Prove the following trigonometric identities:$ \frac{(1+\sin \theta)^{2}+(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta} $

To do:

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

Solution:

We know that,

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

Therefore,

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

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

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

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

$=\frac{1+\sin^2 \theta}{1-\sin^2 \theta}$                     [From (i)]

Hence proved.