TITLE: The area of a right spherical triangle
QUESTION [9 upvotes]: Is there a compact formula for the area (excess angle – assuming a unit sphere)  of a right spherical triangle given its side lengths $a$ and $b$?

As explained in an answer to an earlier question about the area of a generic spherical triangle, the excess angle $E$ is given by
$$\tan\frac E4=\sqrt{\tan\frac{a+b+c}4\tan\frac{-a+b+c}4\tan\frac{a-b+c}4\tan\frac{a+b-c}4}$$
However, I do not have $c$. Of course, I can use Napier's rules for right spherical triangles to find $c$.
I thought about using integration in spherical coordinates (assuming B is the north pole and equating $a$ and $c$ with polar and azimuthal angles respectively). That approach, however, requires knowing the exact representation of the great circle that connects $A$ to $B$ in the coordinate chart.

REPLY [8 votes]: In Theorem 9.2 of this document, it is shown that the area of a spherical right triangle, $E$, is given by
$$
\tan\left(\frac E2\right)=\tan\left(\frac a2\right)\tan\left(\frac b2\right)
$$
where $a$ and $b$ are the two legs of the right spherical triangle.

REPLY [7 votes]: Napier’s Rules do it easily. The Rule that helps you is $\sin b=\tan a\cot A$, in other words,
\begin{align}
\tan A&=\frac{\tan a}{\sin b}&A&=\arctan\left(\frac{\tan a}{\sin b}\right)\,,\\
\tan B&=\frac{\tan b}{\sin a}&B&=\arctan\left(\frac{\tan b}{\sin a}\right)\,,\\
\text{area}&=A+B-\frac\pi2\,.
\end{align}

REPLY [5 votes]: According to Todhunter's Spherical Trigonometry (section VIII example 4):
$$\sin\frac E2=\sin\frac a2\sin\frac b2\sec\frac c2$$
$$\cos\frac E2=\cos\frac a2\cos\frac b2\sec\frac c2$$
Then
$$\tan\frac E2=\tan\frac a2\tan\frac b2$$
According to item 99 of the same source, $E=A+B+C-\pi$ for any triangle on the unit sphere. When $C=\frac\pi2$:
$$E=A+B-\frac\pi2$$