Let PQR be a triangle of area $∆$ with $a=2, b=\frac{7}{2}$ and $c=\frac{5}{2},$ where $a, b$ and $c$ are lengths of the sides of triangle opposite to the angles at $P, Q$ and $R,$respectively. Then, $\frac{2\sin P-\sin 2P}{2\sin P+\sin 2P}$ is equal to 

Concept Involved

If $∆ABC$ has sides$a, b, c.$ Then

   $\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

where, $s=\frac{a+b+c}{2}$

Given, $a=2,b=\frac{7}{2},c=\frac{5}{2}$

Then, $s=\frac{2+\frac{7}{2}+\frac{5}{2}}{2}=4$

$\therefore \frac{2\sin P-\sin 2P}{2\sin P+\sin 2P}=\frac{2\sin P(1-\cos P)}{2\sin P(1+\cos P)}$

                                $=\frac{2{\sin }^2\frac{P}{2}}{2{\cos }^2\frac{P}{2}}={\tan }^2\frac{P}{2}$

                               $\begin{array}{l}=\frac{(s-b)(s-c)}{s(s-a)}\times \frac{(s-b)(s-c)}{(s-b)(s-c)}\\ =\frac{(s-b{)}^2(s-c{)}^2}{{\mathrm{\Delta }}^2}=\frac{{\left(4-\frac{7}{2}\right)}^2{\left(4-\frac{5}{2}\right)}^2}{{\mathrm{\Delta }}^2}\end{array}$

                               $={\left(\frac{3}{4\mathrm{\Delta }}\right)}^2$