又是三角恒等变换

上面明显做烦了

tany=(cosx-sinx)/(cosx+sinx),
将(cosx-sinx)/(cosx+sinx)分子分母同时除以cosx
即:tany=(cosx-sinx)/(cosx+sinx)=(1-tanx)/(1+tanx)
所以tany*(1+tanx)=1-tanx
即：tanx+tany=1-tanx*tany

则tan(x+y)=(tanx+tany)/(1-tanx*tany)=1

tan(x+y)=(tanx+tany)/(1-tanxtany)
tanx+tany
=(sinx/cosx)+[(cosx-sinx)/(cosx+sinx)]
=(sin^2x+sinxcosx+cos^2x-sinxcosx)/cosx(sinx+cosx)
=1/cosx(sinx+cosx)
1-tanxtany
=1-[sinx(cosx-sinx)/cosx(sinx+cosx)]
=(cos^x+cosxsinx-sinxcosx+sin^2x)/cosx(sinx+cosx)
=1/cosx(sinx+cosx)
∴tan(x+y)=(tanx+tany)/(1-tanxtany)
=1