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

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.

sin(x+y)=sinxcosy+cosxsiny=1/2;
sin(x-y)=sinxcosy-cosxsiny=1/3;
所以sinxcosy=5/12,cosxsiny=1/12,两式相比得tanx/tany=5,即tanx=5tany
注意到tan(x+y)=(tanx+tany)/(1-tanxtany)
所以tan(x+y)-tanx-tany=(tanx+tany)(tanxtany)/(1-tanxtany)
(tany)^2tan(x+y)= (tany)^2(tanx+tany)/(1-tanxtany)
从而[tan(x+y)-tanx-tany]/[(tany)^2tan(x+y)]=
(tanx+tany)(tanxtany)/(tany)^2(tanx+tany)=tanx/tany=5