∵tanA+tanB=2sinC/cosA
∴sinA/cosA+sinB/cosB=2sinC/cosA
去分母：
sinAcosB+sinBcosA=2sinCcosB
∴sin(A+B)=2sinCcosB
∵sin(A+B)=sin(π-C)=sinC
∴sinC=2sinCcosB
∵sinC>0
∴2cosB=1 ,cosB=1/2
∴B=π/3