If $${\text{tan }}{9^ \circ } = \frac{p}{q}{\text{,}}$$   then the value of $$\frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{1 + {{\cot }^2}{{81}^ \circ }}}$$   is?

Solution (By Examveda Team)

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