化简并求值(tan10-√3)*cos10/sin50*(2cos10-sin20)/cos20紧急
1个回答

(tan10°-√3)×(cos10°/sin50°)×[(2cos10°-sin20°)/cos20°]

①(tan10°-√3)×(cos10°/sin50°)

=[(tan10°-√3)×cos10°]/sin50°

=(sin10°-√3cos10°)/sin50°

=[2sin(10°-60°)]/sin50°

=(-2sin50°)/sin50°

=-2

②(2cos10°-sin20°)/cos20°

=[2cos(30°-20°)-sin20°]/cos20°

=(2cos30°cos20°+2sin30°sin20°-sin20°)/cos20°

=[2×(√3/2)×cos20°+2×(1/2)×sin20°-sin20°]/cos20°

=(√3cos20°+sin20°-sin20°)/cos20°

=(√3cos20°)/cos20°

=√3

∴原式=(-2)×√3=-2√3