If `A +B = ( pi )/( 3)` and `cos A + cos B =1`, then which of the following are true `:`

Options

A. `cos ( A-B) = ( 1)/( 3)`
B. `cos ( A-B) = - ( 1)/( 3)`
C. `| cos A - cos B | = sqrt( 2//3 )`
D. `| cos A - cos B | = ( 1)/( sqrt( 3))`

Answer

To solve the problem, we need to find the values of \( \cos A \) and \( \cos B \) given that \( A + B = \frac{\pi}{3} \) and \( \cos A + \cos B = 1 \).

### Step 1: Use the cosine addition formula
We know that:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
Substituting \( A + B = \frac{\pi}{3} \):
\[ \cos A + \cos B = 2 \cos\left(\frac{\frac{\pi}{3}}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
This simplifies to:
\[ \cos A + \cos B = 2 \cos\left(\frac{\pi}{6}\right) \cos\left(\frac{A - B}{2}\right) \]

### Step 2: Calculate \( \cos\left(\frac{\pi}{6}\right) \)
We know:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
So we have:
\[ \cos A + \cos B = 2 \cdot \frac{\sqrt{3}}{2} \cos\left(\frac{A - B}{2}\right) = \sqrt{3} \cos\left(\frac{A - B}{2}\right) \]

### Step 3: Set up the equation
Given that \( \cos A + \cos B = 1 \), we can equate:
\[ \sqrt{3} \cos\left(\frac{A - B}{2}\right) = 1 \]
This leads to:
\[ \cos\left(\frac{A - B}{2}\right) = \frac{1}{\sqrt{3}} \]

### Step 4: Solve for \( \frac{A - B}{2} \)
Taking the inverse cosine, we find:
\[ \frac{A - B}{2} = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
This gives us:
\[ A - B = 2 \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Step 5: Find \( \cos(A - B) \)
Using the cosine double angle identity:
\[ \cos(A - B) = 2\cos^2\left(\frac{A - B}{2}\right) - 1 \]
Substituting \( \cos\left(\frac{A - B}{2}\right) = \frac{1}{\sqrt{3}} \):
\[ \cos(A - B) = 2\left(\frac{1}{\sqrt{3}}\right)^2 - 1 = 2 \cdot \frac{1}{3} - 1 = \frac{2}{3} - 1 = -\frac{1}{3} \]

### Step 6: Conclusion
Thus, we find:
1. \( \cos(A - B) = -\frac{1}{3} \)
2. The modulus \( |\cos(A - B)| = \frac{1}{3} \)

The final results are:
- \( \cos(A - B) = -\frac{1}{3} \)
- \( |\cos(A - B)| = \sqrt{\frac{2}{3}} \)