In `DeltaABC`,
If in a `DeltaABC`, a = 6, b = 3 and `cos(A-B)=4/5`, find its area.

Answer

To find the area of triangle ABC given the sides \( a = 6 \), \( b = 3 \), and \( \cos(A - B) = \frac{4}{5} \), we can follow these steps:

### Step 1: Use the formula for the area of a triangle
The area \( K \) of triangle ABC can be expressed using the formula:
\[ K = \frac{1}{2}ab \sin C \]
where \( a \) and \( b \) are the lengths of two sides, and \( C \) is the included angle.

### Step 2: Find \( \sin C \) using the cosine of the angle difference
We know that:
\[ \cos(A - B) = \frac{4}{5} \]
Using the identity:
\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]
we can express \( \sin C \) in terms of \( A \) and \( B \).

### Step 3: Use the tangent half-angle formula
We can use the relation:
\[ \tan\left(\frac{A - B}{2}\right) = \sqrt{\frac{1 - \cos(A - B)}{1 + \cos(A - B)}} \]
Substituting \( \cos(A - B) = \frac{4}{5} \):
\[ \tan\left(\frac{A - B}{2}\right) = \sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}} = \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \]

### Step 4: Relate \( \tan\left(\frac{A - B}{2}\right) \) to \( \cot\left(\frac{C}{2}\right) \)
Using Napier's analogy:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cot\left(\frac{C}{2}\right) \]
Substituting the known values:
\[ \frac{1}{3} = \frac{6 - 3}{6 + 3} \cot\left(\frac{C}{2}\right) = \frac{3}{9} \cot\left(\frac{C}{2}\right) = \frac{1}{3} \cot\left(\frac{C}{2}\right) \]
This implies:
\[ \cot\left(\frac{C}{2}\right) = 1 \]

### Step 5: Find angle \( C \)
Since \( \cot\left(\frac{C}{2}\right) = 1 \), we have:
\[ \frac{C}{2} = \frac{\pi}{4} \implies C = \frac{\pi}{2} \]

### Step 6: Calculate the area of triangle ABC
Now substituting \( C = \frac{\pi}{2} \) into the area formula:
\[ K = \frac{1}{2} \times 6 \times 3 \times \sin\left(\frac{\pi}{2}\right) \]
Since \( \sin\left(\frac{\pi}{2}\right) = 1 \):
\[ K = \frac{1}{2} \times 6 \times 3 \times 1 = \frac{18}{2} = 9 \]

### Final Answer
The area of triangle ABC is:
\[ \boxed{9} \text{ square units} \]