若tan(π/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx)

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 x→1,求lim（tanπx/4）^tanπx/2求极限, 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/4-x)=-1/2.求(2sinx-cosx)/(sinx+3cosx) 设f(x)=(tan(π/4)x-1)(tan(π/4)x²-2).(tan(π/4)x^100-100),求f'(1) 已知tan(π/4+x)=1/2,求tanx 若tan(3x+y)=2,tan(x+y)=-4 求tan2x 2tanα／(1-tanαtanα)=-4／3求tanα 2tanα／(1-tanαtanα)=-4／3求tanα要过程 tan(x/2)=2.求tanx tan(x+π/4)=-1/7求tanx 求y=tan(x/2 -π/4)的周期 y=tan(2x-π/4)求周期RTRTRTRR 若2sinx=1+cosx,求tan x/2 1.已知tanα和tan(π/4-α)是方程x^2+px+q=0的两个根,若3tanα=tan(π/4-α)求q,p2.已知tanα=√3(tanαtanβ+m),又α,β都是钝角,求α+β的值 已知tanα和tan(π/4-α)是方程x^2+px+q=0的两个根,若3tanα=tan(π/4-α)求q,p已知tanα=√3(tanαtanβ+m),又α,β都是钝角,求α+β的值 求函数y=-tan²x+4tan x+1,x∈[-π/4,π/4]的值域 tan(x-π/4)=2,求sin2x+2cos2x/2cos^x-3sin2x-1