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.
 A: $$\tan { \theta -\sec { \theta  } =\sqrt { 3 }  } \\ \frac { \sin { \theta  }  }{ \cos { \theta  }  } -\frac { 1 }{ \cos { \theta  }  } =\sqrt { 3 } \\ \sin { \theta -\sqrt { 3 } \cos { \theta =1 }  } \\$$ divide both side to $2$
$$ \frac { 1 }{ 2 } \sin { \theta -\frac { \sqrt { 3 }  }{ 2 }  } \cos { \theta  } =\frac { 1 }{ 2 } \\ \sin { \frac { \pi  }{ 6 } \sin { \theta  } -\cos { \frac { \pi  }{ 6 } \cos { \theta =\frac { 1 }{ 2\\  }  }  }  } \\ \cos { \left( \frac { \pi  }{ 6 } +\theta  \right) =-\frac { 1 }{ 2 }  } \\ \frac { \pi  }{ 6 } +\theta =\pm \arccos { \left( -\frac { 1 }{ 2 }  \right)+2n\pi =\pm \left( \pi -\frac { \pi  }{ 3 }  \right)  } +2n\pi \\ \theta =\pm \frac { 2\pi  }{ 3 } -\frac { \pi  }{ 6 } +2n\pi,n\in\Bbb Z\  $$
A: $$\tan x-\sec x=\sqrt3\iff\tan x+\sec x=-\dfrac1{\sqrt3}$$
Adding we get $\tan x=\dfrac1{\sqrt3}\implies x=n\pi+\dfrac\pi6\  \ \ \ (1)$
Subtracting we get $\sec x=-\dfrac2{\sqrt3}\iff\cos x=-\dfrac{\sqrt3}2=\cos\left(\pi-\dfrac\pi6\right)$
$\implies x=2m\pi\pm\left(\pi-\dfrac\pi6\right)$
i.e, $=(2m+1)\pi-\dfrac\pi6\  \ \ \ (2)$ or $=(2m-1)\pi+\dfrac\pi6\  \ \ \ (3)$
$(1),(2),(3)\implies x=(2r-1)\pi+\dfrac\pi6$
A: Hint:
the solutions of $\tan x= \dfrac{\sqrt{3}}{3}$ are $x=\dfrac{\pi}{6}+k\pi$ (or in degrees: $x=30°+k 180°$).
Since you have squared to find the final equation, we must test the solutions in the starting equation. 
For 
$x= \dfrac{\pi}{6}$  we find :
$$
\dfrac{\sqrt{3}}{3}-2\dfrac{\sqrt{3}}{3}=-\dfrac{\sqrt{3}}{3} \ne \sqrt{3}
$$
so it is an improper solution.
For 
$x= \dfrac{\pi}{6}+\pi$  we find :
$$
\dfrac{\sqrt{3}}{3}+2\dfrac{\sqrt{3}}{3}= \sqrt{3}
$$
so this is the principal solution.
A: HINT:
Using Weierstrass substitution  $$\dfrac{2t}{1-t^2}-\dfrac{1+t^2}{1-t^2}=\sqrt3$$ where $t=\tan\dfrac\theta2$
Rearrange to form a Quadratic equation in $t$
