Answer
$$\frac{{dy}}{{d\theta }} = \sec \theta $$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\sec \theta + \tan \theta } \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\ln \left( {\sec \theta + \tan \theta } \right)} \right] \cr
& {\text{use the rule }}\frac{d}{{d\theta }}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sec \theta + \tan \theta }}\frac{d}{{d\theta }}\left[ {\sec \theta + \tan \theta } \right] \cr
& {\text{solve the derivatives}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sec \theta + \tan \theta }}\left( {\sec \theta tan\theta + se{c^2}\theta } \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{d\theta }} = \frac{{\sec \theta tan\theta + se{c^2}\theta }}{{\sec \theta + \tan \theta }} \cr
& \frac{{dy}}{{d\theta }} = \frac{{\sec \theta \left( {tan\theta + sec\theta } \right)}}{{\sec \theta + \tan \theta }} \cr
& \frac{{dy}}{{d\theta }} = \sec \theta \cr} $$
