在 $\triangle A B C$ 中, 角 $A, B, C$ 的对边 $a, b, c$ 成等差数列, 且 $A-C=90^{\circ}$, 则 $\cos B=$
$\because a, b, c$ 成等差数列, $\therefore 2 b=a+c$,

又 $\because A-C=90^{\circ}, A+B+C=180^{\circ}$,

$\therefore C=45^{\circ}-\frac{B}{2}$,

由正弦定理可得 $2 \sin B=\sin A+\sin C$ ，

$\therefore 2 \sin B=\sin \left(90^{\circ}+C\right)+\sin C=\cos C+\sin C=\sqrt{2} \sin \left(C+45^{\circ}\right)$

$=\sqrt{2} \sin \left(45^{\circ}-\frac{B}{2}+45^{\circ}\right)=\sqrt{2} \sin \left(90^{\circ}-\frac{B}{2}\right)=\sqrt{2} \cos \frac{B}{2}$,

$\therefore 2 \sin B=4 \sin \frac{B}{2} \cos \frac{B}{2}=\sqrt{2} \cos \frac{B}{2}$,

解得 $\sin \frac{B}{2}=\frac{\sqrt{2}}{4}$,

$\therefore \cos B=1-2 \sin ^{2} \frac{B}{2}=\frac{3}{4}$

故答案为: $\frac{3}{4}$