Question

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

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