$$\frac{{2 + {{\tan }^2}\theta + {{\cot }^2}\theta }}{{\sec \theta \,{\text{cosec}}\,\theta }}$$    is equal to:

Solution (By Examveda Team)

$$\eqalign{ & \frac{{2 + {{\tan }^2}\theta + {{\cot }^2}\theta }}{{\sec \theta .{\text{cosec}}\,\theta }} \cr & = \frac{{{{\left( {\tan \theta + \cot \theta } \right)}^2}}}{{\sec \theta .{\text{cosec}}\,\theta }} \cr & = \frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\left( {\sec \theta .{\text{cosec}}\,\theta } \right)\left( {{{\sin }^2}\theta .{{\cos }^2}\theta } \right)}} \cr & = \sec \theta .{\text{cosec}}\,\theta \cr} $$