Explaining failure of MVT hypotheses for tan(x) on [0, π]

The MVT requires $f$ to be continuous on the closed interval $[0,\pi]$ and differentiable on the open interval $(0,\pi)$. Recall that $\tan(x) = \sin(x)/\cos(x)$, which is undefined wherever $\cos(x) = 0$.

On the interval $[0,\pi]$, $\cos(x) = 0$ at $x = \pi/2$, which lies inside the open interval $(0,\pi)$. At this point $\tan(x)$ has a vertical asymptote, so $f$ is neither continuous nor differentiable there.

Therefore both hypotheses of the MVT fail at $x = \pi/2 \in (0,\pi)$: $f$ is not continuous on $[0,\pi]$ (it is discontinuous at $\pi/2$) and consequently not differentiable on all of $(0,\pi)$. Since the hypotheses of the theorem are not satisfied, the MVT simply does not guarantee the existence of such a $c$ — it cannot be applied to this function on this interval.

The trouble point is exactly halfway between zero and pi, at pi over two, so it is equidistant from both endpoints rather than closer to either one. This is where tangent has its vertical asymptote inside the open interval, causing both the continuity and differentiability hypotheses of the MVT to fail. Since pi over two is the midpoint of the interval from zero to pi, it is equidistant from both endpoints.