The 2R comes from the standard proof of the sine rule.
Draw your triangle ABC with its circumscribing circle, radius R, centre O.
With the usual notation, angle BOC = 2A, so if you drop a perpendicular from O to BC you get a rt-angled triangle with
two sides R and a/2 and an angle A, from which a/2R = sin(A) and a/sin(A) = 2R.
Repeat for B and C and equate the three.
Here's an alternative answer to the original question.
LHS,
\(\displaystyle \sin(C)=2\sin(C/2)\cos(C/2)=2\sin(90-(A+B)/2)\cos(90-(A+B)/2)\\ =2\cos((A+B)/2)\sin((A+B)/2) \dots\dots\dots(1)\)
RHS
\(\displaystyle \frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)}=\frac{2\sin((A+B)/2)\cos((A-B)/2)}{2\cos((A+B)/2)\cos((A-B)/2)}=\frac{\sin((A+B)/2)}{\cos((A+B)/2)}\dots\dots(2)\)
Equate (1) and (2), cancel and cross multiply,
\(\displaystyle \cos^{2}((A+B)/2)=1/2,\\\text{so }\cos((A+B)/2) = 1/\sqrt{2}\\\text{so }(A+B)/2=45\text{ deg}\\A+B=90\text{ deg}\\C=90\text{ deg}.\)
Tiggsy.
