若α+β=π/6,且根号3(tanα-tanβ-2)+2tanα+3tanβ=0,则tanα=?
问题描述：
若α+β=π/6,且根号3(tanα-tanβ-2)+2tanα+3tanβ=0,则tanα=?
答
tan（π/6）=tan(a+b)=(tana+tanb)/(1-tanatanb)=√3/3
所以3(tana+tanb)=√3-√3tanatanb
√3（tanαtanβ+2）+2tanα+3tanβ
=√3tanαtanβ+2√3+3tanα+3tanβ -tana
=√3tanαtanβ+2√3+√3-√3tanatanb-tana
=2√3+√3-tana=0 tana
=2√3+√3 1 则tanα=3√3