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 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