In $\triangle \mathrm{ABC}$, with usual notations, $2 \mathrm{ac} \sin…
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
- $a^2+b^2-c^2$
- $c^2+a^2-b^2$
- $b^2-c^2-a^2$
- $c^2-a^2-b^2$
Solution
$\begin{aligned} 2 \mathrm{ac} \sin \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2} & =2 \mathrm{ac} \sin \frac{\pi-2 \mathrm{~B}}{2} \\ & =2 \mathrm{ac} \cos \mathrm{B}\end{aligned}$
$=2 \mathrm{ac} \frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}$....[By cosine rule $]$
$=c^2+a^2-b^2$
Asked in: MHT CET 2023 (14 May Shift 1)
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