(1+tan π/12)/(1-tan π/12)的值是
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(1+tanπ/12)/(1-tanπ/12)的值是(1+tanπ/12)/(1-tanπ/12)的值是(1+tanπ/12)/(1-tanπ/12)的值是(1+tanπ/12)/(1-tanπ/12
(1+tan π/12)/(1-tan π/12)的值是
(1+tan π/12)/(1-tan π/12)的值是
(1+tan π/12)/(1-tan π/12)的值是
(1+tan π/12)/(1-tan π/12)
=(tanπ/4+tan π/12)/(1-tanπ/4tan π/12)
=tan(π/4+π/12)
=tanπ/3
=√3
tan(π/12)-(1/tan(π/12))
(1-tan^2π/12)/(tanπ/12)化简
计算 tanπ/12+1/tanπ/12
计算tanπ/12 / (1-tan^2π/12)=
tan(π/12)+1/tan(π/12)=啥?
tanπ/12-1/tanπ/12=
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化简:[tan(5π)/4+tan(5π)/12]/[1-tan(5π)/12]
(tanπ/4+tanα)/(1-tanπ/4tanα)怎么解
化简tan+1/tan
tanα+1/tanα
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)
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证明tanαtanβ+tanαtanγ+tanβtanγ=1,α+β+γ=π/2
已知∠α+∠β+∠γ=π/2 求证tanαtanβ+tanαtanγ+tanβtanγ=1