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1.[1/(cos80°)^2-3/(cos10°)^2].1/cos20°
人气:409 ℃ 时间:2020-06-13 06:35:38
解答
[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
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