Find the value of `theta, " if " m^(2) sin . Pi/2 - n^(2) sin . (3pi)/2 + 2 mn sec theta = (m-n)^(2) , 0 le theta le pi ` .

Answer

To solve the equation \( m^2 \sin \frac{\pi}{2} - n^2 \sin \frac{3\pi}{2} + 2mn \sec \theta = (m-n)^2 \), we will follow these steps:

### Step 1: Substitute the values of sine
We know that:
\[ \sin \frac{\pi}{2} = 1 \quad \text{and} \quad \sin \frac{3\pi}{2} = -1 \]
Substituting these values into the equation gives:
\[ m^2(1) - n^2(-1) + 2mn \sec \theta = (m-n)^2 \]
This simplifies to:
\[ m^2 + n^2 + 2mn \sec \theta = (m-n)^2 \]

### Step 2: Expand the right-hand side
Now, we expand \( (m-n)^2 \):
\[ (m-n)^2 = m^2 - 2mn + n^2 \]
So, we can rewrite the equation as:
\[ m^2 + n^2 + 2mn \sec \theta = m^2 - 2mn + n^2 \]

### Step 3: Simplify the equation
Now, we can cancel \( m^2 \) and \( n^2 \) from both sides:
\[ 2mn \sec \theta = -2mn \]

### Step 4: Divide by \( 2mn \)
Assuming \( mn \neq 0 \), we can divide both sides by \( 2mn \):
\[ \sec \theta = -1 \]

### Step 5: Find the value of \( \theta \)
The secant function is defined as:
\[ \sec \theta = \frac{1}{\cos \theta} \]
Thus, \( \sec \theta = -1 \) implies:
\[ \cos \theta = -1 \]
The value of \( \theta \) for which \( \cos \theta = -1 \) in the interval \( [0, \pi] \) is:
\[ \theta = \pi \]

### Final Answer
The value of \( \theta \) is:
\[ \theta = \pi \]