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If tan $(\alpha+\beta)=2$ and $\tan (\alpha-\beta)=1$, then $\tan (2 \alpha)$ is equal
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Verified Answer
The correct answer is:
$-3$
$\tan (\alpha+\beta)=2$
$\tan (\alpha-\beta)=1$
$\tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)]$
$=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)}$
$=\frac{2+1}{1-2 \cdot 1}=\frac{3}{-1}=-3$
$\tan (\alpha-\beta)=1$
$\tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)]$
$=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)}$
$=\frac{2+1}{1-2 \cdot 1}=\frac{3}{-1}=-3$
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