Question

高一数学题（三角函数和向量等），求数学达人快速解答，不胜感激！！！

1.[1/(cos80°)^2-3/(cos10°)^2].1/cos20°

Answer 1

[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