Let 0° < θ < 90°, (1 + cot²θ)(1 + tan²θ) × (sinθ - cosecθ)(cosθ - secθ) is equal to:

Solution (By Examveda Team)

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