Question

Simplify it -

$\sqrt{\frac{{(\sec A+\tan A)}^2}{{\sec }^{2}A-{\tan }^{2}A}}$

Answer 1

Consider the following equation.

$\sqrt{\frac{{(\sec A+\tan A)}^2}{({\sec }^{2}A-{\tan }^{2}A)}}$

Using given formula,

${\sec }^{2}A-{\tan }^{2}A=1$

$\sqrt{\frac{{(\sec A+\tan A)}^2}{({\sec }^{2}A-{\tan }^{2}A)}}\overrightarrow{=}(\sec A+\tan A)$

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

Hence, this is the correct answer.