化简tan(x+π/4)-tan(x-π/4)
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化简tan(x+π/4)-tan(x-π/4)化简tan(x+π/4)-tan(x-π/4)化简tan(x+π/4)-tan(x-π/4)tan(x+π/4)-tan(x-π/4)=(tanx+tan
化简tan(x+π/4)-tan(x-π/4)
化简tan(x+π/4)-tan(x-π/4)
化简tan(x+π/4)-tan(x-π/4)
tan(x+π/4)-tan(x-π/4)
=(tanx+tan(π/4))/(1-tanx*tan(π/4))-(tanx-tan(π/4))/(1+tanx*tan(π/4))
=(tanx+1)/(1-tanx)-(tanx-1)/(1+tanx)
=((tanx+1)²+(tanx-1)²)/(1-tan²x)
=2(tan²x+1)/(1-tan²x)
tan(x+π/4)-tan(x-π/4)=(tanx+tanπ/4)/(1-tanx*tanπ/4)-(tanx-tanπ/4)/(1+tanx*tanπ/4)=(tanx+1)/(1-tanx)-(tanx-1)/(1+tanx)=2(tan^2x+1)/(1-tan^2x)
=(tanx+1)/(1-tanx)+(1-tanx)/(1+tanx)
=[(1+tanx)^2+(1-tanx)^2]/1-(tanx)^2
=2*[1+(tanx)^2]/[1-(tanx)^2]
要化简到什么程度。利用公式拆开，通分不能算结束么？
化简tan(x+π/4)+tan(x+π/4)
化简tan(x+π/4)-tan(x-π/4)
tan(-x+π/4)
tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanxtan(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)-1)^2]/[1-(tan(x
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)-
lim(2-x)tanπ/4x
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