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If $\alpha$ and $\beta$ are such that $\tan \alpha=2 \tan \beta$, then what is $\sin (\alpha+\beta)$ equal to ?
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Verified Answer
The correct answer is:
$3 \sin (\alpha-\beta)
As given :
$\tan \alpha=2 \tan \beta$
$\Rightarrow \frac{\tan \alpha}{\tan \beta}=2 \Rightarrow \frac{\sin \alpha / \cos \alpha}{\sin \beta / \cos \beta}=2$
$\Rightarrow \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta}=2$
Using componendo and dividendo we get
$\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}=\frac{2+1}{2-1}=3$
$\Rightarrow \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$
$\sin (\alpha+\beta)=3 \sin (\alpha-\beta)$
$\tan \alpha=2 \tan \beta$
$\Rightarrow \frac{\tan \alpha}{\tan \beta}=2 \Rightarrow \frac{\sin \alpha / \cos \alpha}{\sin \beta / \cos \beta}=2$
$\Rightarrow \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta}=2$
Using componendo and dividendo we get
$\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}=\frac{2+1}{2-1}=3$
$\Rightarrow \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$
$\sin (\alpha+\beta)=3 \sin (\alpha-\beta)$
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