simplification of the area of a hyperbolic circle (BONOLA, S 53) I'm trying to understand the S-53 of "Non-Euclidean Geometry" (BONOLA, R.) in which the formula for the area of a circle of radius r:
$$2\pi k^2(\cosh\frac rk -1)$$
is somehow reduced by only applying $$\tan(\frac {\pi[x]}{2})=e^{-\frac xk}$$, to: $$\frac {4\pi k^2}{\tan^2(\pi[\frac r2])}$$
But I cannot find a way to achieve this, because it somehow implies some relation between $\tan(\frac {\pi[x]}{2})$ and $\tan(\pi[x])$, that I'm unaware of, or/and using trigonometric transformations that are supposed?
 A: By the Half-Angle Formula for $\sinh$, the original expression can be written
$$4\pi k^2 \sinh^2\frac{r}{2k}$$
Comparing this to the target formula, all we have to show is 
$$\sinh\frac{r}{2k} = \frac{1}{\tan \pi\left[\frac{r}{2}\right]}  \tag{$\star$}$$
Consider that the Double-Angle Formula for $\tan$ allows us to re-write $\tan\pi[x]$ in terms of $\tan\frac{\pi[x]}{2}$: 
$$\begin{align}
\tan\pi[x] = \tan\left( 2\cdot \frac{\pi[x]}{2} \right) = \frac{2\tan\frac{\pi[x]}{2}}{1-\tan^2\frac{\pi[x]}{2}} 
\end{align}$$
Reciprocating, and invoking the formula for $\tan\frac{\pi[x]}{2}$, we have
$$\frac{1}{\tan\pi[x]} = \frac{1-\tan^2\frac{\pi[x]}{2}}{2\tan\frac{\pi[x]}{2}} = \frac{1-e^{-2x/k}}{2 e^{-x/k}} = \frac{e^{x/k}-e^{-x/k}}{2} = \sinh\frac{x}{k}$$
With $x = r/2$, we get $(\star)$.
A: $$   \tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}   $$
In particular, if this $0 < \theta < \pi / 4,$ then $0 < \tan \theta < 1$ and $\tan 2 \theta$ is positive.
So, for $0 < \theta < \pi / 2,$ we find
$$  \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} = \frac{ \sin \theta}{1 + \cos \theta}   $$
