Question

What is the derivative of $\sqrt{\frac{1+\cos x}{1-\cos x}}$?

• A. $\frac{1}{2}{\sec }^2\frac{x}{2}$
• B. $-\frac{1}{2}{\text{cosec}}^2\left(\frac{x}{2}\right)$
• C. $-{\text{cosec}}^2\frac{x}{2}$
• D. None of the above

Answer 1

The correct option is B $-\frac{1}{2}{\text{cosec}}^2\left(\frac{x}{2}\right)$

Applying $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$, we get

$\sqrt{\frac{1+\cos x}{1-\cos x}}$

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

$=\frac{1}{2}\frac{\frac{-\sin x(1-\cos x)-\sin x(1+\cos x)}{\sqrt{1+\cos x}\sqrt{1-\cos x}}}{(1-\cos x)}$

$=\frac{1}{2}\frac{\frac{-2\sin x}{\sqrt{1-{\cos }^{2}x}}}{(1-\cos x)}$

$=\frac{1}{2}\frac{\frac{-2\sin x}{\sin x}}{(1-\cos x)}$

$=\frac{1}{(-1+\cos x)}$

from half angle formula

$\cos x=2{\cos }^2\frac{x}{2}-1$

$\cos x-1=2{\cos }^2\frac{x}{2}-2$

Hence the equation becomes

$\Rightarrow \frac{1}{2({\cos }^2\frac{x}{2}-1)}$

$=\frac{1}{2(-{\sin }^2\frac{x}{2})}$

$=-\frac{{\csc }^2\left(\frac{x}{2}\right)}{2}$