A) \[\frac{\cos B+\sin B}{\cos B-\sin B}\]
B) \[\frac{\cos A+\sin A}{\cos A-\sin A}\]
C) \[\frac{\cos A-\sin A}{\cos A+\sin A}\]
D) None of these
Correct Answer: B
Solution :
\[\frac{\sin \,(B+A)+\cos \,(B-A)}{\sin \,(B-A)+\cos \,(B+A)}\] \[=\frac{\sin \,(B+A)+\sin \,({{90}^{o}}-\overline{B-A})}{\sin \,(B-A)+\sin \,({{90}^{o}}-\overline{A+B})}\] \[=\,\frac{2\,\sin \,(A+{{45}^{o}})\,\cos \,({{45}^{o}}-B)}{2\,\sin \,({{45}^{o}}-A)\,\cos \,({{45}^{o}}-B)}\] \[=\frac{\sin \,(A+{{45}^{o}})}{\sin \,({{45}^{o}}-A)}=\frac{\cos A+\sin A}{\cos A-\sin A}\].
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