Prove the following trigonometric identities.

We know that,

$\sin ^{2} A+\cos ^{2} A=1$

$\sec ^{2} A-\tan ^{2} A=1$

So,

$\sin ^{2} A+\frac{1}{1+\tan ^{2} A}=\sin ^{2} A+\frac{1}{\sec ^{2} A}$

$=\sin ^{2} A+\left(\frac{1}{\sec A}\right)^{2}$

$=\sin ^{2} A+(\cos A)^{2}$

$=\sin ^{2} A+\cos ^{2} A$

$=1$