Question

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

Answer 1

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 ) <= 3 / 4, and when tan B = 1/2 the equality stands.

End of Answer, hope it helps

Answer 2

（2）acosB-bsinA=3/5c可化为sinAcosB-sinBcosA=3/5sinC又sinC=sin(A+B)所以可化为tanA=4tanB所以tan（A-B)=(tanA-tanB)/(1+tanAtanB)=3/(1/tanB+4tanB)所以最大值为3/4当tanB=1/2