Let a triangle A B C be inscribed in a circle of radius 2 units. If the 3 bisectors of the angles A ,  …

Consider the figure below.

The triangle $ABC$ is equilateral.

Therefore, $A=B=C=60°$.

Substitute the value in the given expression,

${\left[\frac{A{A}_1\cos {\displaystyle \frac{A}{2}}+B{B}_1\cos {\displaystyle \frac{B}{2}}+C{C}_1\cos {\displaystyle \frac{C}{2}}}{\sin A+\sin B+\sin C}\right]}^2$

$={\left\{\frac{4\cos {\displaystyle \frac{60°}{2}}+4\cos {\displaystyle \frac{60°}{2}}+4\cos {\displaystyle \frac{60°}{2}}}{\sin 60°+\sin 60°+\sin 60°}\right\}}^2$

$={\left[\frac{4\times 3{\displaystyle \frac{\sqrt{3}}{2}}}{3{\displaystyle \frac{\sqrt{3}}{2}}}\right]}^2$

$=16$