Answer
$${\text{We have proved that }}LHS = RHS$$
Work Step by Step
$$\eqalign{
& 2\csc 2\theta = \sec \theta \csc \theta \cr
& \cr
& LHS = 2\csc 2\theta {\text{ and }}RHS = \sec \theta \csc \theta \cr
& \cr
& {\text{Use the identity }}\csc \alpha = \frac{1}{{\sin \alpha }} \cr
& LHS = 2\left( {\frac{1}{{\sin 2\theta }}} \right) \cr
& {\text{Use the identity }}\sin 2\theta = 2\sin \theta \cos \theta \cr
& LHS = 2\left( {\frac{1}{{2\sin \theta \cos \theta }}} \right) \cr
& LHS = \frac{1}{{\sin \theta \cos \theta }} \cr
& LHS = \left( {\frac{1}{{\sin \theta }}} \right)\left( {\frac{1}{{\cos \theta }}} \right) \cr
& LHS = \csc \theta \sec \theta \cr
& \cr
& {\text{We have proved that }}LHS = RHS \cr} $$
