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Let $P Q R$ be a triangle of area $\Delta$ with $a=2, b=\frac{7}{2}$ and $c=\frac{5}{2} ;$ where $a, b$, and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ and $R$ respectively. Then $\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}$ equals.
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Verified Answer
The correct answer is:
$\left(\frac{3}{4 \Delta}\right)^{2}$
$\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}=\frac{2 \sin P-2 \sin P \cos P}{2 \sin P+2 \sin P \cos P}=\frac{1-\cos P}{1-\sin P}$
$=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}=\tan ^{2} \frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}$ where $s=\frac{a+b+c}{2}$
$=\frac{(s-b)^{2}(s-c)^{2}}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b)^{2}(a+b-c)^{2}}{16 . \Delta^{2}}$
$=\frac{\left(2+\frac{5}{2}-\frac{7}{2}\right)^{2}\left(2+\frac{7}{2}-\frac{5}{2}\right)^{2}}{16 \Delta^{2}}=\frac{1 \times 9}{16 \Delta^{2}}=\left(\frac{3}{4 \Delta}\right)^{2}$
$=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}=\tan ^{2} \frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}$ where $s=\frac{a+b+c}{2}$
$=\frac{(s-b)^{2}(s-c)^{2}}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b)^{2}(a+b-c)^{2}}{16 . \Delta^{2}}$
$=\frac{\left(2+\frac{5}{2}-\frac{7}{2}\right)^{2}\left(2+\frac{7}{2}-\frac{5}{2}\right)^{2}}{16 \Delta^{2}}=\frac{1 \times 9}{16 \Delta^{2}}=\left(\frac{3}{4 \Delta}\right)^{2}$
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