9. Let $\triangle A B C$ have internal angles $A, B, C$ with opposite sides $a, b, c$ respectively, and $A-C=\frac{\pi}{2}, a, b, c$ form an arithmetic sequence, then the value of $\cos B$ is
Given $\frac{3}{4}$.
Analysis $2 b=a+c \Leftrightarrow 2 \sin (A+C)=\sin A+\sin C \Leftrightarrow 2 \sin \left(2 C+\frac{\pi}{2}\right)=\sin \left(C+\frac{\pi}{2}\right)+\sin C$. Therefore,
$$
2 \cos 2 C=\cos C+\sin C \Leftrightarrow 2\left(\cos ^{2} C-\sin ^{2} C\right)=\cos C+\sin C \Rightarrow \cos C-\sin C=\frac{1}{2}
$$

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