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Let $p, q$ and $r$ be the side4s opposite to the angles $P, Q$ and $R,$ respectively in a $\Delta P Q R$. Then, $2 p r \sin \left(\frac{p-Q+R}{2}\right)$ equals
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Verified Answer
The correct answer is:
$p^{2}+r^{2}-q^{2}$
In $\Delta P Q R, P+Q+R=180^{\circ}$
$\begin{aligned} \therefore 2 p r \sin \left(\frac{P-Q+R}{2}\right) &=2 p r \sin \frac{180^{\circ}-Q-Q}{2} \\ &=2 p r \sin \left(90^{\circ}-Q\right) \\ &=2 p r \cos Q \\=&\left(\quad \ln \Delta P Q R \cos Q=\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right) \\=& 2 p r\left(\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right)=p^{2}+r^{2}-q^{2} \end{aligned}$
$\begin{aligned} \therefore 2 p r \sin \left(\frac{P-Q+R}{2}\right) &=2 p r \sin \frac{180^{\circ}-Q-Q}{2} \\ &=2 p r \sin \left(90^{\circ}-Q\right) \\ &=2 p r \cos Q \\=&\left(\quad \ln \Delta P Q R \cos Q=\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right) \\=& 2 p r\left(\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right)=p^{2}+r^{2}-q^{2} \end{aligned}$
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