Question

求值：（√3tan12°-3)/(4(cos12°)^2-2)sin12°

Answer 1

（√3tan12°-3)/[(4(cos12°)^2-2)sin12° ]=√3(tan12°-√3)/[2(cos24°)*sin12° ]

=√3(tan12°-√3)*cos12°/[2(cos24°)*sin12°cos12° ]=2√3(sin12°-√3cos12°)/sin48°

=4√3sin(12°-60°)/sin48°=-4√3