题目

化简 tan70°cos10°(√3tan20°-1)
还有一个：
sin50°(1+√3tan10°)

提问时间：2020-10-20

答案

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