A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) 0
Correct Answer: B
Solution :
Putting \[x=\tan \theta \] \[{{\tan }^{-1}}\frac{1-{{x}^{2}}}{2x}+{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}\] \[={{\tan }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{2\tan \theta } \right)+{{\cos }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)\] \[={{\tan }^{-1}}(\cot 2\theta )+{{\cos }^{-1}}(\cos 2\theta )\]\[=\frac{\pi }{2}-2\theta +2\theta =\frac{\pi }{2}\].
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