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  4. If tan θ ⋅ tan (120°-θ) tan (120°+θ)=(1/√3) then θ is equal to

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Q. 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

AP EAMCETAP EAMCET 2015

Solution:

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