若1+tanx|1-tanx=2008,则1|cos2x+tan2x=?2.已知sinx+siny=sin225,cosx+cosy=cos225,求cos(x-y)的值

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若1+tanx|1-tanx=2008,则1|cos2x+tan2x=?2.已知sinx+siny=sin225,cosx+cosy=cos225,求cos(x-y)的值若1+tanx|1-tanx=

若1+tanx|1-tanx=2008,则1|cos2x+tan2x=?2.已知sinx+siny=sin225,cosx+cosy=cos225,求cos(x-y)的值
若1+tanx|1-tanx=2008,则1|cos2x+tan2x=?
2.已知sinx+siny=sin225,cosx+cosy=cos225,求cos(x-y)的值

若1+tanx|1-tanx=2008,则1|cos2x+tan2x=?2.已知sinx+siny=sin225,cosx+cosy=cos225,求cos(x-y)的值
1/cos2x+tan2x=(1+sin2x)/cos2x
=(sin²x+cos²x+2sinxcosx)/(cos²x-sin²x)
=(sinx+cosx)²/[(cosx+sinx)(cosx-sinx)]
=(sinx+cosx)/(cosx-sinx)
=(tanx+1)/(1-tanx)
=2008
sinx+siny=sin225°=sin(180°+45°)=-sin45°=-√2/2
cosx+cosy=cos225°=cos(180°+45°)=-cos45°=-√2/2,
那么(sinx+siny)²=sin²x+2sinxsiny+sin²y=1/2
(cosx+cosy)²=cos²x+2cosxcosy+cos²y=1/2
两式相加,得:1+1+2(cosxcosy+sinxsiny)=1
那么2+2cos(x-y)=1
所以cos(x-y)=-1/2