Question

高一数学求解：tan70°cos10°(√3tan20°-1)

要过程，得数。重点是过程…

Answer 1

tan70°cos10°(√3tan20°-1)
=sin70°cos10°(√3sin20°-cos20°)/(cos70°cos20°)
=-2cos20°cos10°sin(30°-20°)/(sin20°cos20°)
=-cos20°sin20°/(sin20°cos20°)
= -1

Answer 2

tan70cos10(√3tan20-1)
=2cot20cos10(√3sin20/cos20-1)
=2cot20cos10(√3/2sin20-1/2cos20)/cos20
=2(cos20/sin20)cos10(sin20cos30-cos20sin30)/cos20
=(2/sin20)cos10sin(20-30)
=(-2sin10cos10)/sin20
=-1

sin50(1+√3tan10)
=sin50(1+tan60tan10)
=sin50[1+(sin60/cos60)*(sin10/cos10)]
=sin50[1+(sin60*sin10)/(cos60*cos10)]
=sin50[(sin10*sin60+cos10*cos60)/(cos10*cos60)]
=sin50[cos(60-10)/(cos10*cos60)]
=sin50[cos50/(cos10*cos60)]
=(sin50*cos50)/(cos10*cos60)
=sin100/(2*cos10*cos60)
=sin80/(2*cos10*cos60)
=cos10/(2*cos10*cos60)
=1/(2cos60)
=1

Answer 3

tan70°cos10°(√3tan20°-1)
=2tan70°cos10°[(√3/2)sin20°-(1/2)cos20°]/cos20°
=2tan70°cos10°sin(30°-20°)/cos20°
=2tan70°cos10°sin10°/cos20°
=sin70°sin20°/(cos20°cos70°)
=cos20°cos70°/(cos20°cos70°)
=1