A) \[2\sin 2x\]
B) \[\sin 2x\]
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
E) \[\cos 2x\]
Correct Answer: D
Solution :
\[y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+x}{1-x}}\] Put, \[x=\cos \theta \] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}x\] ...(i) \[y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}\] \[y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{2{{\cos }^{2}}\theta /2}{2{{\sin }^{2}}\theta /2}}\] \[y={{\sin }^{2}}{{\cot }^{-1}}(\cot \theta /2)\] \[y={{\sin }^{2}}\theta /2\] \[y=\frac{1-\cos \theta }{2}\] \[y=\frac{1-x}{2}\] [From Eq.(i)] Differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=-\frac{1}{2}\]
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