Let p, q and r be the sides opposite to the a | vedclass

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(B) In $\Delta PQR$, the sum of angles is $P+Q+R = 180^{\circ}$.Since $P+R = 180^{\circ}-Q$, we substitute this into the expression:$2pr \sin \left(\f

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$p^{2}+r^{2}-q^{2}$

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(B) In $\Delta PQR$, the sum of angles is $P+Q+R = 180^{\circ}$.Since $P+R = 180^{\circ}-Q$, we substitute this into the expression:$2pr \sin \left(\frac{P+R-Q}{2}\right) = 2pr \sin \left(\frac{180^{\circ}-Q-Q}{2}\right)$$= 2pr \sin \left(\frac{180^{\circ}-2Q}{2}\right)$$= 2pr \sin (90^{\circ}-Q)$$= 2pr \cos Q$Using the Law of Cosines, $\cos Q = \frac{p^{2}+r^{2}-q^{2}}{2pr}$.Substituting this value:$= 2pr \left(\frac{p^{2}+r^{2}-q^{2}}{2pr}\right)$$= p^{2}+r^{2}-q^{2}$.

Let $p, q$ and $r$ be the sides opposite to the angles $P, Q$ and $R$ respectively in a $\Delta PQR$. Then, $2pr \sin \left(\frac{P-Q+R}{2}\right)$ equals

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