设ΔABC的内角A,B,C所对的边分别为a,b,c,且acosB-bcosA=3/5c(1)求tanA/tanB的值（2

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设ΔABC的内角A,B,C所对的边分别为a,b,c,且acosB-bcosA=3/5c(1)求tanA/tanB的值（2）求tan(A-B)的最大值

1）acosB-bcosA=3/5c ,所以 (a/c) * cosB - (b/c) * cosA = 3 / 5 .
From Law of Sine (正弦定理) we get (sinA / SinC) * cosB - (sinB/ SinC) * cosA = 3 / 5.
Simplifying this equation we get sin(A - B) / sin(C) = 3 / 5 and further we get sin(A - B) / sin(A + B) = 3 / 5.(2)
Next Expand equation (2),(sinA * cosB - cosA * sin B ) / (sinA * sinB + cosA * sin B) = 3 / 5 and we can get 2 * sinA * cosB = 8 cosA * sin B.
Therefore,sinA cos B / ( cosA * sin B ) = tanA / tanB = 4.End
2) From part (1) tan(A) =4 × tan B ,and now tan( A - B ) = ( tanA - tanB ) / (1 + tanA * tanB) = 3 tanB / (1 +4 tan^2 (B)) = 3 / ( （1 / tanB ）+ 4 tanB )

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