7. Let $\theta_{1} 、 \theta_{2}$ be acute angles, and
$$
\frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}}+\frac{\cos ^{2020} \theta_{1}}{\sin ^{2018} \theta_{2}}=1 \text {. }
$$

Then $\theta_{1}+\theta_{2}=$ $\qquad$ .
7. $\frac{\pi}{2}$.

Notice,
$$
\begin{array}{l}
\frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}}+\underbrace{\cos ^{2} \theta_{2}+\cdots+\cos ^{2} \theta_{2}}_{1009 \uparrow} \\
\geqslant 1010(\frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}} \cdot \underbrace{\cos ^{2} \theta_{2} \cdots \cdots \cdot \cos ^{2} \theta_{2}}_{1009 \uparrow})^{\frac{1}{1010}} \\
=1010 \sin ^{2} \theta_{1} .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
\frac{\cos ^{2020} \theta_{1}}{\sin ^{2018} \theta_{2}}+\underbrace{\sin ^{2} \theta_{2}+\cdots+\sin ^{2} \theta_{2}}_{1009} \\
\geqslant 1010 \cos ^{2} \theta_{1} . \\
\text { Therefore, } \frac{\sin ^{2020} \theta_{1}}{\cos ^{2018} \theta_{2}}+\frac{\cos ^{2020} \theta_{1}}{\sin ^{2018} \theta_{2}} \geqslant 1 .
\end{array}
$$

From the equality, we know $\theta_{1}+\theta_{2}=\frac{\pi}{2}$.