A) \[{{a}^{2}}-{{b}^{2}}\]
B) \[\frac{1}{{{a}^{2}}-{{b}^{2}}}\]
C) \[\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}\]
D) \[{{a}^{2}}+{{b}^{2}}\]
Correct Answer: C
Solution :
\[\frac{\cos \,2A}{{{a}^{2}}}-\frac{\cos \,2\,B}{{{b}^{2}}}\] \[=\frac{(1-2\,{{\sin }^{2}}A)}{{{a}^{2}}}-\frac{(1-2\,{{\sin }^{2}}B)}{{{b}^{2}}}\] \[=\frac{(1-2\,{{a}^{2}}{{k}^{2}})}{{{a}^{2}}}-\frac{(1-2\,{{b}^{2}}{{k}^{2}})}{{{b}^{2}}}\] \[\left( \because \,\frac{\sin A}{a}=\frac{\sin B}{b}=k \right)\] \[=\left( \frac{1}{{{a}^{2}}}-2{{k}^{2}} \right)-\left( \frac{1}{{{b}^{2}}}-2{{k}^{2}} \right)\] \[=\left( \frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}} \right)-2{{k}^{2}}+2{{k}^{2}}\] \[=\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}\]
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