The value of $\frac{\sin (\mathrm{B}+\mathrm{A})+\cos (\mathrm{B}-\mathrm{A})}{\sin (\mathrm{B}-\mathrm{A})+\cos (\mathrm{B}+\mathrm{A})}$ is equal to

$\begin{array}{c}=\frac{\sin (\mathrm{B}+\mathrm{A})+\cos (\mathrm{B}-\mathrm{A})}{\sin (\mathrm{B}-\mathrm{A})+\cos (\mathrm{B}+\mathrm{A})}\\ =\frac{\sin (\mathrm{B}+\mathrm{A})+\sin \left({90}^{\circ }-\left(\mathrm{B}-\mathrm{A}\right)\right)}{\sin (\mathrm{B}-\mathrm{A})+\sin \left({90}^{\circ }-\left(\mathrm{B}+\mathrm{A}\right)\right)}\\ =\frac{2\sin \left(\mathrm{A}+{45}^{\circ }\right)\cos \left({45}^{\circ }-\mathrm{B}\right)}{2\sin \left({45}^{\circ }-\mathrm{A}\right)\cos \left({45}^{\circ }-\mathrm{B}\right)}\\ =\frac{\sin \left(\mathrm{A}+{45}^{\circ }\right)}{\sin \left({45}^{\circ }-\mathrm{A}\right)}=\frac{\frac{1}{\sqrt{2}}\sin \mathrm{A}+\frac{1}{\sqrt{2}}\cos \mathrm{A}}{\frac{1}{\sqrt{2}}\cos \mathrm{A}-\frac{1}{\sqrt{2}}\sin \mathrm{A}}\\ =\frac{\cos \mathrm{A}+\sin \mathrm{A}}{\cos \mathrm{A}-\sin \mathrm{A}}\end{array}$