10. Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, and $\frac{\sin ^{4} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{4} \alpha}{\sin ^{2} \beta}=1$, find the value of $\alpha+\beta$. (Express the answer as a rational multiple of $\pi$)
Solution: Let
$$
p=\frac{\sin ^{2} \alpha}{\cos \beta}, q=\frac{\cos ^{2} \alpha}{\sin \beta}
$$

Then
$$
\left\{\begin{array}{c}
p \cos \beta+q \sin \beta=1 \\
p^{2}+q^{2}=1
\end{array}\right.
$$

Construct the quadratic function $f(x)=[(\cos \beta) x-p]^{2}+[(\sin \beta) x-q]^{2}$, then
$$
f(x)=\left(\cos ^{2} \beta+\sin ^{2} \beta\right) x^{2}-2(p \cos \beta+q \sin \beta) x+\left(p^{2}+q^{2}\right)=(x-1)^{2}
$$

Thus $f(1)=0$
That is, $(\cos \beta-p)^{2}+(\sin \beta-q)^{2}=0$
Therefore
$$
\left\{\begin{array}{l}
p=\cos \beta \\
q=\sin \beta
\end{array}\right.
$$

Since $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, then $\sin \alpha=\cos \beta, \cos \alpha=\sin \beta$, so $\alpha+\beta=\frac{\pi}{2}$