设tanx=3tany(0

此题需利用一下转换这一思想。
设S=tan(x-y)，则利用三角公式化简，S=(tanx-tany)/(1+tanxtany).
又tanx=3tany，于是S=2tany/(1+3tan²y)=2/(3tany+1/tany).
而0≤y≤x因此S{max}=√3/3.
于是x-y{max}=arctan(S{max})=π/6。

因0≤y≤xtan(x-y)=(tanx-tany)/(1+tanxtany)=2tany/((1+3tan²y)=2/[(1/tany)+3tany]
≤2/2根号3=根号3/3=tanπ/6,
由tanx在[0,π/2)是增函数，所以x-y≤π/6,即x-y的最大值是π/6.

tan(x-y)=(tanx-tany)/(1+tanxtany)
=2tany/(1+3tan^2y)
=2/(1/tany+3tany)
≤2/2√(1/tany*3tany)
=1/√3
=√3/3
=tanπ/6
x-y的最大值为：π/6

tan(x-y)
=(tanx-tany)/(1+tanxtany)
=(3tany-tany)/(1+3tanytany)
=2tany/[1+3(tany)^2]
=2/(1/tany+3tany)
1/tany+3tany≥2√[(1/tany)*(3tany)]
当1/tany=3tany时，1/tany+3tany有最小值2√3，tanu有最大值√3/3
在[0,π/2)，tan是增函数
所以x-y有最大值π/6