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