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

(N/A) $L$.$H$.$S$. $= \frac{\tan A}{1+\sec A} + \frac{\tan A}{\sec A-1}$
$= \tan A \left( \frac{\sec A - 1 + 1 + \sec A}{(\sec A + 1)(\sec A - 1)} \right)$
$= \tan A \left( \frac{2 \sec A}{\sec^2 A - 1} \right)$
$= \tan A \left( \frac{2 \sec A}{\tan^2 A} \right) \quad [\because \sec^2 A - 1 = \tan^2 A]$
$= \frac{2 \sec A}{\tan A} = \frac{2 \cdot \frac{1}{\cos A}}{\frac{\sin A}{\cos A}} = \frac{2}{\sin A} = 2 \operatorname{cosec} A = \text{R.H.S.}$