A) \[-1\]
B) \[1/2\]
C) \[-1/2\]
D) \[1\]
Correct Answer: B
Solution :
Let \[y={{\sin }^{2}}\left\{ {{\cot }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}\] \[=\frac{1}{\cos e{{c}^{2}}\left\{ {{\cot }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}}\] \[=\frac{1}{1+{{\cot }^{2}}\left\{ {{\cot }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}}=\frac{1}{1+\frac{1-x}{1+x}}\] \[=\frac{1}{\frac{2}{1\_x}}=\frac{1+x}{2}\] \[\therefore \] \[\frac{dy}{dx}=\frac{1}{2}\]You need to login to perform this action.
You will be redirected in
3 sec