<s>[INST] 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.
 [/INST] Multiply the LHS by $1+\cos\theta$ yields
\begin{align}
\frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}&=\frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}\\
&=\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}\qquad;\qquad\color{red}{\cos^2\theta+\sin^2\theta=1}\\
&=\frac{1+\cos\theta}{\sin\theta}\\
&=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}\\
&=\color{blue}{\csc\theta+\cot\theta}.\qquad\qquad\blacksquare
\end{align}
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