tan(x-π/12)/tan(x+π/12)=1/3,则tanx的值为

设 a = tan(Pi/12) = tan( Pi/3 - Pi/4) = .= 2- √3
tan(x-π/12)/tan(x+π/12)=1/3 => 3(tanx - a)(1 - a*tanx) - (tanx + a)(1 + a* tanx) = 0
令 y= tanx
整理：y^2 - [ (1 + a^2) / (2a)] * y + 1 = 0
a=2- √3,(1 + a^2) / (2a) = 2 ,=> (y-1)^2 = 1
y=1,即 tanx =1