A) 1
B) \[-1\]
C) 0
D) \[\infty \]
Correct Answer: A
Solution :
[a] \[\because \sin \left[ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]\] \[=\sin \left[ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{2x} \right)+{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right) \right]\] \[=\sin \left[ {{\tan }^{-1}}\left\{ \frac{\frac{1-{{x}^{2}}}{2x}+\frac{2x}{1-{{x}^{2}}}}{1-\frac{1-{{x}^{2}}}{2x}\times \frac{2x}{1-{{x}^{2}}}} \right\} \right]\] \[=\sin \left[ {{\tan }^{-1}}\left\{ \frac{\frac{1-{{x}^{2}}}{2x}+\frac{2x}{1-{{x}^{2}}}}{0} \right\} \right]\] \[=sin\,\left[ ta{{n}^{-1}}(\infty ) \right]=sin\,\left[ ta{{n}^{-1}}(tan\frac{\pi }{2}) \right]=sin\frac{\pi }{2}=1\] Hence, option [a] is correct.
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