Question

化简① tan70°cos10°(根号3tan20°-1) ②sin50°(1+根号3tan10°) 求过程

Answer 1

tan70cos10*(√3tan20-1)
=sin70/cos70*cos10*(√3sin20/cos20-1)
=cos20/sin20*cos10*(√3sin20-cos20)/cos20
=cos10*(√3sin20-cos20)/sin20
=(√3sin20-cos20)/(2sin10)
=(1/sin10)*(sin20cos30-cos20sin30)
=1/sin10*sin(20-30)
=-1

原式=sin50°*[1+根号3(sin10°/cos10°)]
=sin50°*[(cos10°+根号3sin10°)/cos10°]
=sin50°*[2sin(30°+10°)/cos10°]
=(2sin50°*sin40°)/cos10°
=(2cos40°*sin40°)/cos10°
=sin80°/cos10°
=cos10°/cos10°
=1

　原式=sin50°[cos10°＋√3sin10°]/(cos10°)
=[(cos40°)/(cos10°)]×[2sin40°]
=sin80°/cos10°
=1

Answer 2

tan70×cos10×(√3tan20-1) =tan70×cos10×(tan60×tan20-1) =tan70×cos10×[(sin60×sin20/cos60×cos20)-1] =tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20) =tan70×cos10×[-cos(60+20)]/(cos60 ×cos20) =-tan70×cos10×cos80/(cos60×cos20) =-tan70×cos10×sin10/(cos60×cos20) =-(sin70/cos70)×(1/2)×sin20/(cos60×cos20) =-(cos20/sin20)×sin20/(2cos60×cos20) =-1/(2cos60) =-1