Let `P Q R` be a triangle of area 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 (2sinP-sin2P)/(2sinP+sin2P) equals

Explanation

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

### Step 1: Use the double angle identity for sine
We know that:
\[ \sin 2P = 2 \sin P \cos P \]
So we can substitute this into our expression:
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