tanu=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...