Question

$\frac{d}{dx}\left\{{\sin }^2\left({\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right)\right\}$ equals

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

Answer 1

Answer: (B)

$\frac{1}{2}$

Let $y=\frac{d}{dx}\left[{\sin }^2\left({\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right)\right]$
$=\frac{d}{dx}\left[\frac{1}{{\text{cosec}}^2\left\{{\cot }^2\sqrt{\frac{1-x}{1+x}}\right\}}\right]$
$=\frac{d}{dx}\left[\frac{1}{1+{\cot }^2\left\{{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}}\right]=\frac{d}{dx}\left[\frac{1}{1+\frac{1-x}{1+x}}\right]$
$=\frac{d}{dx}\left[\frac{1}{\frac{2}{1+x}}\right]=\frac{d}{dx}\left[\frac{1+x}{2}\right]$
Therefore, $\frac{dy}{dx}=\frac{1}{2}$