14 Acute angles $\alpha, \beta$ satisfy the condition $\frac{\sin ^{4} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{4} \alpha}{\sin ^{2} \beta}=1$, which of the following conclusions is correct?
A. $\alpha+\beta>\frac{\pi}{2}$
B. $\alpha+\beta<\frac{\pi}{2}$
C. $\alpha+\beta \neq \frac{\pi}{2}$
D. $\alpha+\beta=\frac{\pi}{2}$
14 D. From $\frac{\sin ^{4} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{4} \alpha}{\sin ^{2} \beta}-1=0$, we get $\left(\frac{\sin ^{2} \alpha}{\cos \beta}\right)^{2}+\left(\frac{\cos ^{2} \alpha}{\sin \beta}\right)^{2}+\left(\cos ^{2} \beta+\sin ^{2} \beta\right)-$ $2\left(\frac{\sin ^{2} \alpha}{\cos \beta}\right) \cos \beta-2\left(\frac{\cos ^{2} \alpha}{\sin \beta}\right) \cdot \sin \beta=0$, which simplifies to $\left(\frac{\sin ^{2} \alpha}{\cos \beta}-\cos \beta\right)^{2}+\left(\frac{\cos ^{2} \alpha}{\sin \beta}-\sin \beta\right)^{2}=$ 0, thus $\sin ^{2} \alpha=\cos ^{2} \beta$. Given that $\alpha, \beta$ are acute angles, we have $\alpha=\frac{\pi}{2}-\beta$.