In Delta ABC, {b^2}cos 2A - {a^2}cos 2B = | vedclass

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(A) Given expression: ${b^2}\cos 2A - {a^2}\cos 2B$ Using the identity $\cos 2	heta = 1 - 2\sin^2	heta$: $= {b^2}(1 - 2\sin^2 A) - {a^2}(1 - 2\sin^2

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

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(A) Given expression: ${b^2}\cos 2A - {a^2}\cos 2B$ Using the identity $\cos 2	heta = 1 - 2\sin^2	heta$: $= {b^2}(1 - 2\sin^2 A) - {a^2}(1 - 2\sin^2 B)$ $= {b^2} - 2{b^2}\sin^2 A - {a^2} + 2{a^2}\sin^2 B$ From the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R$,so $a = 2R\sin A$ and $b = 2R\sin B$. Thus,${b^2}\sin^2 A = (2R\sin B)^2 \sin^2 A = 4R^2 \sin^2 B \sin^2 A$ and ${a^2}\sin^2 B = (2R\sin A)^2 \sin^2 B = 4R^2 \sin^2 A \sin^2 B$. Therefore,${b^2}\sin^2 A = {a^2}\sin^2 B$. Substituting this into the expression: $= {b^2} - {a^2} - 2({b^2}\sin^2 A - {a^2}\sin^2 B)$ $= {b^2} - {a^2} - 2(0) = {b^2} - {a^2}$.

In $\Delta ABC,$ ${b^2}\cos 2A - {a^2}\cos 2B = $

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