作业帮化简tan(x+π/4)-tan(x-π/4)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/25 12:59:11
x){3<
&`.mTO@~
1 Th EZ!]]5hj
3jSSV\Z`d
SCx0=F! p5yv@spx g0pQ&ĵqFHʀ 4N@ h7$(Vb?
)hCm
t4B>^,k$ nx=nhzڱO[<鞆{z^6zdGM{d< %H 1
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
化简tan(x/2+π/4)-tan（π/4-x/2）
化简 tan(x/2 + π/4)-tan(π/4 - x/2)
化简：tan(x+π)-tan(π/4-x)的结果为.第二题
求tan(x+π/4)的定义域
tan(x+π/4)的对称中心
(tan x/2-π/6)
tan(π/2-x),tan(π/2+x),tan(x)有什么关系
(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(18-x)tan(12+x)+根号3[tan(18-x)+tan(12+x)]
tan(x+π/4)+tan(x-π/4)化简得