Prove $(1-\cos x)/\sin x = \tan x/2$

Using double angle and compound angles formulae prove,

$$ \frac{1-\cos x}{\sin x} = \tan\frac{x}{2} $$

Can someone please help me figure this question, I have no idea how to approach it?

Use geometry: $AO= 1$ It is strange that no one has mentioned this drawing yet.

P.S. Note that you can easily extract other trigonometric identites involving $\phi/2, 2\phi $ argument from this picture. For example to get $sin(\phi/2)$ use $ECD$ triangle and Pythagorean theorem to calculate $CD/ED = sin (\phi/2)$

$$\dfrac{1-\cos x}{\sin x}=\dfrac{1-(1-2\sin^2\frac x2)}{2\sin\frac x 2\cos\frac x2}=\dfrac{\sin\frac x2}{\cos\frac x2}=\tan\frac x2$$

Hint: It is easier to show:

$$\frac{1-\cos 2y}{\sin 2y} = \tan y$$

using the formulas for $\cos 2y$ and $\sin 2y$.

$$ \cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2} = 1-2\sin^2\frac{x}{2} $$ and $$ \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} $$ so $$ \frac{1-\cos x}{\sin x} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2} $$