∵tanB/tanA+1=（tanB+tanA)/tanA ,2c/a=2sinC/sinA
∴（tanB+tanA)/tanA = 2sinC/sinA
两边同乘sinA得
cosA(tanB+tanA)=2sinC
∵sinC=sin(A+B)=sinAcosB+cosAsinB
∴cosA(tanB+tanA)=2(sinAcosB+cosAsinB)
两边同除以cosA得
tanB+tanA=2(tanAcosB+sinB)
∵sinB=cosBtanB
∴tanB+tanA=2(tanAcosB+cosBtanB)=2cosB(tanB+tanA)
2cosB=1
cosB=1/2 B=π/3
(2)
∵C=π－A－B=2π/3－A
∴C+π/6=5π/6－A
∴cos(C+π/6)=cos(5π/6－A)=－cos[(5π/6－A)－π]=－cos(－π/6－A)=－cos(A+π/6)
＝－(cosAcosπ/6－sinAsinπ/6)＝sinAsinπ/6－cosAcosπ/6＝½sinA－(√3/2)cosA＝1/3
∴(√3/2)cosA＝½sinA－1/3
两边平方得
3/4*cos²A=1/4*sin²A+1/9－1/3*sinA
∵sin²A+cos²A=0
∴3/4*cos²A=3/4*(1－sin²A)=3/4－3/4*sin²A
3/4－3/4*sin²A＝1/4*sin²A+1/9－1/3*sinA
sin²A－1/3*sinA+1/9-3/4=0
设sinA=x (x>0)
则有x²-1/3*x+1/9-3/4=0
36x²-12x-23=0
Δ=12²-4*36*(-23)=3600
x=1或-2/3(舍去）
∴sinA=1