MCQ
If $\cos(\alpha+\beta)=0,$ then $\sin(\alpha-\beta)$ can be reduced to:
- A$\cos\beta$
- ✓$\cos2\beta$
- C$\sin\alpha$
- D$\sin2\alpha$
✓
Answer
Correct option: B.
$\cos2\beta$
$\cos(\alpha+\beta)=0$
$\Rightarrow\ \alpha+\beta=90^\circ \big[\because\ \cos90^\circ=0\big]$
$\Rightarrow\ \alpha=90^\circ-\beta\ .....(\text{i})$
$\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ \big[\text{using (i)}\big]$
$=\sin\big(90^\circ-2\beta\big)$
$=\cos2\beta\ \big[\because \sin(90^\circ -\theta)=\cos\theta\big]$
$\Rightarrow\ \alpha+\beta=90^\circ \big[\because\ \cos90^\circ=0\big]$
$\Rightarrow\ \alpha=90^\circ-\beta\ .....(\text{i})$
$\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ \big[\text{using (i)}\big]$
$=\sin\big(90^\circ-2\beta\big)$
$=\cos2\beta\ \big[\because \sin(90^\circ -\theta)=\cos\theta\big]$
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