Given an unit sphere and on this sphere a right angled triangle and given the length of the hypotenuse as $c$ and a leg as $a$.
The area $E$ can calculated via the Napier rules as: (using radians as angle unit)
$$ E= \arcsin \left(\frac{\sin a}{\sin c}\right) + \arccos \left(\frac{\tan a}{\tan c}\right) - \frac{\pi}{2} $$
I cannot believe that this formula cannot be simplified, but I don't know how.
Can somebody show me (with steps) how to simplify this formula?
Following the suggestions below: Using $p= sin a/ sin c $ and $ q = tan a / tan c $ the formula gets to
$$ E = \arcsin p + \arccos q - \frac{\pi}{2} $$
To be continued ....
"Raw" areas and lengths tend to be ugly. The trick to an elegant relation is usually to wrap the values in appropriate trig functions.
For instance, here you could write $$\cos E = \frac{\cos^2a + \cos c}{\cos a (1 + \cos c)}$$ This isn't super-terrible; at least everything's in terms of cosines.
Nevertheless, my experience has taught me that non-Euclidean trig seems to prefer half-angle arguments. (In fact, it happens so often that I've taken to using $x_2$ as a compact way to denote $x/2$. I'll use that notation here.) Consider ... $$\begin{align} \tan^2E_2 = \frac{1-\cos E}{1+\cos E} &= \frac{(1-\cos a)(\cos a-\cos c)}{(1+\cos a)(\cos a+\cos c)} \\[6pt] &= \frac{2\sin^2a_2 \cdot 2 \sin(c_2-a_2) \sin(c_2+a_2)}{2\cos^2a_2\cdot 2 \cos(c_2-a_2) \cos(c_2+a_2)} \\[6pt] &= \tan^2a_2 \tan(c_2-a_2) \tan(c_2+a_2) \tag{$\star$} \end{align}$$ This form is rather nice; everything's in terms of (half-angle) tangents! (And, if you like, you can re-solve explicitly for $E$.) In particular, it allows for convenient "infinitesimal sanity-checking" (not to be confused with "infinitesimal-sanity checking", a service I wish my bank would offer).
For vanishingly-small $x$, we have $\tan x \approx x$. So, for an infinitesimal spherical right triangle, relation $(\star)$ indicates $$\left(\frac12E\right)^2 \approx \left(\frac12 a\right)^2 \cdot \frac12 (c-a) \cdot \frac12(c+a) \quad\to\quad E \approx \frac12 a \sqrt{c^2-a^2}$$ which is, of course, the formula for the area of a Euclidean right triangle with legs $a$ and $\sqrt{c^2-a^2}$.
Note. Expanding-upon my comment about getting to my first equation ... Take the cosines of both sides of OP's equation. Defining $p := \sin a/\sin c$, $q := \tan a/\tan c$ for convenience, we have ... $$\begin{align} \cos E &= \cos\left(\arcsin p + \arccos q - \pi/2\right) \\ &=\sin(\arcsin p+\arccos q) \\ &=\sin(\arcsin p)\cos(\arccos q) + \cos(\arcsin p) \sin(\arcsin q) \\ &= p q + \sqrt{ (1-p^2)(1-q^2) } \end{align}$$ Now, observe that $$\begin{align} 1 - p^2 &= \frac{\sin^2c-\sin^2a}{\sin^2c} = \frac{(1-\cos^2c)-(1-\cos^2a)}{\sin^2c} = \frac{\cos^2a-\cos^2c}{\sin^2c} \\[8pt] 1-q^2 &= \frac{\tan^2c-\tan^2a}{\tan^2c} = \frac{(\sec^2c-1)-(\sec^2a-1)}{\tan^2c} = \frac{\sec^2c-\sec^2a}{\tan^2c} \\[4pt] &=\frac{\cos^2a-\cos^2c}{\tan^2c\cos^2c\cos^2a}=\frac{\cos^2a-\cos^2c}{\sin^2c\cos^2a} \end{align}$$ Therefore, we can write $$\begin{align} \cos E &= \frac{\sin a}{\sin c}\cdot\frac{\tan a}{\tan c} + \frac{\cos^2a-\cos^2c}{\cos a\sin^2 c} = \frac{\sin^2 a \cos c+\cos^2a-\cos^2c}{\cos a\sin^2 c} \\[4pt] &= \frac{(1-\cos^2 a) \cos c+\cos^2a-\cos^2c}{\cos a\sin^2 c} = \frac{(1-\cos^2 a) \cos c+\cos^2a-\cos^2c}{\cos a\sin^2 c} \\[4pt] &= \frac{(1 + \cos c) (\cos^2 a + \cos c)}{\cos a (1-\cos^2c)} \\[4pt] &= \frac{\cos^2 a+\cos c}{\cos a(1+\cos c)} \end{align}$$
As I mentioned in my comment: It's a bit of a slog. ... Moreover, it's not entirely obvious at first what trig manipulations should be done. Again, my experience has guided me here: in particular, I've found that a bias for cosines tends to serve me pretty well in simplifications like this, so that gave me a direction to start.
Plus, experience has instilled in me a faith that there must be an elegant relation behind all of this mess ... which is why I wasn't satisfied with the rather clunky cosine formula.
The messy denominator suggested to me that I might want to divide by, say, $\sin E$, whose denominator would match and therefore cancel; but the overall result isn't any better. Recalling that non-Euclidean trigs seem to prefer half-angle arguments, I calculated $1+\cos E$ and $1-\cos E$ (most of the way to computing $\cos^2E_2$ and $\sin^2E_2$), which happened to have nicely-factored numerators (and still-messy-but-cancelable denominators), leading me to the initial $\tan^2 E_2$ formula.
The steps from there to the final form were perhaps a bit less mysterious, and the elegance of that final form told me that that was the proper place to stop.