MCQ
Choose the correct answer from the given four options. 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.
$\cos2\beta$
Given $, \cos(\alpha+\beta)=0=\cos90^\circ$
$[\because\cos90^\circ=0]$
$\Rightarrow\ \alpha+\beta=90^\circ$
$\Rightarrow\ \alpha=90^\circ-\beta\ \ ...(\text{i})$
Now $, \sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ [$put the value from Eq. $(i)]$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
$[\because\sin(90^\circ-\theta)=\cos\theta]$
Hence, $\sin(\alpha-\beta)$ can be reduced to $\cos2\beta.$
$[\because\cos90^\circ=0]$
$\Rightarrow\ \alpha+\beta=90^\circ$
$\Rightarrow\ \alpha=90^\circ-\beta\ \ ...(\text{i})$
Now $, \sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ [$put the value from Eq. $(i)]$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
$[\because\sin(90^\circ-\theta)=\cos\theta]$
Hence, $\sin(\alpha-\beta)$ can be reduced to $\cos2\beta.$
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