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

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