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$A = \frac{π}{7} B = \frac{2 π}{7} C = \frac{4 π}{7} and \cos A \cos B \cos C = \frac{- 1}{8} then$ $\sum \tan A . \tan B =$

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Correct option is B

We have A+B+C=πNow cos(A+B+C)=cos(A+B)cosC-sin(A+B)sinC⇒cos(A+B+C)cosAcosBcosC=1-tanAtanB-tanBtanC-tanCtanA⇒cosπ-1/8=1-∑tanAtanB⇒8=1-∑tanAtanB⇒∑tanAtanB=-7

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A=π7B=2π7C=4π7 and cosAcosBcosC=-18 then ∑tanA.tanB=

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$A = \frac{π}{7} B = \frac{2 π}{7} C = \frac{4 π}{7} and \cos A \cos B \cos C = \frac{- 1}{8} then$ $\sum \tan A . \tan B =$