解
cosA=3/5
∵A∈(0,π)
∴sinA=4/5
∴tanA=4/3
tan[A+(B-A)]
=[tanA+tan(B-A)]/[1-tanAtan(B-A)]
=(4/3+1/2)/(1-2/3)
=(11/6)/(1/3)
=11/2
tanC=-tan(A+B)
=-(tanA+tanB)/(1-tanAtanB)
=-(4/3+11/2)/(1-22/3)
=-(8/6+33/6)/(-19/3）
=41/6*3/19
=41/38不是负的吗=-(8/6+33/6)/(-19/3）后面有一个负号