Trigonometric equations $\tan\theta-\sec\theta=\sqrt3$ For the following problem(s) I cannot get any answer(s). I would appreciate your help very much.
$$\tan { \theta -\sec { \theta  } =\sqrt { 3 }  } $$
TI get 30 degrees as the reference angle. What am I doing wrong because the answer is 210 degrees.
Link for my work since the post is not showing up.
Thanks for the help.

$$\tan { \theta -\sec { \theta  } =\sqrt { 3 }  } \\ \frac { \sin { \theta  }  }{ \cos { \theta  }  } -\frac { 1 }{ \cos { \theta  }  } =\sqrt { 3 } \\ \sin { \theta -\sqrt { 3 } \cos { \theta =1 }  } \\$$ divide both side to $2$
$$ \frac { 1 }{ 2 } \sin { \theta -\frac { \sqrt { 3 }  }{ 2 }  } \cos { \theta  } =\frac { 1 }{ 2 } \\ \sin { \frac { \pi  }{ 6 } \sin { \theta  } -\cos { \frac { \pi  }{ 6 } \cos { \theta =\frac { 1 }{ 2\\  }  }  }  } \\ \cos { \left( \frac { \pi  }{ 6 } +\theta  \right) =-\frac { 1 }{ 2 }  } \\ \frac { \pi  }{ 6 } +\theta =\pm \arccos { \left( -\frac { 1 }{ 2 }  \right)+2n\pi =\pm \left( \pi -\frac { \pi  }{ 3 }  \right)  } +2n\pi \\ \theta =\pm \frac { 2\pi  }{ 3 } -\frac { \pi  }{ 6 } +2n\pi,n\in\Bbb Z\  $$
