Prove the following trigonometric identities.

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

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

Multiplying both numerator and denominator by $(1-\sin \theta)$, we have

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

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

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

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

$=(\sec \theta-\tan \theta)^{2}$