A) \[{{c}^{2}}/{{a}^{2}}{{b}^{2}}\]
B) \[\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}\]
C) \[\frac{1}{ab}\]
D) none of these
Correct Answer: B
Solution :
\[\therefore \] \[\frac{\cos 2A}{{{a}^{2}}}-\frac{\cos 2B}{{{b}^{2}}}\] \[=\frac{1-2{{\sin }^{2}}A}{{{a}^{2}}}-\frac{1-2{{\sin }^{2}}B}{{{b}^{2}}}\] Applying sine rule \[=\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}-2k(1-1)\] \[=\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}\]
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