1. Given that $\alpha, \beta$ are both acute angles, and satisfy $\sin ^{2} \alpha=$ $\cos (\alpha-\beta)$. Then the relationship between $\alpha$ and $\beta$ is ( ).
(A) $\alpha\beta$
(D) $\alpha+\beta=\frac{\pi}{2}$
1.C.

Since $\cos \alpha>0, \cos \beta>0$, therefore, $\cos (\alpha-\beta)=\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta>\sin \alpha \cdot \sin \beta$.
Also, $\sin ^{2} \alpha=\cos (\alpha-\beta)$, hence $\sin ^{2} \alpha>\sin \alpha \cdot \sin \beta$.
Given that $\alpha, \beta$ are acute angles, then $\sin \alpha>\sin \beta, \alpha>\beta$.