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

Answer: Option D

$$\begin{array}{rl}& \left(1+{\cot }^2\theta \right)\left(1+{\tan }^2\theta \right)\times \left(\sin \theta -\text{cosec}\theta \right)\left(\cos \theta -\sec \theta \right)\\ & =\left(\text{cose}{\text{c}}^2\theta \right)\left({\sec }^2\theta \right)\times \left(\frac{{\sin }^2\theta -1}{\sin \theta }\right)\left(\frac{{\cos }^2\theta -1}{\cos \theta }\right)\\ & =\left(\text{cose}{\text{c}}^2\theta \right)\left({\sec }^2\theta \right)\times \left(\frac{-{\cos }^2\theta }{\sin \theta }\right)\left(\frac{-{\sin }^2\theta }{\cos \theta }\right)\\ & =\left(\text{cose}{\text{c}}^2\theta \right)\left({\sec }^2\theta \right)\times \sin \theta \cos \theta \\ & =\sec \theta \text{cosec}\theta \end{array}$$