In a triangle ABC,a^2 sin 2C + c^2 sin 2A is | vedclass

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(B) We know that $a = 2R \sin A$ and $c = 2R \sin C$. Substituting these into the expression:$a^2 \sin 2C + c^2 \sin 2A = (2R \sin A)^2 (2 \sin C \cos

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$4$

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(B) We know that $a = 2R \sin A$ and $c = 2R \sin C$. Substituting these into the expression:$a^2 \sin 2C + c^2 \sin 2A = (2R \sin A)^2 (2 \sin C \cos C) + (2R \sin C)^2 (2 \sin A \cos A)$$= 8R^2 \sin^2 A \sin C \cos C + 8R^2 \sin^2 C \sin A \cos A$$= 8R^2 \sin A \sin C [\sin A \cos C + \cos A \sin C]$$= 8R^2 \sin A \sin C \sin(A + C)$Since $A + B + C = 180^{\circ}$,$\sin(A + C) = \sin B$.$= 8R^2 \sin A \sin B \sin C$Using the formula for the area of a triangle $\Delta = \frac{abc}{4R}$,we have $abc = 4R\Delta$. Also,$\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.$= 8R^2 \left( \frac{a}{2R} \right) \left( \frac{b}{2R} \right) \left( \frac{c}{2R} \right) = \frac{abc}{R} = \frac{4R\Delta}{R} = 4\Delta$.

In a $\triangle ABC$,$a^2 \sin 2C + c^2 \sin 2A$ is equal to (in $\Delta$)

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