If $ an heta \cdot an \left(120^{\circ}- heta ight) an \left(120^{\circ}+ heta ight)=\frac{1}{\sqrt{3}}$, then $ heta$ is equal to

Question

If $	an 	heta \cdot 	an \left(120^{\circ}-	hetaight) 	an \left(120^{\circ}+	hetaight)=\frac{1}{\sqrt{3}}$, then $	heta$ is equal to

Aicharya · Accepted Answer

We have given, $ 	an 	heta \cdot 	an \left(120^{\circ}-	hetaight) 	an \left(120^{\circ}+	hetaight)=\frac{1}{\sqrt{3}} $ Since, we know $ \begin{aligned} & 	an 	heta 	an \left(120^{\circ}-	hetaight) 	an \left(120^{\circ}+	hetaight)=	an 3 	heta \ & 	herefore \quad 	an 3 	heta=\frac{1}{\sqrt{3}} \Rightarrow 	an 3 	heta=	an \frac{\pi}{6} \ & 3 	heta=n \pi+\frac{\pi}{6} \ & 	heta=\frac{n \pi}{3}+\frac{\pi}{18}, n \in Z \end{aligned} $

If $\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan…

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