If cos (alpha+beta)=0, then sin (alpha-beta) | vedclass

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Question

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

Vedclass · Accepted Answer

$\cos 2 \beta$

Vedclass · Answer

(A) Given,$\cos (\alpha+\beta)=0=\cos 90^{\circ}$ $\left[\because \cos 90^{\circ}=0ight]$$\Rightarrow \alpha+\beta=90^{\circ}$$\Rightarrow \alpha=90^{\circ}-\beta$ ......$(i)$Now,$\sin (\alpha-\beta)=\sin (90^{\circ}-\beta-\beta)$ [substituting the value from Eq. $(i)$]$= \sin (90^{\circ}-2 \beta)$$= \cos 2 \beta$ $\left[\because \sin (90^{\circ}-	heta)=\cos 	hetaight]$Hence,$\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.

If $\cos (\alpha+\beta)=0,$ then $\sin (\alpha-\beta)$ can be reduced to

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