Explanation

(b)
$\frac{\sin(B + A) + cos(B - A)}{\sin(B - A) + \cos{(B + A\ )}}$
$= \frac{\sin(B + A) + \sin{(90{^\circ} - \overline{B - A}})}{\sin(B - A) + \sin\left( 90{^\circ} - \overline{A + B} \right)}$
$= \frac{2\ sin(A + 45{^\circ})cos(45{^\circ} - B)}{2\sin(45{^\circ} - A)cos(45{^\circ} - B)}$
$= \frac{sin(A + 45{^\circ})}{sin(45{^\circ} - A)} = \frac{\frac{1}{\sqrt{2}}\sin{A + \frac{1}{\sqrt{2}}}\cos A}{\frac{1}{\sqrt{2}}\cos{A - \frac{1}{\sqrt{2}}}\sin A}$
$= \frac{\cos{A + \sin A}}{\cos{A - \sin A}}$