Given, $\cos (\alpha+\beta)=0=\cos 90^{\circ}$ $\left[\because \cos 90^{\circ}=0\right]$

$\Rightarrow \quad \alpha+\beta=90^{\circ}$

$\Rightarrow \quad \alpha=90^{\circ}-\beta$ ……$(i)$

Now, $\sin (\alpha-\beta)=\sin \left(90^{\circ}-\beta-\beta\right)$ [put the value from Eq. $(i)$]

$=\sin \left(90^{\circ}-2 \beta\right)$

$=\cos 2 \beta$ $\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$

Hence, $\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.