# Prove that:

Question:

Prove that:

$\frac{\cos x}{1-\sin x}=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$

Solution:

$\mathrm{LHS}=\frac{\cos x}{1-\sin x}$

$=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \times \cos \frac{x}{2}} \quad\left[\because \cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}, \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right.$ and $\left.\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}=1\right]$

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

$=\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}$

On dividing the numerator and denominator by $\cos \frac{x}{2}$, we get

$=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$

$=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\mathrm{RHS}$

Hence proved.