Question

tan(60°+a)-tan(30°+a)/{1+tan(60°+a)tan(30°+a)} 具体过程 谢谢

Answer 1

tan(α-β)=(tanα-tanβ)/(1-tanαtanβ)

[tan(60°+α)-tan(30°+α)]/[1+tan(60°+α)tan(30°+α)]

=tan[(60°+α)-(30°+α)]

=tan30°

=√3/3