Question

$\frac{dy}{dx}$ for $y={\tan }^{-1}\left\{\sqrt{\frac{1+\cos x}{1-\cos x}}\right\}$, where $0<x<\pi$, is

• A. $-\frac{1}{2}$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

Answer: (A)

$-\frac{1}{2}$

We have,
$y={\tan }^{-1}\left\{\sqrt{\frac{1+\cos x}{1-\cos x}}\right\}$
$={\tan }^{-1}\left\{\sqrt{\frac{2{\cos }^2\frac{x}{2}}{2{\sin }^2\frac{x}{2}}}\right\}$
$={\tan }^{-1}\left|\cot \frac{x}{2}\right|={\tan }^{-1}\left(\cot \frac{x}{2}\right)$
$={\tan }^{-1}\left\{\tan (\frac{\pi }{2}-\frac{x}{2})\right\}=\frac{\pi }{2}-\frac{x}{2}$
$\therefore y=\frac{\pi }{2}-\frac{x}{2}$
By differentiating w.r. to $x$, we get
$\frac{dy}{dx}=0-\frac{1}{2}=-\frac{1}{2}$