Q:

Area between geodesic and arc of constant lattitude

A geodesic is a line representing the shortest route between two points on a sphere, for example on the Earth treated here as a perfect sphere.
Two points on Earth having the same latitude can be also connected with the line being a part of a circle for selected constant latitude.
Differences between these two lines can be visualized with the use of 
 this Academo program presenting the situation in the context of the map of Earth.
Question:

How to calculate the area between these two lines?

(Assume for example that the starting point is $(\alpha, \beta_1)=(45^\circ, -120^\circ)$ and the destination $(\alpha, \beta_2)=(45^\circ, 0^\circ))$ - the arc of constant latitude $45^\circ$ has length $120^\circ$.
In wikipedia a formula for an area of a spherical polygon is presented, but the polygon is limited in this case with parts of geodesics. Is it possible somehow transform these formulas of spherical geometry into the case of finding the area between geodesic and arc of constant latitude?

A:

Summary: If $A$ and $B$ lie on the latitude line at angle $0 < \alpha < \pi/2$ north of the equator on a sphere of unit radius, and at an angular separation $0 < \theta = \beta_{2} - \beta_{1} < \pi$, then the "digon" bounded by the latitude and the great circle arc $AB$ (in blue) has area
\begin{align*}
  \pi - \theta \sin\alpha - 2\psi
  &= \text{sum of interior angles} - \theta \sin\alpha \\
  &= \pi - \theta \sin\alpha - 2\arccos\left(\frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}\right).
\end{align*}

If $A$ and $B$ have longitude-latitude coordinates $(0, \alpha)$ and $(\theta, \alpha)$, their Cartesian coordinates (on the unit sphere) are
$$
A = (\cos\alpha, 0, \sin\alpha),\quad
B = (\cos\theta\cos\alpha, \sin\theta \cos\alpha, \sin\alpha).
$$
Let $C = (0, 0, 1)$ be the north pole, $G$ the "gore" (shaded) bounded by the spherical arcs $AC$, $BC$, and the latitude through $A$ and $B$, and $T$ the geodesic triangle with vertices $A$, $B$, and $C$.
Lemma 1: The area of $G$ is $\theta(1 - \sin\alpha)$.
Proof: The spherical zone bounded by the latitude through $A$ and $B$ and containing the north pole has height $h = 1 - \sin\alpha$ along the diameter through the north and south poles. By a theorem of Archimedes, this zone has area $2\pi h = 2\pi(1 - \sin\alpha)$. The area of the gore $G$, which subtends an angle $\theta$ at the north pole, is
$$
(\theta/2\pi)2\pi(1 - \sin\alpha) = \theta(1 - \sin\alpha).
$$
Lemma 2: The area of $T$ is $\theta - \pi + 2\arccos\dfrac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}$.
Proof: If $\psi$ denotes the interior angle of $T$ at either $A$ or $B$, the area of $T$ is the angular defect, $\theta + 2\psi - \pi$. To calculate $\psi$, note that the unit vector $n_{1} = \frac{A \times C}{\|A \times C\|} = (0, -1, 0)$ is orthogonal to the great circle $AC$, the unit vector
$$
n_{2} = \frac{A \times B}{\|A \times B\|}
  = \frac{(-\sin\theta \sin\alpha, \sin\alpha(\cos\theta - 1), \cos\alpha \sin\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}
$$
is orthogonal to the great circle $AB$, and
$$
\cos\psi = n_{1} \cdot n_{2}
  = \frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}.
$$
This completes the proof of Lemma 2.
The area of the digon is the difference,
\begin{align*}
  A &= \theta(1 - \sin\alpha) - (\theta + 2\psi - \pi)
  = \pi - \theta \sin\alpha - 2\psi \\
  &= \pi - \theta \sin\alpha - 2\arccos\left(\frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}\right).
\end{align*}
When $\alpha = 0$, the area vanishes for $0 < \theta < \pi$ (because the latitude through $A$ and $B$ coincides with the great circle arc), while if $\alpha$ is small and positive, the area is close to $\pi$ when $\theta = \pi$ (because $A$ and $B$ are nearly antipodal and the great circle arc passes through the north pole).

