8. Given $\frac{\sin ^{4} \alpha}{\sin ^{2} \beta}+\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}=1, \sin \alpha \cos \alpha \sin \beta \cos \beta \neq 0$, then $\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}+\frac{\cos ^{4} \beta}{\cos ^{2} \alpha}=$
$\qquad$ .
8. 1
$$
\begin{array}{l}
\frac{\sin ^{4} \alpha}{\sin ^{2} \beta}+\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}=1 \\
\text { i.e., } \frac{\cos ^{2} \beta \sin ^{2} \alpha\left(1-\cos ^{2} \alpha\right)+\sin ^{2} \beta \cos ^{2} \alpha\left(1-\sin ^{2} \alpha\right)}{\sin ^{2} \beta \cos ^{2} \beta}=1
\end{array}
$$

i.e., $\cos ^{2} \beta \sin ^{2} \alpha+\sin ^{2} \beta \cos ^{2} \alpha-\left(\cos ^{2} \beta \sin ^{2} \alpha \cos ^{2} \alpha+\sin ^{2} \beta \cos ^{2} \alpha \sin ^{2} \alpha\right)=$
$$
\sin ^{2} \beta \cos ^{2} \beta
$$

i.e., $\cos ^{2} \beta \sin ^{2} \alpha+\sin ^{2} \beta \cos ^{2} \alpha=\sin ^{2} \alpha \cos ^{2} \alpha+\sin ^{2} \beta \cos ^{2} \beta$, if we swap $a, \beta$ in this equation, the equation still holds. Therefore, if we swap $\alpha, \beta$ in the original equation, the original equation still holds, i.e., $\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}+$ $\frac{\cos ^{4} \beta}{\cos ^{2} \alpha}=1$.