Prove the following

Question:

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

(a) $\cos \beta$

(b) $\cos 2 \beta$

(c) $\sin \alpha$

(d) $\sin 2 a$

Solution:

(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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