Question

高中数学。解三角，化简。求上图几道题的详细过程。

Answer 1

1/sin10-√3/cos10
=2*(sin30/sin10-cos30/cos10)
=2(sin30cos10-sin10cos30)/(sin10cos10)
=4(sin(30-10))/sin20
=4
2.
sin40°（tan10°-根号3）
=sin40°(sin10°/cos10°-根号3)
=sin40°（sin10°-根号3cos10°)/cos10°
=2sin40°sin(10°-60°)/cos10°
=-2sin40°cos40°/cos10°
=-sin80°/sin80°
=-1
3.
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
4.
sin50（1+√3tan10）
=sin50[1+√3(sin10/cos10)]
=sin50[(cos10+√3sin10)/cos10]
=sin50[2sin(30+10)/cos10]
=(2sin50sin40)/cos10°
=(2cos40°*sin40°)/cos10°
=sin80°/cos10°
=cos10°/cos10°
=1