In a $\triangle \mathrm{ABC}$, we have $\sin A-\cos B=\cos C \Rightarrow \sin A=\cos B+\cos C$
$\Rightarrow 2 \sin \frac{A}{2} \cdot \cos \frac{A}{2}=2 \cos \left(\frac{B+C}{2}\right) \cdot \cos \left(\frac{B-C}{2}\right)$
$[\because \sin 2 \mathrm{~A}=2 \sin \mathrm{A} \cdot \cos \mathrm{A}]$
and $\cos B+\cos C=2 \cos \left(\frac{B+C}{2}\right) \cdot \cos \left(\frac{B-C}{2}\right)$
$\Rightarrow 2 \sin \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{A}}{2}=2 \cos \left(90^{\circ}-\frac{\mathrm{A}}{2}\right) \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\left[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \Rightarrow\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=90^{\circ}-\frac{\mathrm{A}}{2}\right]$
$\Rightarrow 2 \sin \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{A}}{2}=2 \sin \frac{\mathrm{A}}{2} \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
$\Rightarrow \quad \cos \frac{\mathrm{A}}{2}=\cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\Rightarrow \frac{A}{2}=\frac{B-C}{2}$
$\Rightarrow \quad \mathrm{A}+\mathrm{C}=\mathrm{B}$ ....(i)
Also, $A+C=180^{\circ}-B$ ....(ii)
So, $180^{\circ}-\mathrm{B}=\mathrm{B}$
$\Rightarrow \quad 2 \mathrm{~B}=180^{\circ}$
$\therefore \quad \mathrm{B}=90^{\circ}$