7. Let $\theta_{1}, \theta_{2}$ be acute angles, and $\frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}}+\frac{\cos ^{2020} \theta_{1}}{\sin ^{2018} \theta_{2}}=1$, then $\theta_{1}+\theta_{2}=\triangle$,
$$
\begin{array}{l}
\frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}}+\underbrace{\cos ^{2} \theta_{2}+\cos ^{2} \theta_{2}+\cdots+\cos ^{2} \theta_{2}}_{1009 \uparrow} \\
\geq 1010\left(\frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}} \underbrace{\cos ^{2} \theta_{2} \cdots \cos ^{2} \theta_{2}}_{1009}\right)^{\frac{1}{1010}} \\
\quad=1010 \sin ^{2} \theta_{1} ;
\end{array}
$$

Similarly, $\frac{\cos ^{2020} \theta_{1}}{\sin ^{2018} \theta_{2}}+\underbrace{\sin ^{2} \theta_{2}+\sin ^{2} \theta_{2}+\cdots+\sin ^{2} \theta_{2}}_{1009 \uparrow} \geq 1010 \cos ^{2} \theta_{1}$.
Therefore, $\frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}}+\frac{\cos ^{2020} \theta_{1}}{\sin ^{2018} \theta_{2}} \geq 1$, with equality if and only if $\theta_{1}+\theta_{2}=\frac{\pi}{2}$.