若tan (x+π/4)=2007 1+cos (2x)/cos (2x) + tan 2x 的值答案2008

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)- 若tan(π-α)=1/3,则tan(x+π/4)等于多少, 若tan (x+π/4)=2007 1+cos (2x)/cos (2x) + tan 2x 的值答案2008 tan(π-x）=-1/2,则tan(π/4-x)= 设f(x)=(tan(π/4)x-1)(tan(π/4)x²-2).(tan(π/4)x^100-100),求f'(1) tan(-x+π/4) 1+(tan x)=? 求函数y=-tan²x+4tan x+1,x∈[-π/4,π/4]的值域 若tan(π+x+y)=3/5,tan(y-π/3)=1/3则tan(x+π/3)=是π/3 不是整体除以3！ 证明1+sinx/cosx=tan(π/4+x/2) 求证：1+sin2a/cos2a=tan(π/4+x) (1+sin2x)/cos2x=tan(π/4+x) tan(x+π/4)=-1/7求tanx 求证：(1-sin2x)/cos2x=tan(π/4-x) 已知tan(π/4+x)=1/2,求tanx 高一数学y=tan(x+π/4)+1/tan(x+π/4)y=tan(x+π/4)+1/tan(x+π/4)求该函数的值域及单调递增区间 求证tan(x+y)*tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^2y