Question

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

• A. $\frac{1}{2}{\sec }^2\frac{x}{2}$
• B. $\frac{1}{2}{\text{cosec}}^2\frac{x}{2}$
• C. ${\sec }^2\frac{x}{2}$
• D. ${\text{cosec}}^2\frac{x}{2}$

Answer 1

Answer: (A)

$\frac{1}{2}{\sec }^2\frac{x}{2}$

$y=\sqrt{\frac{1-\cos x}{1+\cos x}}$
$\Rightarrow y=\sqrt{\frac{1-\cos x}{1+\cos x}}\times \sqrt{\frac{1-\cos x}{1-\cos x}}$
$=\frac{1-\cos x}{\sqrt{1-{\cos }^{2}x}}=\frac{1-\cos x}{\sin x}=\csc x-\cot x$

$\therefore \frac{dy}{dx}=\tan \left(\frac{x}{2}\right)\csc x$
$=\frac{\sin \left(\frac{x}{2}\right)}{\cos \frac{x}{2}\left(2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right)}$
$=\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)$