91Ó°ÊÓ

Given trigonometric function

${\sin }^2{20}^{∘}+\frac{1}{se{c}^2{20}^{∘}}$

We will use standard formula of $\sin \mathrm{θ}$ and $\cos \mathrm{θ}$ at different angles for calculating the final result of given expression.

Since,

$$\begin{gathered} sec\mathrm{θ}=\frac{1}{\cos \mathrm{θ}} \\ now, \\ {\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}=1 \\ {\sin }^2\mathrm{θ}=1-{\cos }^2\mathrm{θ} \\ Nowsubstitutingthesevaluesinexpressionwehave \\ 1-{\cos }^2{20}^{∘}+\frac{1}{se{c}^2{20}^{∘}} \\ =1-{\cos }^2{20}^{∘}+{\cos }^2{20}^{∘} \\ =1 \end{gathered}$$