tanπ/12-1/tanπ/12=

tan(π/12)-(1/tan(π/12)) 计算tanπ/12 / (1-tan^2π/12)= tan(π/12)+1/tan(π/12)=啥? tanπ/12-1/tanπ/12= (1-tan^2π/12)/(tanπ/12)化简 计算 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 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] tan(x-π/12)/tan(x+π/12)=1/3,则tanx的值为 化简(tan5π/4+tan5π/12)/(1-tan5π/12)(tan 5π/4 + tan 5π/12)/(1-tan 5π/12) =(tan π/4 + tan π/6+π/4 )/(1-tan π/4*tan π/6+π/4 ) =tan(π/4 + π/6+π/4 )=tan(π/2+π/6) 请问我这样算错在哪呢?tan π/2不是无意义吗? 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)=?