求证.(tanx+tany)/(tanx-tany)=(sinx+siny)/(si

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求证.(tanx+tany)/(tanx-tany)=(sinx+siny)/(si
求证.(tanx+tany)/(tanx-tany)=(sinx+siny)/(sinx-siny)

1、sin(x+y）/sin(x-y)
=(sinxcosy+cosxsiny)/(sinxcosy-cosxsiny) 上下都除以cosxcosy
=(tanx+tany)/(tanx-tany)
2、(tanx+tany)/(tanx-tany)
=(sinx/cosx+siny/cosy)/(sinx/cosx-siny/cosy)
=[(sinxcosy+cosxsiny)/cosxcosy]/[(sinxcosy-cosxsiny)/cosxcosy]
=(sinxcosy+cosxsiny)/(sinxcosy-cosxsiny)
=sin(x+y)/sin(x-y).

求证tanx+tany/tanx-tany=sin（x+y）/sin（x-y）（详细步骤） (1)已知sinx=a,a为正常数,求tanx的值 (2)已知tany=b,b为常数,求siny的值已知sinx+siny=2sin(x+y)≠0,则tanx／2tany／2等于是三角函数和差化积部分的证明 (tanX+tanY)/(tanX-tanY)=(sin(X+Y))/(sin(X-Y)) 已知sinx+cosy=3/4,cosx+cosy+1/2,求tanx*tany 若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz成立.用此结论来证明恒等式 tanx+tany=25 cotx+coty=30,那么tan(x+y)=? 已知tanx=2,tany=1/3,则tan2(x+y)= tanx+tany=25,cotx+coty=30.tan(x+y)=? cos(x+y)=1/5,cos(x-y)=3/5则tanx.tany=? tanX=2分之1,tan（x+y）=5分之2,那么tanY=? 已知tanx=2,tany=3,x,y∈(0,π/2),求x+y.