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

Solution(By Examveda Team)

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

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