求[3/(sin140)^2 -1/(cos140)^2]*[1/(但老师说是-16
问题描述：
求[3/(sin140)^2 -1/(cos140)^2]*[1/(
但老师说是-16
答
[3/(sin140)^2 -1/(cos140)^2]*[1/(2sin10)]
原式=(3cos²40-sin²40)/[2sin²40cos²40sin10]
=2(√3cos-sin40)(√3cos40+sin40)/[sin²80sin10]
=8sin(60-40)sin(60+40)/(cos²10sin10)
=16sin20sin100/(cos10sin20)
=16cos10/cos10
=16