求(√3tan12-3)/sin12(4cos12-2)的值
问题描述：
求(√3tan12-3)/sin12(4cos12-2)的值
答
是(√3tan12-3)/[sin12*(4(cos12)^2-2)]吧
原式=(√3sin12-3cos12)/[sin12*cos12*(4cos12-2)]=(√3sin12-3cos12)/[sin24*(2(cos12)^2-1)]=(√3sin12-3cos12)/(sin24*cos24)
=4√3(1/2*sin12-√3/2*cos12)/sin48
=4√3sin48/sin48=4√3