tan(x+y)+tan(x-y)
=sin(x+y)/cos(x-y)+sin(x-y)/cos(x-y)
=[sin(x+y)*cos(x-y)+sin(x-y)*cos(x+y)]/[cos(x+y)*cos(x-y)]
=sin(x+y+x-y)/[ (cosxcosy)^2-(sinxsiny)^2]
=sin2x/[(cosx)^2(1- (siny)^2)-(1-(cosx)^2) (siny)^2]
= sin2x/[(cosx)^2-(cosx)^2(siny)^2-(siny)^2+(cosx)^2(siny)^2]
= sin2x/[(cosx)^2-(siny)^2]
所以腊源慎裂猜原式成立。轮敬