If tan 9^ = p/q , then the value of se c^2 8 1^ 1 + ^281^ is?

Answer: Option D

$$\begin{array}{rl}& \text{ tan }{9}^{\circ }=\frac{p}{q}\\ & \Rightarrow \frac{\text{se}{\text{c}}^2\text{8}{\text{1}}^{\circ }}{1+{\cot }^2{81}^{\circ }}\\ & \Rightarrow \frac{\text{se}{\text{c}}^2\text{8}{\text{1}}^{\circ }}{{\mathrm{cosec}}^2{81}^{\circ }}\\ & \Rightarrow \frac{1}{\text{co}{\text{s}}^2{81}^{\circ }}\times {\sin }^{2}81^{\circ }\\ & \Rightarrow \text{ta}{\text{n}}^{2}81^{\circ }\\ & \Rightarrow \text{ta}{\text{n}}^2\left({90}^{\circ }-9^{\circ }\right)\\ & \Rightarrow \text{co}{\text{t}}^2{9}^{\circ }\\ & \Rightarrow \frac{q^2}{p^2}\end{array}$$