tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?

三角函数算值题1-tan^2(π/8) /tan(π/8)=tan(15)/1-tan^2(15)= tan(-14π/15)=-tan(π/15)=tan(-π/15)对吗? tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗? 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 证明tanαtanβ+tanαtanγ+tanβtanγ=1,α+β+γ=π/2 已知∠α+∠β+∠γ=π／2 求证tanαtanβ+tanαtanγ+tanβtanγ=1 tan(π/2+1)=?tan(1-π/2)=? 计算tanπ/12 / (1-tan^2π/12)= tan(π/12)+1/tan(π/12)=啥? tanπ/12-1/tanπ/12= 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(π/12)-(1/tan(π/12)) tan=3,π tan(π/8)=? （tanπ/4+tanα)/(1-tanπ/4tanα)怎么解 tan ( )=1/3