A) \[\frac{{{a}^{2}}-{{b}^{2}}}{{{c}^{2}}}\]
B) \[\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}}\]
C) \[\frac{{{c}^{2}}}{{{a}^{2}}-{{b}^{2}}}\]
D) \[\frac{{{c}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
Correct Answer: A
Solution :
\[\frac{\sin (A-B)}{\sin (A+B)}=\frac{\sin A\cos B-\sin B\cos A}{\sin C}\] \[=\frac{a}{c}\cos B-\frac{b}{c}\cos A\] But \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac},\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[\Rightarrow \frac{a}{c}\cos B-\frac{b}{c}\cos A=\frac{1}{2{{c}^{2}}}\] \[({{a}^{2}}+{{c}^{2}}-{{b}^{2}}-{{b}^{2}}-{{c}^{2}}+{{a}^{2}})\] \[=\frac{{{a}^{2}}-{{b}^{2}}}{{{c}^{2}}}\].
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