Proving $\sin x + \sin x \cdot \cot^2 x = \csc x $ The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process:
$$\begin{align*}
\tan \theta &= \frac{\sin \theta}{\cos \theta}\\
(\sin \theta)^2 + (\cos \theta)^2 &= 1\\
\csc \theta &= \frac{1}{\sin \theta}\\
\sec \theta &= \frac{1}{\cos \theta}\\
\cot \theta &= \frac{1}{\tan \theta}.
\end{align*}$$
The trig identity given is:
$$\sin x + \sin x \cdot \cot^2 x = \csc x $$
I simplified it to:
$$\sin x + \sin x \cdot \left(\frac{\cos x}{\sin x}\right)^2 = \frac{1}{\sin x} $$
After which I get lost in a mess of term rewriting that never seems to lead anywhere fruitful. Any hints would be greatly appreciated.

Taking the left hand side, try cancelling the $\sin^2{x}$ with the $\sin{x}$ in the numerator 
(i.e. $\sin{x} .(\frac{\cos^2{x}}{\sin^2{x}}) = \frac{\cos^2{x}}{\sin{x}}$)
Then add the remaining two terms ($\sin{x} + \frac{\cos^2{x}}{\sin{x}}$) and use the identity $\sin^2{x}+\cos^2{x}=1$. That should do the trick :)
