In $△ A B C, b^{2} c o s 2 A - a^{2} c o s 2 B$ =

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 $b^{2} - a^{2}$
$b^{2} c o s 2 A - a^{2} c o s 2 B = b^{2} (1 - 2 s i n^{2} A) - a^{2} (1 - 2 s i n^{2} B)$
$= b^{2} - a^{2} - 2 (b^{2} s i n^{2} A - a^{2} s i n^{2} B) = b^{2} - a^{2} .$

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Similar questions

△ A B C

\frac{cos 2 A}{a^{2}} - \frac{cos 2 B}{b^{2}} = (\frac{1}{a^{2}} - \frac{1}{b^{2}})

A B C,

\frac{cos 2 A}{a^{2}} - \frac{cos 2 B}{b^{2}}

A B C,

cos (A - C) cos B + cos 2 B = 0

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$\frac{c o s 2 A}{a^{2}} - \frac{c o s 2 B}{b^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}$

\frac{cos 2 A}{a^{2}} - \frac{cos 2 B}{b^{2}} = (\frac{1}{a^{2}} - \frac{1}{b^{2}})

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