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