What is the value of $$\frac{{2\left( {1 - {{\sin }^2}\theta } \right){\text{cose}}{{\text{c}}^2}\theta }}{{{{\cot }^2}\theta \left( {1 + {{\tan }^2}\theta } \right)}} - 1$$

Solution (By Examveda Team)

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