Answer
$$\frac{{1 - \sin \theta }}{{1 + \sin \theta }} = {\sec ^2}\theta - 2\sec \theta \tan \theta + {\tan ^2}\theta $$
Work Step by Step
$$\eqalign{
& \frac{{1 - \sin \theta }}{{1 + \sin \theta }} = {\sec ^2}\theta - 2\sec \theta \tan \theta + {\tan ^2}\theta \cr
& {\text{We transform the more complicated left side to match the right side}}. \cr
& \frac{{1 - \sin \theta }}{{1 + \sin \theta }} = \frac{{1 - \sin \theta }}{{1 + \sin \theta }} \cdot \frac{{1 - \sin \theta }}{{1 - \sin \theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {1 - \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 - 2\sin \theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{{{\cos }^2}\theta }} - \frac{{2\sin \theta }}{{{{\cos }^2}\theta }} + \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\sec ^2}\theta - 2\left( {\frac{1}{{\cos \theta }}} \right)\left( {\frac{{\sin \theta }}{{\cos \theta }}} \right) + \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\sec ^2}\theta - 2\sec \theta \tan \theta + {\tan ^2}\theta \cr
& {\text{Thus have verified that the given equation is an identity}} \cr} $$
