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In $\triangle \mathrm{ABC}$, with usual notations, $2 \mathrm{ac} \sin \left(\frac{1}{2}(\mathrm{~A}-\mathrm{B}+\mathrm{C})\right)$ is equal to
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The correct answer is:
$c^2+a^2-b^2$
$\begin{aligned} 2 \mathrm{ac} \sin \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2} & =2 \mathrm{ac} \sin \frac{\pi-2 \mathrm{~B}}{2} \\ & =2 \mathrm{ac} \cos \mathrm{B}\end{aligned}$
$$
=2 \mathrm{ac} \frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}
$$....[By cosine rule $]$
$=c^2+a^2-b^2$
$$
=2 \mathrm{ac} \frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}
$$....[By cosine rule $]$
$=c^2+a^2-b^2$
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