设tan(π+α)=2,则sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)=
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设tan(π+α)=2,则sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)=设tan(π+α)=2,则sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)=设t
设tan(π+α)=2,则sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)=
设tan(π+α)=2,则sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)=
设tan(π+α)=2,则sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)=
tan(π+α)= tan(α) =2,sinα≠0,cosα≠0
sin(α-π)+cos(π-α)/sin(π+α)-cos(π+α)
=-sin(α)-cos(α)/-sin(α)+cos(α)
=-tan(α)-1/-tan(α)+1
=-2-1/-2+1
=3
全部化简。。
先化简,再÷cosa试试
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