The value of $$\sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } \times \sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } $$       is equal to:

Solution (By Examveda Team)

$$\eqalign{ & \sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } \times \sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \cr & = \sqrt {1 + {{\tan }^2}\theta + 1 + {{\cot }^2}\theta } \times \sqrt {{{\sin }^2}\theta \left( {\frac{1}{{{{\cos }^2}\theta }} - 1} \right)} \cr & = \sqrt {2 + {{\tan }^2}\theta + {{\cot }^2}\theta } \times \sqrt {\frac{{{{\sin }^2}\theta .{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \cr & = \left( {\tan \theta + \cot \theta } \right) \times \left( {\frac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right) \times \left( {\frac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \frac{1}{{\sin \theta .\cos \theta }} \times \frac{{{{\sin }^2}\theta }}{{\cos \theta }} \cr & = \sin \theta .{\sec ^2}\theta \cr} $$