Question

If $y=\frac{1-\cos x}{1+\cos x}$, then find the value of $\frac{dy}{dx}$.

Answer 1

Given: $y=\frac{1-\cos x}{1+\cos x}$
$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right)$
$=\frac{(1+\cos x)\frac{d}{dx}(1-\cos x)-(1-\cos x)\frac{d}{dx}(1+\cos x)}{{(1+\cos x)}^2}$
$=\frac{(1+\cos x)\left(\sin x\right)+(1-\cos x)\left(\sin x\right)}{{(1+\cos x)}^2}=\frac{\sin x+\sin x\cos x+\sin x-\sin x\cos x}{{(1+\cos x)}^2}$
$\frac{dy}{dx}=\frac{\sin x+\sin x}{{(1+\cos x)}^2}$
$=\frac{2\times 2\sin \frac{x}{2}\cos \frac{x}{2}}{{\left(2{\cos }^2\frac{x}{2}\right)}^2}=\frac{4}{4}\times \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\times \frac{1}{{\cos }^2\frac{x}{2}}$
$=\tan \frac{x}{2}{\sec }^2\frac{x}{2}$
$\frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right)=\tan \frac{x}{2}{\sec }^2\frac{x}{2}$