作业帮tan(x+π/4)+tan(x-π/4)化简得
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tan(x+π/4)+tan(x-π/4)化简得
tan(x+π/4)+tan(x-π/4)化简得
tan(x+π/4)+tan(x-π/4)化简得
tan(x+π/4)+tan(x-π/4)
设a=x+π/4,则x-π/4=a-π/2
原式=tana+tan(a-π/2)
=tana-cota
=sina/cosa-cosa/sina
=-2cos2a/sin2a
=-2cot2a
=-2cot(2x+π/2)
=2cot(-2x-π/2)
=2cot(π/2-2x)
=2tan2x
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
化简tan(x+π/4)+tan(x+π/4)
化简tan(x+π/4)-tan(x-π/4)
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
求tan(x+π/4)的定义域
tan(x+π/4)的对称中心
(tanπx/4)^(tanπx/2)当x趋向于1时的极限?
x→1,求lim（tanπx/4）^tanπx/2求极限,
求极限.(tan x)^(tan 2x) x→π/4
化简tan(x/2+π/4)-tan（π/4-x/2）
化简 tan(x/2 + π/4)-tan(π/4 - x/2)
tan(x+π/4)+tan(x-π/4)化简得
tan(x/2+ π4)+tan(x/2- π/4)=2tanx证明
求证:tan（x/2+π/4）+tan(x/2-π/4)＝2tanx
解方程：tan(x+π/4)+tan(x-π/4)=2cotx
化简：tan(x+π)-tan(π/4-x)的结果为.第二题