Question

高一三角函数化简 tan70°cos10°（√3tan20°-1)

如题

Answer 1

tan70°cos10°（√3tan20°-1)
=(1/tan20°)cos10°(√3sin20°-cos20°)/cos20°
=(cos20°/sin20°)cos10°/cos20° *(√3sin20°-cos20°)
=cos10°*2sin(-10°)/sin20°
=-2sin10°cos10°/sin20°
=-sin20°/sin20°
=-1

Answer 2

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

Answer 3

记a=根号3
atan30-1=2(a/2sin20-1/2cos20)/cos20=-2sin10/cos20
所以原式=tan70*-tan20=-1