Prove the following trigonometric identities.

We have to prove $\sec A(1-\sin A)(\sec A+\tan A)=1$

We know that, $\sec ^{2} A-\tan ^{2} A=1$

So,

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

$=(\sec A-\sec A \sin A)(\sec A+\tan A)$

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

$=\left(\sec A-\frac{\sin A}{\cos A}\right)(\sec A+\tan A)$

$=(\sec A-\tan A)(\sec A+\tan A)$

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

$=1$