求值:tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)=
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求值:tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)=
求值:tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)=
求值:tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)=
tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)=2010/2=1005
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求值:tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)=
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这道题为什么第二步的+到了第三步变成了-1.tan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-
已知tanα/2=2,求tanα与tan(α+π/4)