Answer
$${\text{We have proved that }}LHS = RHS$$
Work Step by Step
$$\eqalign{
& \frac{{\sin 2\theta }}{{\sin \theta }} - \frac{{\cos 2\theta }}{{\cos \theta }} = \sec \theta \cr
& \cr
& LHS = \frac{{\sin 2\theta }}{{\sin \theta }} - \frac{{\cos 2\theta }}{{\cos \theta }}{\text{ and }}RHS = \sec \theta \cr
& \cr
& {\text{Use the identities}} \cr
& {\text{ sin2}}\theta = 2\sin \theta \cos \theta {\text{ and cos2}}\theta = {\cos ^2}\theta - {\sin ^2}\theta \cr
& \cr
& LHS = \frac{{2\sin \theta \cos \theta }}{{\sin \theta }} - \frac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{\cos \theta }} \cr
& LHS = 2\cos \theta - \frac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{\cos \theta }} \cr
& LHS = \frac{{2{{\cos }^2}\theta - {{\cos }^2}\theta + {{\sin }^2}\theta }}{{\cos \theta }} \cr
& LHS = \frac{{{{\cos }^2} + {{\sin }^2}\theta }}{{\cos \theta }} \cr
& LHS = \frac{1}{{\cos \theta }} \cr
& LHS = \sec \theta \cr
& \cr
& {\text{We have proved that }}LHS = RHS \cr} $$
