If ∠A and ∠B are complementary to each other, then the value of sec2A + sec2B - sec2A.sec2B is?

Answer: Option D

$$\begin{array}{rl}& \text{A + B}=90^{\circ }\\ & \text{B}=90^{\circ }-\text{A}\\ & {\sec }^{2}A+se{c}^2\text{B}-\text{se}{\text{c}}^2\text{A}\text{.se}{\text{c}}^2\text{B}\end{array}$$
  $$={\sec }^{2}A+se{c}^2\left({90}^{\circ }-\text{A}\right)-$$       $$\text{se}{\text{c}}^2\text{A}\text{.se}{\text{c}}^2$$   $$\left({90}^{\circ }-\text{A}\right)$$
$$\begin{array}{rl}& ={\sec }^{2}A+{\mathrm{cosec}}^2\text{A}-\text{se}{\text{c}}^2\text{A}\text{.cose}{\text{c}}^2\text{A}\\ & =\frac{1}{{\cos }^2\text{A}}+\frac{1}{si{n}^2\text{A}}-\frac{1}{\text{co}{\text{s}}^2\text{A}}\times \frac{1}{\text{si}{\text{n}}^2\text{A}}\\ & =\frac{{\sin }^2\text{A + co}{\text{s}}^2\text{A}}{\text{co}{\text{s}}^2\text{A}\text{.}{\sin }^2\text{A}}-\frac{1}{\text{co}{\text{s}}^2\text{A}\text{.}{\sin }^2\text{A}}\\ & =\frac{1}{\text{co}{\text{s}}^2\text{A}.{\sin }^2\text{A}}-\frac{1}{\text{co}{\text{s}}^2\text{A}.{\sin }^2\text{A}}\\ & =0\end{array}$$