Prove the following trigonometric identities:$ \frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta $

To do:

We have to prove that \( \frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta \).

Solution:

We know that,

$\sec \theta=\frac{1}{\cos \theta}$.....(i)

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

Therefore,

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

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

$=\frac{2+2\sin \theta}{\cos \theta(1+\sin \theta)}$             [From (ii)]                

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

$=\frac{2}{\cos \theta}$           

$=2 \sec \theta$                       [From (i)]

Hence proved.