Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:

Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$

Here are the two solutions that I found:

I think the problem with the second solution may have to do with the assumption that:
$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$
is equivalent to
$$A + \frac{\pi}{6} = -B + \frac{\pi}{6}$$
But can't you say that $\cos N = \cos M$ is equivalent to $\pm N = \pm M$ for all values of $N$ and $M$ (because cosine is an even function)?

$$\sqrt3\cos A-\sin A=\sqrt3\cos B+\sin B$$
$$\implies\cos(A+30^\circ)=\cos(B-30^\circ)$$
$$A+30^\circ=360^\circ n\pm(B-30^\circ) $$
$$-\implies A+B=360^\circ n$$ which is impossible as  $0<A+B<180^\circ$
$+\implies A-B=360^\circ n-60^\circ$
As $-90<A-B<90,n=0$
Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html in  anyone of the given relation
