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