Based on the definition of the hyperbolic trig functions (in terms of $e^x$ and $e^{-x}$) it's easy to show that the point $\big(\cosh(\alpha),\sinh(\alpha)\big)$ falls on the unit hyperbola ($x^2-y^2=1$). But one may ask: what is $\alpha$?
It turns out that $\alpha$ is the area of the region bounded by the unit hyperbola and the line segment joining the origin and $\big(\cosh(\alpha),\sinh(\alpha)\big)$ and the line segment joining the origin and $\big(\cosh(\alpha),-\sinh(\alpha)\big)$.
I just tried and failed to prove that this is true. Maybe I just gave up too soon or was lacking the proper insight. Does anyone know how to prove this?
Mirroring the polar coordinates proof for the area of a circular sector, introduce "hyperbolic polar coordinates" $$ P(r, t) = (r\cosh t, r\sinh t),\quad 0 \leq r,\quad -\infty < t < \infty, $$ which cover the quarter plane $0 \leq x$, $-x < y < x$ containing the right branch of the hyperbola $x^{2} - y^{2} = 1$, $x > 0$. The Jacobian is easily calculated to be $$ \det DP(r, t) = r, $$ and the region $R$ bounded by the hyperbola, the ray from the origin through $(\cosh\alpha, \sinh\alpha)$, and the ray from the origin through $(\cosh\alpha, -\sinh\alpha)$ is the rectangle $R^{*}$ in the $(r, t)$-plane defined by $0 \leq r \leq 1$, $-\alpha \leq t \leq \alpha$. By the change of variables theorem, $$ \text{area of $R$} = \int_{R} dx\, dy = \int_{R^{*}} |\det DP(r, t)|\, dr\, dt = \int_{-\alpha}^{\alpha} \int_{0}^{1} r\, dr\, dt = \int_{-\alpha}^{\alpha} \frac{1}{2}\, dt = \alpha. $$
(Diagrams can be found in my answer to Parametric equation - of a hyperbola.)