Trouble simplifying this Trigonometric Identity I am having trouble trying to solve this question:
$\cos \theta \cot\theta(\sec\theta - 2\tan\theta)$ 
I have had 2 different answers but not the right one.
It's supposed to be: $\cot\theta - 2\cos\theta$ 
but my skills are rusty and I am having trouble trying to determine exactly what to do. I keep simplifying the identities then multiplying the fractions but I am just super lost..
I am just looking for some help and guidance for this. Thank you.
 A: $$\cos  \theta \cot  \theta (\sec  \theta -2\tan  \theta )=\cos  \theta \frac { \cos { \theta  }  }{ \sin { \theta  }  } \left( \frac { 1 }{ \cos { \theta  }  } -\frac { 2\sin { \theta  }  }{ \cos { \theta  }  }  \right) =\\ =\frac { \cos ^{ 2 }{ \theta  }  }{ \sin { \theta  }  } \frac { \left( 1-2\sin { \theta  }  \right)  }{ \cos { \theta  }  } =\frac { \cos { \theta  }  }{ \sin { \theta  }  } \left( 1-2\sin { \theta  }  \right) =\frac { \cos { \theta  } -2\sin { \theta \cos { \theta  }  }  }{ \sin { \theta  }  } =\frac { \cos { \theta  }  }{ \sin { \theta  }  } -\frac { 2\sin { \theta  } \cos { \theta  }  }{ \sin { \theta  }  } =\cot { \theta  } -2\cos { \theta  } $$
A: You can write
$$
\cos\theta\cot\theta(\sec\theta-2\tan\theta)=
\cot\theta\cos\theta\sec\theta-2\cos\theta\cot\theta\tan\theta=
\cot\theta-2\cos\theta
$$
because $\cos\theta\sec\theta=1$ and $\cot\theta\tan\theta=1$.
A: $\require {cancel}$
We have $$\cos \theta \cot \theta \ (\sec \theta - 2 \tan \theta)$$
Multiplying everything out (distributive property) we have$$\cot \theta \cos \theta \sec \theta - 2 \cos \theta \cot \theta \tan \theta$$
Note that $\cos \theta \sec \theta = 1$ and $\cot \theta \tan \theta = 1$, so these can cancel...
$$\cot \theta \cancelto {1}{\cos \theta \sec \theta} - 2 \cos \theta \cancelto {1}{\cot \theta \tan \theta}$$
And we're left with what we were looking for...$$\cot \theta - 2 \cos \theta$$
