If `p sintheta+qcostheta=3&qsintheta-pcostheta=2`, then `p^(2)+q^(2)=?`

Answer

To solve the equations \( p \sin \theta + q \cos \theta = 3 \) and \( q \sin \theta - p \cos \theta = 2 \), we will follow these steps:

### Step 1: Square both equations
We start by squaring both equations.

1. **First equation**:
\[ (p \sin \theta + q \cos \theta)^2 = 3^2 \]
This expands to:
\[ p^2 \sin^2 \theta + 2pq \sin \theta \cos \theta + q^2 \cos^2 \theta = 9 \]

2. **Second equation**:
\[ (q \sin \theta - p \cos \theta)^2 = 2^2 \]
This expands to:
\[ q^2 \sin^2 \theta - 2pq \sin \theta \cos \theta + p^2 \cos^2 \theta = 4 \]

### Step 2: Combine the equations
Now we have two equations:
1. \( p^2 \sin^2 \theta + 2pq \sin \theta \cos \theta + q^2 \cos^2 \theta = 9 \) (Equation 1)
2. \( q^2 \sin^2 \theta - 2pq \sin \theta \cos \theta + p^2 \cos^2 \theta = 4 \) (Equation 2)

### Step 3: Add the two equations
Adding Equation 1 and Equation 2:
\[ (p^2 \sin^2 \theta + q^2 \sin^2 \theta) + (p^2 \cos^2 \theta + q^2 \cos^2 \theta) + (2pq \sin \theta \cos \theta - 2pq \sin \theta \cos \theta) = 9 + 4 \]
This simplifies to:
\[ (p^2 + q^2)(\sin^2 \theta + \cos^2 \theta) = 13 \]

### Step 4: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ p^2 + q^2 = 13 \]

### Final Result
Thus, the value of \( p^2 + q^2 \) is:
\[ \boxed{13} \]