Proving identities using Pythagorean, Reciprocal and Quotient Back again, with one last identity that I cannot solve:
$$\frac{\cos \theta}{\csc \theta - 2 \sin \theta} = \frac{\tan\theta}{1-\tan^2\theta}$$.
The simplest I could get the left side to, if at all simpler, is $$\frac{\cos\theta}{\csc^2\theta-2}$$
As for the right side, it has me stumped, especially since the denominator is so close to a identity yet so far away. I've tried rationalizing the denominators (both sides) to little success. From my last question, where multiplying by a '1' worked, I didn't see the association here.
Thanks!

HINT:
$$\begin{align*}
\frac{\tan\theta}{1-\tan^2\theta}&=\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}
\end{align*}$$
$$\begin{align*}
\frac{\cos\theta}{\csc\theta-2\sin\theta}&=\frac{\sin\theta\cos\theta}{1-2\sin^2\theta}
\end{align*}$$
