1/cos80°[(3/sin^2 40°)-(1/cos^2 40°)]

2个回答

  • (1/cos80°)(3/sin²40-1/cos²40)

    =(1/cos80°)(√3/sin40°+1/cos40°)(√3/sin40°-1/cos40°)

    =(1/cos80°)·[(√3cos40°+sin40°)/sin40°cos40°]·[(√3cos40°-sin40°)/sin40°cos40°]

    =(1/cos80°)·[4sin(40°+60°)/sin80°]·[4sin(60°-40°)/sin80°]

    =(1/cos80°)(4sin100°/sin80°)·(4sin20°/sin80°)

    =(1/cos80°)·(4sin80°/sin80°)·(4sin20°/sin80°)

    =32sin20°/sin160°

    =32sin20°/sin20°

    =32.