I whipped up a quick Matlab program that computes area stochastically (generate 20000 points; count how many are inside either triangle), and then ran it on 10,000 randomly generated triangle pairs (i.e., 6 random points on the unit circle; the first 3 define one triangle; the next three define the other. WLOG, I made the first angle be 0). Each time a new "largest area" was found, I printed out the triangles. It instantly found an area of ~1.25; then steadily improved this to 1.87 over the next couple of minutes. A previous instance of the program reached 1.97 or so...and this one just jumped to ~1.92, with a solution that's very nearly the two 45-90-45 triangles erected on a diameter.
That suggests to me that the correct answer is indeed this pair of triangles, with area 2 as the optimum.
>> testCircles
area = 1.2587; angles are [0 4.13265878886387 1.69578766384379 2.78030964099446 4.4988452659303 2.81878425776269]
area = 1.4833; angles are [0 1.03997694447012 4.97128628805005 5.53558413650189 4.31034051247746 2.03741390656017]
area = 1.5065; angles are [0 2.583714405482 3.95028117360417 3.41185810372066 4.81492678793123 1.47312279640955]
area = 1.8173; angles are [0 4.99055083947817 1.93427156493409 1.86260140259647 3.04043013930809 5.0276959720494]
area = 1.8517; angles are [0 3.01875721392096 1.30210531664363 4.12564514448203 2.38462146306761 5.8391459560257]
area = 1.87; angles are [0 1.78060248058588 4.66075033844476 2.08372315689924 5.4697587833186 3.76689571284346]
area = 1.9184; angles are [0 4.43108946197308 1.4092884007527 4.31113371053254 2.84429076671212 1.32017294630991]
Here's a picture of the most recent success:
The associated angles/area are
area = 1.9428; angles are [0 1.51926494165392 3.23093144636552 0.0773804130776986 3.34779844425289 4.92681116635565]
The area $T$ in question can be found as
the difference between the area of $\triangle AID$
and a circular segment $IDM$:
\begin{align}
T&=
\tfrac12\,|AI|\cdot|DI|\sin\angle DIA
-
\frac12\angle DIA\cdot|DI|^2
,
\end{align}
where $\angle DIA$ is measured in radians.
To use the standard route to find the radius,
consider this circle as inscribed
in isosceles $\triangle AFB$, for which
\begin{align}
|AC|& = 2+\sqrt3
,\\
|BC|& = 3+2\sqrt3
,\\
|AB| &=\sqrt{|AC|^2+|BC|^2}
\\
&=\sqrt{28+16\sqrt3}=4+2\sqrt3
,\\
|FB|&=2|BC|=6+4\sqrt3
,\\
S_{\triangle AFB}&=|AC|\cdot|BC|
=12+7\,\sqrt3
,\\
\rho&=\tfrac12(2|AB|+|FB|)
=|AB|+|BC|=7+4\sqrt3
,
\end{align}
\begin{align}
r&=|DI|=|MI|=|EI|
\\
r
&=\frac{S_{\triangle AFB}}{\rho}
=\frac{12+7\sqrt3}{7+4\sqrt3}
=
\frac{(12+7\sqrt3)(7-4\sqrt3)}{(7+4\sqrt3)(7-4\sqrt3)}
\\
&=\sqrt3
,\\
\end{align}
hence,
\begin{align}
|AI|&=|AC|-|CI|=2
,\\
\cos\angle DIA&=\frac{|DI|}{|AI|}=\frac{\sqrt3}2
,\\
\angle DIA&=\frac \pi 6
,
\end{align}
and the answer is
\begin{align}
T&=\tfrac12\cdot2\cdot\sqrt3\cdot\tfrac12
-\tfrac12\cdot\tfrac{\pi}{6}\cdot(\sqrt3)^2
\\
&=\tfrac14(2\sqrt3-\pi)
\approx 0.08062724
.
\end{align}
Best Answer
Let $r$ be the radius of the circle. The area of the triangle is $\frac12 (base \times height)$, if you split each of the triangles down the middle (to give a right angled triangle) then the height will be given by $$height = r \cos(\tfrac\pi n),$$ notice how we have halfed the angle as we split the triangle in 2. Similarly the base will be given by $$ base = 2 r \sin(\tfrac\pi n), $$ where we have remembered to multiply by two as the base of the right angled triangle is only half that of the isoceles.
Putting these quantities into the formula for the area of the isoceles triangle we get \begin{align*} A_{triangle} &= \frac 12 \times 2 r \sin(\tfrac\pi n) \times r \cos(\tfrac\pi n) \\ &= r^2 \sin(\tfrac\pi n)\cos(\tfrac\pi n) \end{align*}
Now there are n triangles so the total area of all the triangles would be given by $$ A_{total} = r^2 n\sin(\tfrac\pi n)\cos(\tfrac\pi n) , $$ which can be rewritten using the double angle formula as $$ A_{total} = r^2 n \frac 1 2 \sin(\tfrac{2\pi} {n}). $$ The area of the circle can then be found if we take the limit as $n\rightarrow\infty$. $$ A_{circ} = \lim \limits_{n \rightarrow \infty} r^2 n \frac 1 2 \sin(\tfrac{2\pi} {n}), $$ and using the fact that for very small $x$, $\sin(x) \approx x$ this can be seen to be \begin{align*} A_{circ} &= \lim \limits_{n \rightarrow \infty} r^2\frac{2\pi n} {2n} \\ &=\pi r^2, \end{align*} which is precisely what we want.