求值[1/cos^2(80°)-3/cos^2(10°)]1/cos20°
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/25 23:35:33
求值[1/cos^2(80°)-3/cos^2(10°)]1/cos20°求值[1/cos^2(80°)-3/cos^2(10°)]1/cos20°求值[1/cos^2(80°)-3/cos^2(10
求值[1/cos^2(80°)-3/cos^2(10°)]1/cos20°
求值[1/cos^2(80°)-3/cos^2(10°)]1/cos20°
求值[1/cos^2(80°)-3/cos^2(10°)]1/cos20°
[1/(cos80°)^2-3/(cos10°)^2] 1/cos20°
=[(1/cos80 + √3/cos10) * (1/cos80 - √3/cos10)] *[1/cos20]
=[(1/sin10 + √3/cos10) * (1/sin10 - √3/cos10)] *[1/cos20]
=[(cos10+√3sin10)/sin10cos10 * (cos10-√3sin10)/sin10cos10] *[1/cos20]
=[4sin40/sin20 * 4cos70/sin20] *[1/cos20]
=[16sin40/sin20] *[1/cos20]
=[32cos20] *[1/cos20]
=32