# How do you prove 1/(sec D - tan D) + 1/(sec D + tan D) = 2 sec D?

Prove : $\frac{1}{\sec D - \tan D} + \frac{1}{\sec D + \tan D} = 2 \sec D$
Left side --> $\frac{\left(\sec D + \tan D\right) + \left(\sec D - \tan D\right)}{{\sec}^{2} D - {\tan}^{2} D} =$
$= \frac{2 \sec D}{1}$
${\sec}^{2} D - {\tan}^{2} D = \frac{1}{{\cos}^{2} D} - \frac{{\sin}^{2} D}{{\cos}^{2} D} =$
$= \frac{1 - {\sin}^{2} D}{{\cos}^{2} D} = \frac{{\cos}^{2} D}{{\cos}^{2} D} = 1$