For this one we will need the Pythagorean identity \(1 + \tan^2A=\sec^2A\) in its equivalent form \(\sec^2A-\tan^2A=1\).
\(\begin{eqnarray*} \dfrac{1}{\sec A + \tan A} &=& \dfrac{1}{\sec A+\tan A}\cdot\dfrac{\sec A-\tan A}{\sec A - \tan A}\\ &=& \dfrac{\sec A-\tan A}{\sec^2A-\tan^2A}\\ &=& \dfrac{\sec A-\tan A}{1}\\ &=&\sec A-\tan A \end{eqnarray*}\)