In any triangle ABC,the simplified form of fr | vedclass

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(C) Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R = k$,so $\sin A = ak$ and $\sin B = bk$.We know $\cos 2A = 1 - 2\sin^2 A$ and $\cos

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$\frac{1}{a^2} - \frac{1}{b^2}$

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(C) Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R = k$,so $\sin A = ak$ and $\sin B = bk$.We know $\cos 2A = 1 - 2\sin^2 A$ and $\cos 2B = 1 - 2\sin^2 B$.Substituting these into the expression:$\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2}$$= \frac{1 - 2(ak)^2}{a^2} - \frac{1 - 2(bk)^2}{b^2}$$= \frac{1 - 2a^2k^2}{a^2} - \frac{1 - 2b^2k^2}{b^2}$$= (\frac{1}{a^2} - 2k^2) - (\frac{1}{b^2} - 2k^2)$$= \frac{1}{a^2} - \frac{1}{b^2}$

In any $\triangle ABC$,the simplified form of $\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2}$ is

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