9. Let $\triangle A B C$ have internal angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and $\angle A - \angle C = \frac{\pi}{2}, a, b, c$ form an arithmetic sequence. Then the value of $\cos B$ is . $\qquad$
II. $\frac{3}{4}$.
It is easy to know,
$$
\begin{array}{l}
2 b=a+c \\
\Leftrightarrow 2 \sin (A+C)=\sin A+\sin C \\
\Leftrightarrow 2 \sin \left(2 C+\frac{\pi}{2}\right)=\sin \left(C+\frac{\pi}{2}\right)+\sin C \\
\Leftrightarrow 2 \cos 2 C=\cos C+\sin C \\
\Rightarrow \cos C-\sin C=\frac{1}{2} .
\end{array}
$$

Therefore, $\cos B=-\cos (A+C)$
$$
=-\cos \left(2 C+\frac{\pi}{2}\right)=\sin 2 C=\frac{3}{4} \text {. }
$$