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Prove that b^(2) cos 2 A - a^(2) cos 2B ...

Prove that `b^(2) cos 2 A - a^(2) cos 2B = b^(2) -a^(2)`

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`b^(2) cos 2A - a^(2) cos 2B`
`=b^(2) (1-2 sin^(2)A) - a^(2) (1 - 2 sin^(2) B)`
`= b^(2) - a^(2) - 2 (b^(2) sin^(2) B)`
`= b^(2) -a^(2) -2 (b^(2) sin^(2) A - a^(2) sin^(2) B) = b^(2) - a^(2) " " ( :' a sin B = b sin A)`
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