Prove $\frac{\sin\theta}{1-\cos\theta} \equiv \csc\theta + \cot\theta$ This must be proved using elementary trigonometric identities.
I have not been able to come to any point which seems useful enough to include in this post.

Lets get rid of the trigonometry stuff first:
$$s=\sin(\theta),~~c=\cos(\theta),~~\csc(\theta)=\frac{1}{s},~~\cot(\theta)=\frac{c}{s}$$
Now we are solving this equation:
$$\frac{s}{1-c}=\frac{1}{s}+\frac{c}{s}$$
$$\Leftrightarrow$$
$$\frac{s}{1-c}-\frac{1}{s}-\frac{c}{s}=0$$
multiply by $(1-c)\neq 0$
$$s-\frac{1-c}{s}-\frac{(1-c)c}{s}=0$$
multiply by $s\neq 0$
$$s^2-(1-c)-(1-c)c = s^2-1+c-c+c^2 =0\Leftrightarrow s^2+c^2=1$$
The only solutions to this algebraic equation are $s=\sin(\theta)$ and $c=\cos(\theta)$.
