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).
