tanπ/12-1/tanπ/12=
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tanπ/12-1/tanπ/12=tanπ/12-1/tanπ/12=tanπ/12-1/tanπ/12==(sinπ/12)/(cosπ/12)-(cosπ/12)/(sinπ/12)=（sin&
tanπ/12-1/tanπ/12=
tanπ/12-1/tanπ/12=
tanπ/12-1/tanπ/12=
=(sinπ/12)/(cosπ/12)-(cosπ/12)/(sinπ/12)
=（sin²π/12-cos²π/12）/(sinπ/12*cosπ/12)
=2*（sin²π/12-cos²π/12）/(sinπ/6)
=2*(-cosπ/6)/(sinπ/6)
=-2/(tanπ/6)
=-2√3
(公式：tanA=sinA/cosA,sin2A=2sinAcosA,cos2A=2cos²A-1=1-2sin²A=cos²A-sin²A)
=sin15°/cos15°-cos5°/sin15°
=（sin²15°-cos²15°）/sin15°cos15°
=2（sin²15°-cos²15°）/sin30°
=-4cos30°=-2√3
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)=?