y'=y/x+tany/x解：令u=y/x u=(u)x y=xu

tan（x-y）=(tanx-tany)/(1+tanxtany)=-1/7

tan²x=2tan²y+1
1+tan²x=2(1+tan²y)
sec²x=2sec²y
cos²x=2cos²y
1-sin²x=2(1-sin²y)
sin²x=2sin²y-1

cos(x+y)=cosxcosy-sinxsiny=3/5 (1)
cos(x-y)=cosxcosy+sinxsiny=4/5 (2)
(1)+(2) 2cosxcosy=7/5 (3)
(2)-(1) 2sinxsiny= 1/5 (4)
（4）/ (3)
tanx*tany=1/7

sin (x+y) =1
那么：x+y = 2kπ +π/2
tan(2x+y)+tany = tan [(x +y)+x] + tany
= tan( 2kπ +π/2 +x) +tan y
= tan (π/2 +x) + tan y
= tan(π/2 +x) + tan(2kπ +π/2 -x) (x+y = 2kπ +π/2)
= - cotx + cotx
=0

2sinxcosy=sin(x+y)+sin(x-y)=5/6
2cosxsiny=sin(x+y)-sin(x-y)=1/6
两式相除得所求=5

答案为2 tan（x+y)=-1,tan(x＋y）＝(tanx+tany)/1-tanxtany:所以tanx+tany=tanxtany-1,又因为原式=1-tanx-tany+tanxtany,用tanx+tany=tanxtany-1替代,原式=1+1-tanxtany+tanxtany=2.

(1-tanx)(1-tany)
=1-(tanx+tany)+tanxtany.(1)
因为x+y=3π/4
而且tan(x+y)=(tanx+tany)/(1-tanxtany)
tan(x+y)=tan(3π/4)=-1
(tanx+tany)/(1-tanxtany)=-1
tanx+tany=tanxtany-1.(2)
将(2)式子代入(1)式子
得(1-tanx)(1-tany)=2

cos(x+y)+cos(x-y)=2cos[(x+y+x-y)/2]cos[(x+y-x+y)/2]=2cosx*cosy=4/5
cos(x+y)-cos(x-y)=-2sin[(x+y+x-y)/2]sin[(x+y-x+y)/2]=-2sinx*siny=-2/5
cosx*cosy=2/5
sinx*siny=1/5
tanx.tany=(sinx*siny)/(cosx*cosy)=1/2

x+y=π/4,
tanx+tany=1-tanxtany,
(1+tanx)(1+tany)
=1+(tanx+tany)+tanxtany
=2.
积化和差公式
sin(π/4+x）*sin(π/4-x）
=－1/2[cos π/2-cos2x]
=1/6.
cos2x=1/3>0,
又x∈（－2π,π）,

由题易得
tanx+tany=7/3
tanx*tany=2/3
所以
tanx=1/3
tany=2
sin x =1/根号10
sin y =2/根号5
原式=3/2*2/3*cos x *cos y +sin x *sin y
=3/2*tan x *tan y *cos x *cos y +sin x *sin y
=5/2*sin x *sin y
=根号2/2

sin（x-y）/sin(x+y)=(sinx*cosy-cosx*siny)/(sinx*cosy+cosx*siny) 1
tanx/tany=(sinx/cosx)*(siny/cosy)=(sinx*cosy)/(cosx*siny) 2
1式各项同除cosx*siny=[(sinx*cosy)/(cosx*siny)-1]/[(sinx*cosy)/(cosx*siny)+1] 3
假设z=tanx/tany=(sinx*cosy)/(cosx*siny) 则3式等于 (z-1)/(z+1)=2009/2010 4
求解4式得到 z=tanx/tany=4019