Show that along the cycloid, the tangent vector is not well-definded when $\theta=2k\pi$. 
Show that along the cycloid, the tangent vector is not well-defined when $\theta=2k\pi$.
p.s. : A cycloid has the parametrization $\bf{r}$$(\theta)=(\theta-\sin\theta,1-\cos \theta)$.

Actually I don't understand what "not well-defined" means in this case. I understand there'll be a sharp bent on that point, but how should I present it? Thanks.
 A: 
I find that the slope of the tangent is cotπ, which is undefined, but seems this is not enough.

Correct: this is not enough. There is nothing wrong with vertical tangent lines. In general, it's advisable to avoid the slope-intercept equation of lines when studying differential geometry. It concerns statements that are independent of coordinates chosen on a plane, or on a manifold. The slope being infinite depends on coordinate system 

how should I present it?

In a manner consistent with how tangent vectors were presented to you. There are multiple ways to present the material of differential geometry. Those not taking the same class do not know which one was chosen by the professor. 
To say that an something is not-defined means: it is not true that there exists a (unique) object that satisfies the definition of something. So, the first thing to locate is the definition of a tangent vector of a curve used in the course. 
Here is one of several possible definitions: if the limit 
$$\lim_{t\to t_0} \frac{r(t)-r(t_0)}{|r(t)-r(t_0)|}\operatorname{sign}(t-t_0)  \tag1$$
exists as $t\to t_0$, the value of this limit is the unit tangent vector of at $r(t_0)$. 
The above definition agrees with $r'(t_0)/|r'(t_0)|$ when the derivative exists and is nonzero. However, (1) is more general: it allows us to find the tangent vector even when the parametrization has zero derivative -- provided that the tangent vector exists. 
It is not hard to check that (1) does not exist for the cycloid: the one-sided limits have different values $(0,1)$ and $(0,-1)$.
