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.
 A: Start with $\sin x + \sin x\cot^2 x$. Factoring out $\sin x$ we get
$$\sin x + \sin x\cot^2 x = \sin x (1 + \cot^2 x).$$
Now use the definition of $\cot x$ and rewrite the expression in the parenthesis as a single fraction:
$$\begin{align*}
\sin x + \sin x\cot^2x &= \sin x\Bigl( 1 + \cot^2 x\Bigr)\\
&= \sin x \left( 1 + \frac{\cos^2 x}{\sin^2 x}\right)\\
&= \sin x \left(\frac{\sin^2 x}{\sin ^2x} + \frac{\cos^2 x}{\sin^2x}\right)\\
&= \sin x\left(\frac{\sin^2 x + \cos^2x}{\sin^2x}\right)\\
&= \sin x\left(\frac{1}{\sin^2 x}\right).
\end{align*}$$
Note that I am not manipulating the identity as a whole. One has to be careful when doing that, because you can start with a false equality and end with a true one if not all your steps are reversible. Instead, we start with one side, and apply known identities to it until it becomes the other side.
(I am guessing you can take it home from where I left it, of course...)
A: 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 :)
