In a △ABC, if cosAcosBcosC=√3−18 and sinAsinBsinC=3+√38, then
The value of tanAtanB+tanBtanC+tanCtanA is
A
5−4√3
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B
5+4√3
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C
6+√3
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D
6−√3
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Solution
The correct option is B5+4√3 cos(A+B+C)=cosAcosBcosC[1−tanAtanB−tanBtanC−tanCtanA]⇒−1=√3−18×[1−tanAtanB−tanBtanC−tanCtanA]⇒1−tanAtanB−tanBtanC−tanCtanA=81−√3⇒tanAtanB+tanBtanC+tanCtanA=1−81−√3=−7−√31−√3=7+√3√3−1×√3+1√3+1=5+4√3