<s>[INST] 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 $(2)$?

 [/INST] Replace $\alpha$ by $2\beta$ so that the right side is
$${{1-\cos 2\beta}\over {\sin 2\beta}}=
{{1-(\cos^2\beta-\sin^2\beta)}\over {2\sin\beta\,\cos\beta}}$$
$$={{2\sin^2\beta}\over{2\sin\beta\cos\beta}}=
{{\sin\beta}\over {\cos\beta}}=\tan\beta$$
</s>