Question

If $\quad \tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}$, then $\theta$ is equal to

• A. $\frac{n \pi}{3}+\frac{\pi}{18}, n \in Z$
• B. $\frac{n \pi}{3}+\frac{\pi}{12}, n \in Z$
• C. $\frac{n \pi}{12}+\frac{\pi}{12}, n \in Z$
• D. $\frac{n \pi}{3}+\frac{\pi}{6}, n \in Z$

Answer 1

Answer: (A)

$\frac{n \pi}{3}+\frac{\pi}{18}, n \in Z$

Answer 2

Given that, $\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}$
$\because \tan \theta \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)$
$=\tan 3 \theta$
$\therefore \tan 3 \theta=\frac{1}{\sqrt{3}} \Rightarrow \tan 3 \theta=\tan \frac{\pi}{6}$
$\Rightarrow 3 \theta=n \pi+\frac{\pi}{6}$
$\Rightarrow \theta=\frac{n \pi}{3}+\frac{\pi}{18}, n \in Z$