Question

两道高中数学题火速求解QAQ

1/sin10-根号3/cos10 sin50(1+根号3tan10)
请给过程谢谢

Answer 1

①1/sin10-根号3/cos10
原式=(cos10-√3sin10)/sin10cos10
=4[(1/2)cos10-(√3/2)sin10]/2sin10cos10
=4(cos10cos60-sin60sin10)/sin20
=4cos(10+60)/sin20
=4cos70/sin20
=4sin20/sin20
=4

②sin50(1+根号3tan10)
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

Answer 2

采纳可见

Answer 3

fcwe