Question

If $y={\sin }^2\left[{\cot }^{-1}\left(\frac{1}{\sqrt{\frac{1+x}{1-x}}}\right)\right]$ then find $\frac{dy}{dx}$

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

$y={\sin }^2\left({\cot }^{-1}\left(\frac{1}{\sqrt{\frac{1+x}{1-x}}}\right)\right)$

let $x=\cos 2\theta$

$-1\le \cos 2\theta \le 1$

$\pi \le 2\theta \le \theta$

$0\le \theta \le \pi 2$

$y={\sin }^2\left({\cot }^{-1}\left(\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\right)\right)$

$\theta \in (0,\frac{\pi }{2})\{\begin{array}{c}1-\cos 2\theta =2{\sin }^2\theta \\ 1+\cos 2\theta =2{\cos }^2\theta \end{array}$

$={\sin }^2\left({\cot }^{-1}\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}\right)={\sin }^2{\cot }^{-1}\left(\tan \theta \right)$

$={\sin }^2{\cot }^{-1}\left(\cot (\frac{\pi }{2}-0)\right)$

$={\sin }^2(\frac{\pi }{2}-0)={\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

$=\frac{1-x}{2}$

$\frac{dy}{dx}=0-\frac{1}{2}=\frac{-1}{2}$