Let PQR be a triangle of area `Delta` with `a = 2, b = 7//2`, and `c = 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 `(2 sin P - sin 2P)/(2 sin P + sin 2P)` equals

Options

A. `(3)/(4Delta)`
B. `(45)/(4Delta)`
C. `((3)/(4Delta))^(2)`
D. `((45)/(4Delta))^(2)`

Explanation

To solve the problem, we need to find the value of \((2 \sin P - \sin 2P)/(2 \sin P + \sin 2P)\) for the triangle PQR with given side lengths \(a = 2\), \(b = \frac{7}{2}\), and \(c = \frac{5}{2}\).

### Step 1: Use the double angle formula for sine
We know that:
\[ \sin 2P = 2 \sin P \cos P \]
Substituting this into the expression, we have:
\[ \frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P} = \frac{2 \sin P - 2 \sin P \cos P}{2 \sin P + 2 \sin P \cos P} \]

### Step 2: Factor out \(2 \sin P\)
Now, we can factor \(2 \sin P\) from both the numerator and the denominator:
\[ = \frac{2 \sin P (1 - \cos P)}{2 \sin P (1 + \cos P)} \]
Since \(2 \sin P\) is common in both numerator and denominator, we can cancel it out (assuming \(\sin P \neq 0\)):
\[ = \frac{1 - \cos P}{1 + \cos P} \]

### Step 3: Use the half-angle identities
Using the half-angle identities:
\[ 1 - \cos P = 2 \sin^2 \left(\frac{P}{2}\right) \]
\[ 1 + \cos P = 2 \cos^2 \left(\frac{P}{2}\right) \]
Substituting these into our expression gives:
\[ = \frac{2 \sin^2 \left(\frac{P}{2}\right)}{2 \cos^2 \left(\frac{P}{2}\right)} \]
The 2's cancel out:
\[ = \tan^2 \left(\frac{P}{2}\right) \]

### Step 4: Relate \(\tan^2 \left(\frac{P}{2}\right)\) to the area of the triangle
Using the formula for \(\tan^2 \left(\frac{P}{2}\right)\):
\[ \tan^2 \left(\frac{P}{2}\right) = \frac{(s-b)(s-c)}{s(s-a)} \]
where \(s\) is the semi-perimeter of the triangle.

### Step 5: Calculate the semi-perimeter \(s\)
The semi-perimeter \(s\) is given by:
\[ s = \frac{a + b + c}{2} = \frac{2 + \frac{7}{2} + \frac{5}{2}}{2} = \frac{2 + 6}{2} = 4 \]

### Step 6: Calculate \(s-a\), \(s-b\), and \(s-c\)
Now we calculate:
\[ s - a = 4 - 2 = 2 \]
\[ s - b = 4 - \frac{7}{2} = \frac{1}{2} \]
\[ s - c = 4 - \frac{5}{2} = \frac{3}{2} \]

### Step 7: Substitute into the formula for \(\tan^2 \left(\frac{P}{2}\right)\)
Now substituting these values into the formula:
\[ \tan^2 \left(\frac{P}{2}\right) = \frac{(s-b)(s-c)}{s(s-a)} = \frac{\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)}{4 \cdot 2} = \frac{\frac{3}{4}}{8} = \frac{3}{32} \]

### Final Answer
Thus, the final value of \(\frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P}\) is:
\[ \frac{3}{32} \]