Question

If $y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+x}{1-x}},$ then $\frac{dy}{dx}$ is equal to

• A. $2\sin 2x$
• B. $\sin 2x$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

The Correct Option is D

$y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+x}{1-x}}$ 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 E(i)] Differentiating w.r.t. $x,$ we get $\frac{dy}{dx}=-\frac{1}{2}$