Question : If $\tan (\alpha+\beta)=\sqrt{3}, \tan (\alpha-\beta)=1$ where $(\alpha+\beta)$ and $(\alpha-\beta)$ are acute angles, then what is $\tan$ $(6 \alpha) ?$
Option 1: $-1$
Option 2: $0$
Option 3: $1$
Option 4: $\sqrt{2}-1$
Correct Answer: $-1$
Solution :
Given:
$\tan (\alpha+\beta)=\sqrt{3}=\tan 60^{\circ}$
$\tan (\alpha-\beta)=1=\tan 45^{\circ}$
⇒ $\alpha+\beta=60^{\circ}$
and $\alpha-\beta=45^{\circ}$
⇒ $2\alpha=105^{\circ}$
⇒ $\alpha=\frac{105}{2}^{\circ}$
⇒ $6\alpha=6×\frac{105}{2}^{\circ}=315^{\circ}$
Now, $\tan 6\alpha$
= $\tan 315^{\circ}$
= $\tan (360^{\circ}-45^{\circ})$
= $-\tan45^{\circ}$
= $-1$
Hence, the correct answer is $-1$.
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