Prove the following trigonometric identities.

We have to prove $\frac{\cot ^{2} A(\sec A-1)}{1+\sin A}=\sec ^{2} A\left(\frac{1-\sin A}{1+\sec A}\right)$.

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

So,

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

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

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

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

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

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

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

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

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

$=\frac{1}{(\sec A+1)(1+\sin A)}$

Multiplying both the numerator and denominator by $(1-\sin A)$, we have

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

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

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

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

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

Hence proved.