In any triangle ABC,frac{cos 2A}{a^2} - frac{ | vedclass

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(B) We know that $\cos 2A = 1 - 2\sin^2 A$ and $\cos 2B = 1 - 2\sin^2 B$. Substituting these into the expression: $\frac{1 - 2\sin^2 A}{a^2} - \frac{1

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

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(B) We know that $\cos 2A = 1 - 2\sin^2 A$ and $\cos 2B = 1 - 2\sin^2 B$. Substituting these into the expression: $\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} = \frac{1}{a^2} - \frac{2\sin^2 A}{a^2} - \left(\frac{1}{b^2} - \frac{2\sin^2 B}{b^2}\right)$ $= \frac{1}{a^2} - \frac{1}{b^2} - 2 \left( \frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2} \right)$ By the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = k$,so $\frac{\sin^2 A}{a^2} = \frac{\sin^2 B}{b^2} = k^2$. Thus,$\frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2} = 0$. Therefore,the expression simplifies to $\frac{1}{a^2} - \frac{1}{b^2}$.

In any $\triangle ABC$,$\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} =$

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