Trigonometric Expression in Triangle Opposite Angles

To evaluate the given trigonometric expression, we first simplify it using standard trigonometric identities.

Given expression: $E = \frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P}$

Recall the double-angle identity $\sin 2P = 2 \sin P \cos P$. Substituting this into $E$: $E = \frac{2 \sin P - 2 \sin P \cos P}{2 \sin P + 2 \sin P \cos P} = \frac{2 \sin P (1 - \cos P)}{2 \sin P (1 + \cos P)} = \frac{1 - \cos P}{1 + \cos P}$

Using the half-angle identities $1 - \cos P = 2 \sin^2 \left(\frac{P}{2}\right)$ and $1 + \cos P = 2 \cos^2 \left(\frac{P}{2}\right)$: $E = \frac{2 \sin^2 \left(\frac{P}{2}\right)}{2 \cos^2 \left(\frac{P}{2}\right)} = \tan^2 \left(\frac{P}{2}\right)$

Next, we calculate the semi-perimeter $s$ of triangle $PQR$: $s = \frac{a + b + c}{2} = \frac{2 + \frac{7}{2} + \frac{5}{2}}{2} = \frac{8}{2} = 4$

The terms $s - a$, $s - b$, and $s - c$ are: $s - a = 4 - 2 = 2$ $s - b = 4 - \frac{7}{2} = \frac{1}{2}$ $s - c = 4 - \frac{5}{2} = \frac{3}{2}$

By Heron's formula, the area of the triangle $\Delta$ satisfies: $\Delta^2 = s(s-a)(s-b)(s-c) = 4 \times 2 \times \frac{1}{2} \times \frac{3}{2} = 6$

Using the half-angle formula for tangent in terms of area $\Delta$ and sides: $\tan \left(\frac{P}{2}\right) = \frac{\Delta}{s(s-a)}$

Substituting $s = 4$ and $s - a = 2$: $\tan \left(\frac{P}{2}\right) = \frac{\Delta}{4 \times 2} = \frac{\Delta}{8}$

Squaring both sides yields: $E = \tan^2 \left(\frac{P}{2}\right) = \frac{\Delta^2}{64}$

Substituting $\Delta^2 = 6$: $E = \frac{6}{64} = \frac{3}{32}$

Now, let's evaluate Option C: $\left(\frac{3}{4\Delta}\right)^2 = \frac{9}{16\Delta^2} = \frac{9}{16 \times 6} = \frac{9}{96} = \frac{3}{32}$

Thus, the given expression is equal to $\left(\frac{3}{4\Delta}\right)^2$.

Correct Option: C