Prove $\frac{\sin\theta}{1-\cos\theta} \equiv \csc\theta + \cot\theta$

Question

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.

Answer 1

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}

Answer 2

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 $c=\cos(\theta)$.