△ABC的内角A,B,C所对的边分别为a,b,c,且acosB-bcosA=3/5c,则tan(A-B)的最大值是_.

问题描述:

△ABC的内角A,B,C所对的边分别为a,b,c,且acosB-bcosA=

3
5
c,则tan(A-B)的最大值是______.

∵a=2RsinA,b=2RsinB,c=2RsinC,∴2RsinAcosB-2RsinBcosA=352RsinC,即sinAcosB-sinBcosA=35sinC,①∵sinC=sin[π-(A+B)]=sin(A+B)=sinAcosB+cosAsinB,②将②代入①中,整理得sinAcosB=4cosAsinB,∴sinAcos...