（tan12度-√3）/[2(cos12度)^2-1]sin12度

(tan12-√3)/{[2(cos12)^2-1]sin12}
[2(cos12)^2-1]sin12
=[2*(1+cos24)/2-1]sin12
=cos24sin12
tan12-√3
=(sin12-√3cos12)/cos12
=2sin(12-60)/cos12
=-2sin48/cos12
=-4sin24cos24/cos12
=-8sin12cos12cos24/cos12
=-8cos24sin12
所以(tan12-√3)/{[2(cos12)^2-1]sin12}=-8