$\displaystyle 1 + tan^{2}\theta = sec^{2} \theta $Write the following in their simplest form, involving only one trigonometric function:

$\displaystyle \frac{tan \theta}{sec^{2} \theta-2} $

$\displaystyle \frac{tan \theta}{ 1 + tan^{2} \theta - 2} $

$\displaystyle \frac{tan\theta}{tan^2 \theta-1} $

I am stuck at this point, I know I have to get it to look like the $\displaystyle tan2\theta $ expansion,

any hints/help appreciated.