sin（x+y)=1\2,sin(x—y）=1\3,求[tan（x+y）-tanx-tany]\[tany的平方tan（

sin(x+y)=sinxcosy+cosxsiny=1/2 sin(x-y)=sinxconsy-cosxsiny=1/3 sinxcosy=5/12,cosxsiny=1/12 tanx/tany=sinxcosy/cosxsiny=5。[tan（x+y）-tanx-tany]/[tany的平方tan（x+y)] =[(tanx+tany)/(1-tanxtany)-(tanx+tany)]/[tany^2(tanx+tany)/(1-tanxtany)] =[tanxtany(tanx+tany)]/[tany^2(tanx+tany)] =tanx/tany =5。