If α + β = 90°, then the value of (1 - sin2α)(1 - cos2α) × (1 + cot2β)(1 + tan2β) is?

Answer: Option A

Shortcut method :
$$\left(1-{\sin }^2\alpha \right)\left(1-{\cos }^2\alpha \right)\times$$     $$\left(1+{\cot }^2\beta \right)$$  $$\left(1+{\tan }^2\beta \right)$$
$$\begin{array}{rl}& \Rightarrow \left({\cos }^2\alpha \right)\left(si{n}^2\alpha \right)\left({\mathrm{cosec}}^2\beta \right)\left({\sec }^2\beta \right)\\ & \text{Put }\alpha =\beta =45^{\circ }\\ & \Rightarrow \text{co}{\text{s}}^{2}45^{\circ }.{\sin }^{2}45^{\circ }.\text{cose}{\text{c}}^{2}45^{\circ }.{\mathrm{sce}}^{2}45^{\circ }\\ & \Rightarrow \frac{1}{2}.\frac{1}{2}.2.2\\ & \Rightarrow 1\end{array}$$

Alternate :
$$\left(1-{\sin }^2\alpha \right)\left(1-{\cos }^2\alpha \right)\times$$     $$\left(1+{\cot }^2\beta \right)$$  $$\left(1+{\tan }^2\beta \right)$$
$$\begin{array}{rl}& \Rightarrow \left({\cos }^2\alpha \right)\left(si{n}^2\alpha \right)\times \left({\mathrm{cosec}}^2\beta \right)\left({\sec }^2\beta \right)\\ & \Rightarrow \text{co}{\text{s}}^2\left({90}^{\circ }-\beta \right).{\sin }^2\alpha .\text{cose}{\text{c}}^2\beta .\text{se}{\text{c}}^2\left({90}^{\circ }-\alpha \right)\\ & \Rightarrow {\sin }^2\beta .\text{cose}{\text{c}}^2\beta .{\sin }^2\alpha .\text{cose}{\text{c}}^2\alpha \\ & \Rightarrow 1\end{array}$$