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 $(2)$?


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.
