已知△ABC的内角A,B及其对边a,b满足a+b=a／tanA+b／tanB,求内角C.

a+b=a/tanA+b/tanB
a（1- 1/tanA）+b(1-1/tanB)=0
a(1-cosA/sinA)+b(1-cosB/sinB)=0
根据正弦定理a/sinA=b/sinB
那么
sinA(1-cosA/sinA) +sinB(1-cosB/sinB)=0
(sinA-cosA)+ sinB-cosB=0
根号2sin（A-π/4）+根号2sin（B-π/4）=0
sin(A-π/4)=-sin(B-π/4)
∴A-π/4=π/4-B
A+B=π/2
C=π/2