# Prove the following trigonometric identities.

Question:

Prove the following trigonometric identities.

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

Solution:

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

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

So,

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

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

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

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

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