How to derive Lobachevsky's formula for the angle of parallelism I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:
$$\Pi(x)=2\tan^{-1}\left(e^{-x}\right)$$
Any help? Thanks
P.S. I couldn't find anything by google search either.
 A: Hint An interesting relation between arcus tangent and exponential function are made clear in one complex variable
$$\arctan(z) = \frac{1}{2}i\left[\ln(1-iz) -\ln(1+iz)\right]$$
Maybe it can help you somehow.
A: This formula is defining hyperbolic geometry that originated at first definitively from Bolyai, Lobachevsky and Gauss, so they cannot be derived from any other earlier results.
The angle of parallelism is more clearly seen in the Beltrami 3d model compared to the other models.
There is no need to to derive afresh imho but there is a need to recognize/reconcile with earlier models with the Beltrami 3d view.
Let $$ x=\theta, \Pi (x)= \psi(\theta)\;, \Pi(x)= 2 \tan^{-1}(e^{-x})\tag1 $$
$$  \rightarrow \psi(\theta)/2= \psi/2= \tan^{-1}(e^{-\theta})\tag2$$
Take reciprocals, add and simplify using trigonometric and hyperbolic function relations.
$$ \sin \psi = \text{sech} (\theta) \tag3$$
We are here introducing Beltrami's interpretation of hyperbolic geometry.
$ \psi$ is the angle of parallelism when the asymptotic line is everywhere parallel to the axis of rotation.
$$ \sin \psi = \text{sech} (\theta)=\dfrac{r}{a} \tag4$$
Recognize that $\psi$ is angle made by asymptotic line or hyperbolic geodesic to the meridian of a pseudo-sphere of cuspidal radius $a$, the $r$ is radius at any point in cylindrical coordinates as
$$ (r, \theta,z) \tag5$$
where $\theta$ is polar angle reckoned from cuspid equator, $\phi$ angle tangent makes to axis of symmetry.
we have
$$ \psi= \phi \; ; \tag 6$$
Since
$$ \tan \psi= \dfrac{r\;d\theta}{dz} \tag7$$
we get by integration 3d parameterization:
$$ r/a= \sin \psi = \text{sech} (\theta) ;\; z/a = \theta- \tanh(\theta)\; ; \tag 8 $$
that define the geodesic parallel lines shown in red for one half of the pseudosphere. Note that at the cuspidal equator ( maximum radius)
$$ \psi= \pi/2 \tag 9 $$

and as $ \theta \rightarrow \infty, r\rightarrow 0\;\tag{ 10 }$
The above are in agreement with Wiki: Angle of Parallelism.
There can be two parallels through one point, with another one symmetrical to the plane of meridian.
The angles of parallelism $  \Pi (x)$ or $ \psi(\theta)$ vary from point to point between $ 0 - \pi$, being $\pi/2$ at cusp.
A: Lemma 1: Small scale Lobachevsky is Euclidean.
Lemma 2: The area of a polygon is $$\rho^2\times defect$$
First, let us derive a helpful small scale formula. [Picture 1][1]
This is small scale so $d\approx\tan\alpha \Delta x$.
The trapezoid's area is $$-\rho^2\Delta\alpha\approx xd\approx x\tan\alpha \Delta x $$
Let's solve this differential equation:$$\int\cot\alpha\ d\alpha=-\int\frac{x}{\rho^2}dx+C\Rightarrow\ln\sin\alpha=-\frac{x^2}{2\rho^2}\Rightarrow\sin\alpha=e^{-\frac{x^2}{2\rho^2}}\Rightarrow\underline{\alpha\approx\frac{\pi}{2}-\frac{x}{\rho}} $$
Now we can find the general formula. [Picture 2][2]
The smallest triangle's area is $\rho^2(\phi-\frac\pi2-\Delta\alpha)$. It is negligible compared to $\Delta x$ and $\Delta\alpha$ so we can write that it is zero. Using our first, small-scale formula we get the following differential equation:$$0=\phi-\frac\pi2-\Delta\alpha\approx-\frac{\Delta x\sin\alpha}{\rho}-\Delta\alpha$$ It only remains to solve it:$$\int{d\alpha\over\sin\alpha}=-\int{dx\over\rho}+C\Rightarrow\ln\tan\frac{\alpha}{2}=-{x\over\rho}+C\Rightarrow\underline{\alpha=2\arctan(\cot{\beta\over 2}e^{-{x\over\rho}})}$$
This can be used to derive everything in Lobachevskian Geometry. It's the first and the most important formula and it can be written in many ways:$$\cot{\alpha\over 2}\cot{\beta\over 2}=e^\frac{x}{\rho}$$ $$\tanh{x\over\rho}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}$$ $$\tanh{x\over 2\rho}=\frac{\cos\alpha+\cos\beta}{1+\cos(\alpha-\beta)}=\frac{1+\cos(\alpha+\beta)}{\cos\alpha+\cos\beta}$$ $$\sinh{x\over\rho}=\frac{\cos\alpha+\cos\beta}{\sin\alpha\sin\beta}$$ $$\cosh{x\over\rho}=\frac{1+\cos\alpha\cos\beta}{\sin\alpha\sin\beta}$$
[1]: https://i.stack.imgur.com/SkWSa.jpg
[2]: https://i.stack.imgur.com/szDHH.jpg
