Q) Prove that $\frac {\tan A} {1+ \sec A} - \frac{\tan A}{1- \sec A}$ = 2 cosec A

Let’s start from LHS LHS = $\frac {\tan A} {1+ \sec A} - \frac{\tan A}{1- \sec A}$ = $\frac {\tan A (1 - \sec A) - \tan A (1 + \sec A)} {(1+ \sec A) (1 - \sec A)}$ = $\frac {- 2 \tan A \sec A} {1 - \sec ^2 A}$ = $\frac {- 2 \tan A \sec A} {- \tan ^2 A}$ = $\frac {2 \sec A} {\tan A}$ = $\frac {2} {\sin A}$ = 2 cosec A = RHS ........... Hence, proved !