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In a $\triangle A B C, 2 a c \sin \frac{1}{2}(A-B+C)$ is equal to
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The correct answer is:
$a^2+c^2-b^2$
$\begin{aligned} & \text { In a } \triangle A B C \\ & \quad A+B+C=180^{\circ} \\ & \text { Then, } 2 a c \sin \frac{1}{2}(A-B+C) \\ & \Rightarrow 2 a c \sin \frac{1}{2}(180-B-B) \quad\left[\because A+C=180^{\circ}-B\right] \\ & \Rightarrow \quad 2 a c \sin \left(90^{\circ}-B\right) \\ & =2 a c \cos B [\because \sin (90-\theta)=\cos \theta]
\\ & =2 a c\left\{\frac{a^2+c^2-b^2}{2 a c}\right\} \quad\left[\because \cos B=\frac{a^2+c^2-b^2}{2 a c}\right] \\ & =a^2+c^2-b^2\end{aligned}$
\\ & =2 a c\left\{\frac{a^2+c^2-b^2}{2 a c}\right\} \quad\left[\because \cos B=\frac{a^2+c^2-b^2}{2 a c}\right] \\ & =a^2+c^2-b^2\end{aligned}$
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