化简 tan(x/2 + π/4)-tan(π/4 - x/2)
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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)
化简tan(x/2+π/4)-tan(π/4-x/2)
化简 tan(x/2 + π/4)-tan(π/4 - x/2)
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
tan(-x+π/4)
已知函数f(x)=tan(2x+π/4),(1)求f(x)的定义域与最小正周期 (2)设α∈(0,π/4),已知函数f(x)=tan(2x+π/4) (1)求f(x)的定义域与最小正周期(2)设α∈(0,π/4),若f(α/2)=2cos2α,求α的大小(1)解析:∵函数f(x)=ta
(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+π)-tan(π/4-x)的结果为.第二题
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
tan(π-x)=-1/2,则tan(π/4-x)=
∫ ( tan^2 x + tan^4 x )dx