7. Given acute angles $\alpha, \beta$ satisfy the condition: $\frac{\sin ^{4} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{4} \alpha}{\sin ^{2} \beta}=1$, then $\alpha+\beta=$
Let $\frac{\sin ^{2} \alpha}{\cos \beta}=\cos \theta, \frac{\cos ^{2} \alpha}{\sin \beta}=\sin \theta$, where $\theta$ is an acute angle,
then $\sin ^{2} \alpha+\cos ^{2} \alpha=\cos \beta \cos \theta+\sin \beta \sin \theta=\cos (\beta-\theta)=1 \Rightarrow \theta=\beta$, so $\sin ^{2} \alpha=\cos ^{2} \beta \Rightarrow \sin \alpha=\cos \beta \Rightarrow \alpha+\beta=\frac{\pi}{2}$.