Edit: Added some tex tags

This is a question on a review I am doing, my answer doesn't match up with the answer key and I am looking for some input on what I am doing wrong.

If $\displaystyle cos\theta = \frac{1}{6}$ with $\displaystyle \theta$ in quadrant IV find:

$\displaystyle sin(-\theta)-6cos(-\theta)+3tan(2\theta)$

I solved for $\displaystyle sin\theta = \frac{-\sqrt{35}}{6}$ $\displaystyle cos\theta = \frac{1}{6}$ and $\displaystyle tan\theta = -\sqrt{35}$

Using negative angle identities and the double angle identity I end up with:

$\displaystyle -sin\theta-6cos\theta+3(\frac{2tan\theta}{1-tan^2\theta})$

Plugging in my values gives me:

$\displaystyle \frac{\sqrt{35}}{6} - \frac{6}{6} + 3[\frac{2(-\sqrt{35})}{1-35}]$

Simplified to:

$\displaystyle \frac{35(\sqrt{35}) -102}{102}$

The answer key gives: $\displaystyle \frac{42+25\sqrt{35}}{42}$

Not sure where I am messing up =\ any help is much appreciated.