是由tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)得出的原式求tan(π/6 -θ )+tan(π/6+θ)+√3tan(π/6-θ)tan(π/6+θ) 先看前一部分 由 tan(A+B) = (tanA+tanB)/(1-tanAtanB) tan(π/6 -θ )+tan(π/6+θ)=ta...
tan(π/6 -θ )+tan(π/6+θ)+根号3tan(π/6-θ)tan(π/6+θ)化简这个式子,
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