Prove the following trigonometric identities.

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

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

Multiplying the denominator and numerator of the second term by $(1-\sin \theta)$, we have

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

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

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

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

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

$=\frac{2}{\cos \theta}$

$=2 \sec \theta$