Question

[紧急求助，高中数学！！！] 化简:tan70度cos10度((根3)tan20度-1) 要详细步骤 辛苦

Answer 1

√3tan20-1=(√3sin20/cos20)-1=(√3sin20-cos20)/cos20=2(sin20cos30-cos20sin30)/cos20
=-2sin10/cos20
原式=-tan70cos10×2sin10/cos20
=-(sin20/cos20)tan70
=-tan20tan70
=-tan(90-70)tan70
=-1

Answer 2

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

Answer 3

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

Answer 4

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