9. Given that the three interior angles $A, B, C$ of $\triangle A B C$ satisfy $A+C=2 B$, and $\frac{1}{\cos A}+\frac{1}{\cos C}=$ $-\frac{\sqrt{2}}{\cos B}$, find $\cos \frac{A-C}{2}$.
9. From $A+B+C=180^{\circ}$, we get $B=60^{\circ}, A+C=120^{\circ}, \frac{1}{\cos A}+\frac{1}{\cos C}=\frac{\cos A+\cos C}{\cos A \cos C}=$ $-\frac{\sqrt{2}}{\cos B}=-2 \sqrt{2}$. Let $\cos \frac{A-C}{2}=t$, then $\cos A+\cos C=2 \cos \frac{A+C}{2} \cos \frac{A-C}{2}=t, \cos A \cos C=$ $\frac{1}{2}[\cos (A+C)+\cos (A+C)]=\frac{1}{2}\left[-\frac{1}{2}+2 t^{2}-1\right]$, hence $\frac{\cos A+\cos C}{\cos A \cdot \cos C}=\frac{t}{\frac{1}{2}\left(2 t^{2}-\frac{3}{2}\right)}=$ $-2 \sqrt{2}$, which gives $t^{2}+\frac{\sqrt{2}}{4} t-\frac{3}{4}=0$. Therefore, $t=+\frac{\sqrt{2}}{2}, t=-\frac{3}{4} \sqrt{2}$, and since $t \in[-1,1]$, we have $\cos \frac{A-C}{2}=$ $t=-\frac{\sqrt{2}}{2}$.