$${\left( {\frac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3} - \cos \theta }}} \right)^2} + 1,$$     θ ≠ 45° is equal to:

Solution (By Examveda Team)

$$\eqalign{ & {\left( {\frac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3} - \cos \theta }}} \right)^2} + 1 \cr & = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}{\left( {\frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)^2} + 1 \cr & = {\tan ^2}\theta + 1 \cr & = {\sec ^2}\theta \cr} $$