Question

sin50°(1+√3tan10°)=?

Answer 1

1+√3tan10
=1+√3sin10/cos10
=(cos10+√3sin10)/cos10
=2sin(10+30)/cos10
=2sin40/cos10

sin50(1+√3tan10)
=(2sin40sin50)/cos10
=[cos(50-40)-cos(50+40)]/cos10
=cos10/cos10
=1

Answer 2

sin50(1+√3tan10)
=2sin50[(1/2)+[ (√3/2)(sin10/cos10)]
=2sin50[sin30+cos30(sin10/cos10)]
=2sin50/cos10[sin30cos10+cos30sin10]
=2sin50/cos10*sin40
=(2cos40sin40)/cos10
=sin80/cos10
=cos10/cos10
=1

Answer 3

解sin50°(1+√3tan10°)
=sin50°(cos10°/cos10°+√3sin10°/cos10°)
=sin50°(cos10°+√3sin10°)/cos10°
=sin50°2(1/2cos10°+√3/2sin10°)/cos10°
=sin50°2(sin30°cos10°+cos30°sin10°)/cos10°
=sin50°2(sin(30°+10°))/cos10°
=sin50°2sin40°/cos10°
=2cos40°sin40°/cos10°
=sin80°/cos10°
=cos10°/cos10°
=1