求助一道三角函数化解求值题3/sin二20°－1/cos二20°＋64sin二20°(打不出平方,姑且用二代替吧,希望各

3/(sin20°)^2-1/(cos20°)^2
=[3(cos20°)^2-(sin20°)^2]/(sin20°cos20°)^2
=4(√3cos20°-sin20°)(√3cos20°+sin20°)/(sin40°)^2
=16cos(20°+30°)sin(20°+60°)/(sin40°)^2
=16sin40°cos10°/(sin40°)^2
=16cos10°/sin40°
原式=16cos10°/sin40°+64(sin20°)^2
=16cos10°/sin40°-32cos40°+32
=16(cos10°-2sin40°cos40°)/sin40°+32
=16(cos10°-sin80°)/sin40°+32
=32