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