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 )