Prove the following trigonometric identities.

We have to prove $\frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta}$.

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

Multiplying both numerator and denominator by $(1+\cos \theta)$, we have

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

$=\frac{1-\cos ^{2} \theta}{\sin \theta(1+\cos \theta)}$

$=\frac{\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}$

$=\frac{\sin \theta}{1+\cos \theta}$