Question

If $y=\log \sqrt{\frac{1+\cos x}{1-\cos x}}$, then $\frac{dy}{dx}$ equals

• A. ${\csc }^{2}x$
• B. $-{\csc }^{2}x$
• C. $-\csc x$
• D. $\csc x$

Answer 1

The correct option is C $-\csc x$

Given, $y=\log \sqrt{\frac{1+\cos x}{1-\cos x}}$
$y=\log \sqrt{\frac{2{\cos }^2\frac{x}{2}}{2{\sin }^2\frac{x}{2}}}$
$y=\log \cot \frac{x}{2}$
On differentiating with respect to $x$, we get
$\frac{dy}{dx}=\frac{1}{2}\cdot \frac{1}{\cot \frac{x}{2}}(-{\csc }^2\frac{x}{2})$
$=-\frac{1}{2}\cdot \frac{\sin x/2}{\cos x/2}\cdot \frac{1}{{\sin }^{2}x/2}$
$=-\frac{1}{2}\cdot \frac{1}{\cos \frac{x}{2}\cdot \sin \frac{x}{2}}$
$=-\frac{1}{\sin x}=-\csc x$