If $\cos (\alpha+\beta)=0$, then $\sin (\alpha-\beta)$ can be reduced to
(a) $\cos \beta$
(b) $\cos 2 \beta$
(c) $\sin \alpha$
(d) $\sin 2 a$
(b) 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, $\quad \sin (\alpha-\beta)=\sin \left(90^{\circ}-\beta-\beta\right) \quad$ [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 $\cos 2 \beta$.
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