1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx（2）（1-2sinαcosα/cos²α-sin²α）=（1-tanα）/（1+tanα）

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)- 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, lim(2-x)tanπ/4x 极限(1-x)tanπx/2 tan(π-x）=-1/2,则tan(π/4-x)= 求证tan(x+y)*tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^2y 设f(x)=(tan(π/4)x-1)(tan(π/4)x²-2).(tan(π/4)x^100-100),求f'(1) 化简tan(x+π/4)+tan(x+π/4) 化简tan(x+π/4)-tan(x-π/4) ∫ ( tan^2 x + tan^4 x )dx 求不定积分∫(tan^2x+tan^4x)dx 不定积分(tan^2x+tan^4x)dx (tan x+1/tan)cos^2 x等于? ￡tan x(tan x+1)dx (tan x/2-π/6)