If sin A + sin B = 1 and cos A + cos B = (3)^{1/2 }A and B < 90^{o }Find a value for (A+B) ?
Let's make a silly assumption and assume $a=b$ (this would be a special case, but who knows, it might pan out).
Then we have:
$2\sin A = 1$ and $2\cos A = \sqrt{3}$
So, $\sin A = \dfrac{1}{2}$ and $\cos A = \dfrac{\sqrt{3}}{2}$
This is a well known angle. Plug it in for $A$, and since we set $B$ equal to the same, it is trivial to find $A+B$.
$\dfrac{\sin{A}+\sin{B}}{\cos{A}+\cos{B}} = \dfrac{2\sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2} \right)}{2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)} = \tan\left(\dfrac{A+B}{2}\right) = \dfrac{1}{\sqrt{3}}$
note that $\tan(30^\circ) = \dfrac{1}{\sqrt{3}}$