The area of a right spherical triangle

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.

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}

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 $C=\frac\pi2$: $$E=A+B-\frac\pi2$$