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In any triangle ABC , the simplified form of cos 2A /a^2 - cos 2B /b^2

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Solution: Here,

$fraccos 2Aa^2-fraccos 2Bb^2$

$=frac1-2sin^2Aa^2-frac1-2sin^2Bb^2\ \ Rightarrow frac1a^2-frac1b^2-frac2sin^2Aa^2+frac2sin^2Bb^2\ \ Rightarrow frac1a^2-frac1b^2-2(fracsin^2Aa^2-fracsin^2Bb^2)=frac1a^2-frac1b^2.$

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Deependra Verma

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