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  4. In Δ ABC,( cos 2A/a2)-( cos 2B/b2) is equal to:

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Q. In $ \Delta \,ABC,\frac{\cos 2A}{{{a}^{2}}}-\frac{\cos 2B}{{{b}^{2}}} $ is equal to:

Bihar CECEBihar CECE 2002

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}}} $