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In any $\triangle A B C$, the simplified form of $\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}$ is
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$\frac{1}{a^{2}}-\frac{1}{b^{2}}$
$\begin{aligned} \frac{\cos 2 A}{a^{2}} &-\frac{\cos 2 B}{b^{2}} \\ &=\frac{\left(1-2 \sin ^{2} A\right)}{a^{2}}-\frac{\left(1-2 \sin ^{2} B\right)}{b^{2}} \\ &=\frac{\left(1-2 a^{2} k^{2}\right)}{a^{2}}-\frac{\left(1-2 b^{2} k^{2}\right)}{b^{2}} \\ &=\left(\frac{1}{a^{2}}-2 k^{2}\right)-\left(\frac{1}{b^{2}}-2 k^{2}\right) \\ &=\left(\frac{1}{a^{2}}-\frac{1}{b^{2}}\right)-2 k^{2}+2 k^{2} \\ &=\frac{1}{a^{2}}-\frac{1}{b^{2}} \end{aligned}$
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