Prove this trigonometric identity $\frac{1 - \cosθ}{\sinθ} = \frac{\sinθ}{1 + \cosθ}$

Question

$$\frac{1 - \cosθ}{\sinθ} = \frac{\sinθ}{1 + \cosθ}$$

using $$\sin^2θ + \cos^2θ = 1$$

any hints would be much appreciated!

Answer 1

$$\begin{align} \frac{\sin \theta}{1+\cos \theta} &= \frac{(1-\cos \theta )\sin \theta}{(1+\cos \theta)(1-\cos \theta)} \\ &= \frac{(1-\cos \theta )\sin \theta}{1-\cos^2\theta} \\ &= \frac{(1-\cos \theta )\sin \theta}{\sin^2 \theta} \\ &= \frac{1-\cos \theta }{\sin \theta} \end{align}$$ whenever $\theta \notin \{ k\pi :\, k\in \mathbb{Z} \}$.

Answer 2

$$\frac{1-\cos(\theta)}{\sin(\theta)} = \frac{1-\cos(\theta)}{\sin(\theta)}\cdot\frac{\sin(\theta)}{\sin(\theta)} = \frac{\big(1-\cos(\theta)\big)\sin(\theta)}{\sin^2(\theta)} = \frac{\big(1-\cos(\theta)\big)\sin(\theta)}{1-\cos^2(\theta)} = \frac{\sin(\theta)}{1+\cos(\theta)}$$

But as a small note, for instance $\theta = 2\pi$ does not satisfy that equality. So this is not true in general. I just posted this answer to show you how to manipulate one expression to get another.

Answer 3

Hint:

For non-zero $b,d$, one has $$\frac ab=\frac cd\iff ad=bc.$$

Answer 4

$$\frac{1 - \cosθ}{\sinθ} = \frac{\sinθ}{1 + \cosθ}\iff \frac{1 - \cos^2θ}{(1+\cos\theta)\sinθ} = \frac{\sin^2θ}{(1+\cos\theta)\sinθ}.$$