Question

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

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

Answer 1

Answer: (C)

$\frac{-1}{2}$

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

Let, $\cot \theta =\sqrt{\frac{1+x}{1-x}}$

${\cot }^2\theta =\frac{1+x}{1-x}$

$\Rightarrow x=\frac{{\cot }^2\theta -1}{{\cot }^2\theta +1}=\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta +{\sin }^2\theta }=\cos 2\theta$

Now,

$y={\sin }^2\theta$

$\Rightarrow \frac{dy}{dx}=\frac{dy}{d\left(\theta \right)}\times \frac{d\left(\theta \right)}{dx}$

$\Rightarrow \frac{dy}{dx}=2\sin \theta \cos \theta \times \frac{-1}{2\sin \left(2\theta \right)}$

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