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$In Δ A B C \cos A \cos B \cos C = \frac{\sqrt{3} - 1}{8} and \sin A \sin B \sin C = \frac{3 + \sqrt{3}}{8}$ , then the value of $\tan A + \tan B + \tan C =$

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Correct option is A

We havesinAsinBsinCcosAcosBcosC=3+33-1⇒tanAtanBtanC=3+33-1 ⇒tanA+tanB+tanC=3+33-1 ∵A+B+C=π⇒tanA+tanB+tanC=tanAtanBtanC

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In ΔABC cosAcosBcosC=3-18 and sinAsinBsinC=3+38, then the value of tanA+tanB+tanC=

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$In Δ A B C \cos A \cos B \cos C = \frac{\sqrt{3} - 1}{8} and \sin A \sin B \sin C = \frac{3 + \sqrt{3}}{8}$ , then the value of $\tan A + \tan B + \tan C =$