化简：tan(π/8)+1/tan(π/12)

化简：tan(π/8)+1/tan(π/12) (1-tan^2π/12)/(tanπ/12)化简 (tan(5π/8)*tan(π/8))为什么等于-1 tanπ/8/(1-tan^2π/8) tan(π/8)+1/(tan(π/12))值,要过程 tan(π/12)-(1/tan(π/12)) tanπ/8-1/tanπ/8是（tanπ/8）-〔1/（tanπ/8）〕。 三角函数算值题1-tan^2(π/8) /tan(π/8)=tan(15)/1-tan^2(15)= 化简：[tan(5/4)π+tan(5/12)π]/[1-tan(5/12)π] 化简：[tan（5π）/4+tan(5π)/12]/[1-tan(5π)/12] 求tan(π/8) tan(π/8)=? （tanπ/4+tanα)/(1-tanπ/4tanα)怎么解 化简tan＋1/tan 利用正切函数单调性比较 函数值大小tan [(75/11)π]&tan [(-58/11)π]这样可不可以 下面算式有错吗tan[(75/11)π] = tan[(9/11)π] =tan[(1-2/11)π] = -tan 2/11 πtan [(-58/11)π] = tan [-(8/11)π] = -tan[(1-3/11)π] = tan 3/11 π 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 求证（1）1+tanθ/1-tanθ=tan(π/4+θ) （2）1-tanθ/1+tanθ=tan(π/4-θ） 证明tan(α)*tan(β)+tan(β)*tan(γ)+tan(α)*tan(γ)=1 (α+β+γ=π/2)详细一点