求证：tan(x+y)+tan(x-y)=sin2x/cosx*cosx-siny*siny
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tan(x+y)+tan(x-y)
=sin(x+y)/cos(x+y)+sin(x-y)/cos(x-y)
=[sin(x+y)cos(x-y)+cos(x+y)sin(x-y)]/[cos(x+y)cos(x-y)]
=sin2x/[(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)]
=sin2x/[(cosx)^2(cosy)^2-(sinx)^2(siny)^2]
=sin2x/[(cosx)^2(cosy)^2-(1-(cosx)^2)(1-(cosy)^2)]
=sin2x/[(cosx)^2+(cosy)^2-1]
=sin2x/[(cosx)^2-(siny)^2]
得证
=sin(x+y)/cos(x+y)+sin(x-y)/cos(x-y)
=[sin(x+y)cos(x-y)+cos(x+y)sin(x-y)]/[cos(x+y)cos(x-y)]
=sin2x/[(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)]
=sin2x/[(cosx)^2(cosy)^2-(sinx)^2(siny)^2]
=sin2x/[(cosx)^2(cosy)^2-(1-(cosx)^2)(1-(cosy)^2)]
=sin2x/[(cosx)^2+(cosy)^2-1]
=sin2x/[(cosx)^2-(siny)^2]
得证
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