Prove the following trigonometric identities.

We have to prove $\frac{1+\sec \theta}{\sec \theta}=\frac{\sin ^{2} \theta}{1-\cos \theta}$

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

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

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

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

Multiplying the numerator and denominator by $(1-\cos \theta)$, we have

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

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

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