Prove the following trigonometric identities.

We need to prove $\frac{\sec A-\tan A}{\sec A+\tan A}=\frac{\cos ^{2} A}{(1+\sin A)^{2}}$

Here, we will first solve the LHS.

Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get

$\frac{\sec A-\tan A}{\sec A+\tan A}=\frac{\frac{1}{\cos A}-\frac{\sin A}{\cos A}}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}}$

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

$=\frac{1-\sin A}{1+\sin A}$

Further, multiplying both numerator and denominator by $1+\sin A$, we get

$\frac{1-\sin A}{1+\sin A}=\left(\frac{1-\sin A}{1+\sin A}\right)\left(\frac{1+\sin A}{1+\sin A}\right)$

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

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

Now, using the property $\cos ^{2} \theta+\sin ^{2} \theta=1$, we get

So,

$\frac{1-\sin ^{2} A}{(1+\sin A)^{2}}=\frac{\cos ^{2} A}{(1+\sin A)^{2}}=\mathrm{RHS}$

Hence proved.