What is the value of $$1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}}?$$

Solution (By Examveda Team)

$$\eqalign{ & 1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}} \cr & = 1 + \frac{{{{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{1 + \sec A + {{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{\sec A\left( {1 + \sec A} \right)}}{{1 + \sec A}} \cr & = \sec A \cr} $$