Mathematics - Properties of Triangles Question with Solution | TestHub

$\frac{2\sin P-\sin 2P}{2\sin P+\sin 2P}=\frac{2\sin P-2\sin P\cos P}{2\sin P+2\sin P\cos P}=\frac{1-\cos P}{1-\sin P}$ $=\frac{2{\sin }^2\frac{P}{2}}{2{\cos }^2\frac{P}{2}}={\tan }^2\frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}$ where $s=\frac{a+b+c}{2}$ $=\frac{(s-b{)}^2(s-c{)}^2}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b{)}^2(a+b-c{)}^2}{16.{\Delta }^2}$ $=\frac{{\left(2+\frac{5}{2}-\frac{7}{2}\right)}^2{\left(2+\frac{7}{2}-\frac{5}{2}\right)}^2}{16{\Delta }^2}=\frac{1\times 9}{16{\Delta }^2}={\left(\frac{3}{4\Delta }\right)}^2$