tan(π/6-α)+tan(π/6+α)+√3tan(π/6-α)tan(π/6+α)=?

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tan(π/6-α)+tan(π/6+α)+√3tan(π/6-α)tan(π/6+α)=?tan(π/6-α)+tan(π/6+α)+√3tan(π/6-α)tan(π/6+α)=?tan(π/6-

tan(π/6-α)+tan(π/6+α)+√3tan(π/6-α)tan(π/6+α)=?
tan(π/6-α)+tan(π/6+α)+√3tan(π/6-α)tan(π/6+α)=?

tan(π/6-α)+tan(π/6+α)+√3tan(π/6-α)tan(π/6+α)=?
(π/6-α)+(π/6+α)=π/3
tan(π/3)=√3

即 √3=[tan(π/6-α)+tan(π/6+α)]/[1-tan(π/6-α)tan
(π/6+α)]
所以为√3

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