tan((π/2)+x),tan((π/2)-x)与tanx之间的关系怎么来的?具体一些

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 (tan x/2-π/6) tan(π/2-x),tan(π/2+x),tan(x)有什么关系 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)- tan(-x+π/4) 极限(1-x)tanπx/2 lim(2-x)tanπ/4x y=tan(π/2-x)定义域 tanx+tan(π/2 -x)=? 满足不等式tan(2x+π/3) 已知tan(x-y)=2,tan(y+π/5)=3,则tan(x+π/5)等于 若a属于(0,π/2)试比较tanα、tan(tanα)、tan(sinα) 化简tan(x+π/4)+tan(x+π/4) 化简tan(x+π/4)-tan(x-π/4) (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 tan(x/2+ π4)+tan(x/2- π/4)=2tanx证明