In $△ A B C, b^{2} cos 2 A - a^{2} cos 2 B$ iis equal to

b^{2} - a^{2}

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b^{2} - c^{2}

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c^{2} - a^{2}

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a^{2} + b^{2} + c^{2}

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Solution

The correct option is A $b^{2} - a^{2}$

$solution: b^{2} cos 2 A - a^{2} cos 2 B$

$= b^{2} ({cos}^{2} A - {sin}^{2} A) - a^{2} ({cos}^{2} B - {sin}^{2} B)$

$= b^{2} {cos}^{2} A - b^{2} {sin}^{2} A - a^{2} {cos}^{2} B + a^{2} {sin}^{2} B - (1)$

$Now using sine Rule, we know that$

$\frac{a}{sin A} = \frac{b}{sin B}$

$\Rightarrow a sin B = b sin A$

$\Rightarrow a^{2} {sin}^{2} B = b^{2} {sin}^{2} A$

so from equation. (1)

$b^{2} cos 2 A - a^{2} cos 2 B = b^{2} {cos}^{2} A - b^{2} {sin}^{2} A - a^{2} {cos}^{2} B + a^{2} {sin}^{2} B$

$= b^{2} {cos}^{2} A - a^{2} {cos}^{2} B$

$= (b cos A)^{2} - (a cos B)^{2}$

$= {[b (\frac{b^{2} + c^{2} - a^{2}}{2 b c})]}^{2} - {[a (\frac{a^{2} + c^{2} - b^{2}}{2 a c})]}^{2}$

$= \frac{1}{4 c^{2}} {[(b^{2} + c^{2} - a^{2})]}^{2} - \frac{1}{4 c^{2}} {[(a^{2} + c^{2} - b^{2})]}^{2}$

$= \frac{1}{4 c^{2}} [(b^{2} + c^{2} - a^{2} + a^{2} + c^{2} - b^{2}) (b^{2} + c^{2} - a^{2} - a^{2} - c^{2} + b^{2})]$

$= \frac{1}{4 c^{2}} (2 c^{2}) (2 b^{2} - 2 a^{2})$

$= b^{2} - a^{2}$

$∴$ Answer: option (A)

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