If tan theta cdot tan left(120^{circ}-thetari | vedclass

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(A) We know the identity: $	an 	heta \cdot 	an \left(60^{\circ}-	hetaight) \cdot 	an \left(60^{\circ}+	hetaight) = 	an 3	heta$. Given the

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

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(A) We know the identity: $	an 	heta \cdot 	an \left(60^{\circ}-	hetaight) \cdot 	an \left(60^{\circ}+	hetaight) = 	an 3	heta$. Given the expression: $	an 	heta \cdot 	an \left(120^{\circ}-	hetaight) \cdot 	an \left(120^{\circ}+	hetaight) = \frac{1}{\sqrt{3}}$. Using the identity $	an(180^{\circ} - A) = -	an A$,we have $	an(120^{\circ} - 	heta) = -	an(60^{\circ} + 	heta)$ and $	an(120^{\circ} + 	heta) = -	an(60^{\circ} - 	heta)$. Substituting these: $	an 	heta \cdot [-	an(60^{\circ} + 	heta)] \cdot [-	an(60^{\circ} - 	heta)] = 	an 	heta \cdot 	an(60^{\circ} + 	heta) \cdot 	an(60^{\circ} - 	heta) = 	an 3	heta$. Thus,$	an 3	heta = \frac{1}{\sqrt{3}}$. Since $	an 3	heta = 	an \frac{\pi}{6}$,the general solution is $3	heta = n\pi + \frac{\pi}{6}$. Dividing by $3$,we get $	heta = \frac{n\pi}{3} + \frac{\pi}{18}, n \in Z$.

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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