A) \[1\]
B) \[0\]
C) \[\frac{x-1}{x+1}\]
D) \[\frac{x+1}{x-1}\]
E) \[\frac{{{x}^{2}}+1}{{{x}^{2}}-1}\]
Correct Answer: B
Solution :
\[\because \]\[y={{\sec }^{-1}}\left( \frac{x+1}{x-1} \right)+{{\sin }^{-1}}\left( \frac{x-1}{x+1} \right)\] \[={{\cos }^{-1}}\left( \frac{x-1}{x+1} \right)+{{\sin }^{-1}}\left( \frac{x-1}{x+1} \right)\] \[\Rightarrow \] \[y=\frac{\pi }{2}\] \[\therefore \] \[\frac{dy}{dx}=0\]
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