计算tanπ/12 / (1-tan^2π/12)=

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计算tanπ/12 / (1-tan^2π/12)=
计算tanπ/12 / (1-tan^2π/12)=

计算tanπ/12 / (1-tan^2π/12)=
tanπ/12 / (1-tan^2π/12)
=1/2*(2tanπ/12) / (1-tan^2π/12)
=1/2*tan(π/12+π/12)
=1/2*tanπ/6
=1/2*√3/3
=√3/6

tanx/(1-tan^2x)
=(sinx/cosx)/(1-sin^2x/cos^2x)
=(sinx/cosx)/((cos^2x-sin^2x)/cos^2x)
=sinxcosx/(1-sin^2x-sin^2x)
=2sinxcosx/2cos2x
=sin2x/2cos2x
=(1/2)tan2x
将x=π/12代入
原式=(1/2)tan2*π/12
=(1/2)tanπ/6
=√3/6

计算tanπ/12 / (1-tan^2π/12)= 计算 tanπ/12+1/tanπ/12 (1-tan^2π/12)/(tanπ/12)化简 tan(π/12)-(1/tan(π/12)) 计算：1-tan（5π/12）tan(π/4） ------------------------------- tan(5π/12）+tan(π/4）那条横杠是分数线计算 tanπ/12-tan5π/12 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 计算tanπ/12的值求证（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/12)+1/tan(π/12)=啥? tanπ/12-1/tanπ/12= 化简：tan(π/8)+1/tan(π/12) tan(π/8)+1/(tan(π/12))值,要过程 tan(π/2+1)=?tan(1-π/2)=? tanπ/8/(1-tan^2π/8)