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$\frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}=$
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Verified Answer
The correct answer is:
$\frac{\cos A+\sin A}{\cos A-\sin A}$
$\begin{aligned}
& \frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)} \\
&=\frac{\sin (B+A)+\sin \left(90^{\circ}-B-A\right)}{\sin (B-A)+\sin \left(90^{\circ}-A+B\right)} \\
&=\frac{2 \sin \left(A+45^{\circ}\right) \cos \left(45^{\circ}-B\right)}{2 \sin \left(45^{\circ}-A\right) \cos \left(45^{\circ}-B\right)}
\end{aligned}$
$=\frac{\sin \left(A+45^{\circ}\right)}{\sin \left(45^{\circ}-A\right)}=\frac{\cos A+\sin A}{\cos A-\sin A}$
& \frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)} \\
&=\frac{\sin (B+A)+\sin \left(90^{\circ}-B-A\right)}{\sin (B-A)+\sin \left(90^{\circ}-A+B\right)} \\
&=\frac{2 \sin \left(A+45^{\circ}\right) \cos \left(45^{\circ}-B\right)}{2 \sin \left(45^{\circ}-A\right) \cos \left(45^{\circ}-B\right)}
\end{aligned}$
$=\frac{\sin \left(A+45^{\circ}\right)}{\sin \left(45^{\circ}-A\right)}=\frac{\cos A+\sin A}{\cos A-\sin A}$
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