In a Delta ABC,{a^2}sin 2C + {c^2}sin 2A = | vedclass

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(D) Given expression: ${a^2}\sin 2C + {c^2}\sin 2A$$= {a^2}(2\sin C \cos C) + {c^2}(2\sin A \cos A)$Using the area formula $\Delta = \frac{1}{2}ab \si

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$4\Delta$

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(D) Given expression: ${a^2}\sin 2C + {c^2}\sin 2A$$= {a^2}(2\sin C \cos C) + {c^2}(2\sin A \cos A)$Using the area formula $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A$,we have $\sin C = \frac{2\Delta}{ab}$ and $\sin A = \frac{2\Delta}{bc}$.Substituting these values:$= 2{a^2}\left( \frac{2\Delta}{ab} \cos C \right) + 2{c^2}\left( \frac{2\Delta}{bc} \cos A \right)$$= 4\Delta \left( \frac{a \cos C + c \cos A}{b} \right)$Since $a \cos C + c \cos A = b$ (Projection Rule):$= 4\Delta \left( \frac{b}{b} \right) = 4\Delta$.

In a $\Delta ABC$,${a^2}\sin 2C + {c^2}\sin 2A = $

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