计算[3/(sin^2)20°-[1/(cos^2)20°]+64(sin^2)20°
问题描述：
计算[3/(sin^2)20°-[1/(cos^2)20°]+64(sin^2)20°
计算[3/(sin^2)20°]-[1/(cos^2)20°]+64(sin^2)20°
答
原式
=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2
=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos20)^2+64(sin20)^2
=[2sin(60+20)*2sin(60-20)]/[(sin40)/2]^2+64(sin20)^2
=16sin80sin40/sin40*sin40+64(sin20)^2
=16sin80/sin40+64(sin20)^2
=32sin40cos40/sin40+64(sin20)^2
=32cos40+64(sin20)^2
=32[1-2(sin20)^2]+64(sin20)^2
=32