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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
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Verified Answer
The correct answer is:
$\frac{n \pi}{3}+\frac{\pi}{18}, n \in \boldsymbol{Z}$
We have given,
$$
\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}
$$
Since, we know
$$
\begin{aligned}
& \tan \theta \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\tan 3 \theta \\
& \therefore \quad \tan 3 \theta=\frac{1}{\sqrt{3}} \Rightarrow \tan 3 \theta=\tan \frac{\pi}{6} \\
& 3 \theta=n \pi+\frac{\pi}{6} \\
& \theta=\frac{n \pi}{3}+\frac{\pi}{18}, n \in Z
\end{aligned}
$$
$$
\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}
$$
Since, we know
$$
\begin{aligned}
& \tan \theta \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\tan 3 \theta \\
& \therefore \quad \tan 3 \theta=\frac{1}{\sqrt{3}} \Rightarrow \tan 3 \theta=\tan \frac{\pi}{6} \\
& 3 \theta=n \pi+\frac{\pi}{6} \\
& \theta=\frac{n \pi}{3}+\frac{\pi}{18}, n \in Z
\end{aligned}
$$
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