Prove that tanθ tan(π/3+θ) tan(π/3-θ)=tan3θ

$\tan\theta tan\left(\frac{\pi }{3}+\theta \right)\tan \left(\frac{\pi }{3}-\theta \right)$=$tan3\theta$ Prove that tanθ tan(π/3+θ) tan(π/3-θ)=tan3θ Solution: $\tan \theta tan\left(\frac{\pi }{3} + \theta \right)\tan …

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