Very basic question about how we define frequency in signal processing
For the tangent function a cycle is indeed done in $\pi$ radians, because you want 

\begin{align}\tan(x) &amp;= \tan(x+T)\\
\implies \frac{\sin(x)}{\cos(x)} &amp;= \frac{\sin(x+T)}{\cos(x+T)}
\end{align}
$$
\implies \sin(x+T)\cos(x) - \sin(x)\cos(x+T) =0
$$
$$\implies \sin\big[\left(x + T\right)-x\big] =0\equiv \sin\left(T\right)=0$$
$$\implies T = \pi k, \quad \text{with}\quad k\in \mathbb Z$$

Smallest strictly positive $k$ (i.e. $k= 1$) gives us $T_0 = \pi$. Anyway, we know that. 

Regarding the interpretations, let's use $\Omega \ [\text{rad/sec}]$ to avoid ambiguity with $\omega \ [\text{rad/sample}]$ mostly used in the discrete case.  From the definition of the angular frequency $\Omega = 2\pi F$ with the fundamental period $T_0 = 1/F$, as you have already described, we have 
$$
\Omega_0 = \frac{2\pi}{T_0}\tag{1}
$$ 
The interpretation should go from $(1)$ and read as the number of cycles of the periodic signal in $\mathbf{2\pi}$ radians. As examples, look at the plot below, 

As seen above,


$\tan(x)$ has $T_0 = \pi$ implying $\Omega_0 = 2\pi/\pi=2 \ \text{rad/sec}$ or 2 cycles in $\mathbf{2\pi}$ radians (or equivalently one cycle in $\pi$ radians)
$\sin(x)$ has $T_0 = 2\pi$ implying $\Omega_0 = 2\pi/2\pi=1 \ \text{rad/sec}$ or 1 cycle in $\mathbf{2\pi}$ radians
$\sin(2\pi x)$ has $T_0 = 1$ (see $x$-axis) implying $\Omega_0 = 2\pi/1=2\pi \ \text{rad/sec}$ or $\mathbf{2\pi\approx 6.2832}$ (see $x$-axis) cycles in $\mathbf{2\pi}$ radians. 


See it this way, the ratio in equation $(1)$ is simply stating how many  $T_0$ (i.e. fundamental periods or full cycles) you have in $2\pi$. You have $T_0 = 2\pi \implies \Omega_0 = 1 \ [\text{rad/sec}]$ meaning you have $1$ fundamental period (full cycle) in $2\pi$. Similarly, when $T_0 = 1 \implies \Omega_0 = 2\pi \ [\text{rad/sec}]$ meaning you have $2\pi$ fundamental periods (full cycles) in $2\pi$. Etc. for other cases.