Is $\tan(\alpha/2)=(1-\cos\alpha)/\sin \alpha, \,?$ Is it true that
$$\bbox[5px,border:2px solid #138D75]{\tan\left(\frac {\alpha}2\right)=\frac{1-\cos\alpha}{\sin \alpha}, \quad ?} \tag1$$
My solution:
First, we note that the expression $\tan \left ( \frac{\alpha}{2} \right )$ makes sense when $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$: henceforth, we will assume that this condition is met.
By definition of tangent function, we have: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right )}$$
When $\alpha \neq 2k \pi, k \in \mathbb{Z}$, we can multiply numerator and denominator by $\sin \left ( \frac{\alpha}{2} \right )$, because this quantity is non-zero. We therefore get:
$$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin^2 \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right )}. $$
For what we saw earlier: $$\sin^2 \left ( \frac{\alpha}{2} \right ) = \frac{1 - \cos \alpha}{2}$$and instead, applying the sine duplication formula: $$\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right ) = \frac{1}{2} \sin \alpha$$
In conclusion we get the following formula: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ 1 - \cos \alpha}{\sin \alpha}.$$
If in the expression of the tangent we had multiplied numerator and denominator by $\cos \left ( \frac{\alpha}{2} \right )$ then with steps similar to what we have done we get another formula:$$\bbox[5px,border:2px solid #118D45]{\tan \left ( \frac{\alpha}{2} \right ) = \frac{\sin \alpha}{1 + \cos \alpha}}. \tag 2 $$
We note that it always holds $\cos \left ( \frac{\alpha}{2} \right ) \neq 0$ since we have imposed $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$ (and this entitles us to multiply numerator and denominator by this term).

Is there a faster and more immediate proof of the $(1)$ or $(2)$?

 A: Here are geometric proofs, which I believe are "faster and more immediate", as the OP desired.
All you need is the "circle theorem" that "the angle at the circumference is half the angle at the center".
First, the (somewhat easier) proof for $$\tan\left(\frac{\alpha}{2}\right)=\frac{\sin{\alpha}}{1+\cos\alpha}$$

Second, the one OP requested, that $$\tan\left(\frac{\alpha}{2}\right)=\frac{1-\cos\alpha}{\sin\alpha}$$

I recognise that the picture works only for acute $\alpha$, but it is easy to see how it can work for obtuse or reflex $\alpha$.
A: We could multiply both sides by $\sinα$
Then on the LHS use the double angle for $\sinα = 2\sin(α/2) \cos(α/2)$ and put $\tan(α/2)=\sin(α/2)/\cos(α/2)$  then the LHS becomes $2\sin^2(α/2)$.
The RHS is the same as $1-\cosα = 1-(1-2\sin^2(α/2))$ from a double angle formula for $\cosα$, so it's true.
A: Replace $\alpha$ by $2\beta$ so that the right side is
$${{1-\cos 2\beta}\over {\sin 2\beta}}=
{{1-(\cos^2\beta-\sin^2\beta)}\over {2\sin\beta\,\cos\beta}}$$
$$={{2\sin^2\beta}\over{2\sin\beta\cos\beta}}=
{{\sin\beta}\over {\cos\beta}}=\tan\beta$$
A: By request, I am writing my comment as an answer:
I think there is a small technicality: you only assumed in your proof that $α≠π+2πk$, but you used that $α≠2πk$. It would be better to remark at the start of the proof that the LHS only makes sense when $α≠π+2πk$ (as you did) and that the RHS only makes sense when $\alpha≠πk$ (and therefore we should assume that $α≠πk$ for this proof).
