Question

(√3tan12º-3)/sin12º(cos12º的平方-2)

Answer 1

= [√3*tan12° - 3]/{sin12° * 2 * [2(cos12°)^2 - 1]}
= √3*(tan12° - √3) /{sin12° * 2* cos24°}
= √3*(tan12° - tan60°)/ (2*sin12°*cos24°)
= √3* (sin12°/cos12° - sin60°/cos60°) /(2*sin12° *cos24°)
= √3 *[(sin12° *cos60° - sin60° *cos12°) /(cos12° *cos60°)]/(2*sin12° * cos24°)
= √3 * sin (12° - 60°) /[cos60° * (2*sin12° *cos12°) * cos24°]
= √3 * sin (-48°) /[cos60° * sin24° * cos24°]
= -√3 * sin48° /[1/2 * 1/2 * (2*sin24° * cos24°)]
= -√3 * sin48° /[sin48° * 1/4]
= -4√3