\tan }^{-1}}\frac{1-{{x}^{2}}}{2x}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}=\]

\[{{\tan }^{-1}}\frac{1-{{x}^{2}}}{2x}+{{\cos }^{-1}}\frac{1-{{x}^{2}}

1.	\[{{\tan }^{-1}}\frac{1-{{x}^{2}}}{2x}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}=\]
A.	\[\frac{\pi }{4}\]
B.	\[\frac{\pi }{2}\]
C.	\[\pi \]
D.	0
Answer» C. \[\pi \]

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\[{{\tan }^{-1}}\frac{1-{{x}^{2}}}{2x}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}=\]

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