由正弦定理,题目式子化为
sinAcosB-sinBcosA=3/5sinC
sinAcosB-sinBcosA=3/5sin(A+B)
则sinAcosB-sinBcosA=3/5sinAcosB+3/5sinBcosA
2/5sinAcosB=8/5sinBcosA
sinAcosB=4sinBcosA
tanA=4tanB
tan(A-B)=3tanB/(1+4tanBtanB)
tanC=-tan(A+B)=5tanB/(4tanBtanB-1)
好吧...上面都是扯淡.看下面
((2tanB)-1)^2≥0
4tan²B+1-4tanB≥0
4tan²B+1≥4tanB
tanB/4tan²B+1≤¼
3tanB/(1+4tan²B)≤¾