Let a,b, c are the sides opposite to angles A, B , C respectively in a `Delta ABC tan""(A-B)/(2)=(a-b)/(a+b)cot ""C/2 and (a)/(sin A)=(b)/(sin B) =(c)/(sin C),`
If `a=6,b=3 and cos (A-B) =4/5`
Area of the trianlge is equal to

Options

A. 8
B. 9
C. 10
D. 11

Answer

To solve the problem, we need to find the area of triangle ABC given the sides \( a = 6 \), \( b = 3 \), and \( \cos(A - B) = \frac{4}{5} \). We also use the relation \( \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cot\left(\frac{C}{2}\right) \) and the law of sines.

### Step-by-Step Solution:

1. **Calculate \( \tan\left(\frac{A - B}{2}\right) \)**:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{\sin(A - B)}{\cos(A - B)} \]
We know \( \cos(A - B) = \frac{4}{5} \). To find \( \sin(A - B) \), we use the identity:
\[ \sin^2(A - B) + \cos^2(A - B) = 1 \]
Thus,
\[ \sin^2(A - B) = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
Therefore,
\[ \sin(A - B) = \frac{3}{5} \]
Now, we can calculate \( \tan\left(\frac{A - B}{2}\right) \):
\[ \tan\left(\frac{A - B}{2}\right) = \frac{\sin(A - B)}{\cos(A - B)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

2. **Use the relation \( \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cot\left(\frac{C}{2}\right) \)**:
\[ \frac{3}{4} = \frac{6 - 3}{6 + 3} \cot\left(\frac{C}{2}\right) \]
Simplifying the right side:
\[ \frac{3}{4} = \frac{3}{9} \cot\left(\frac{C}{2}\right) \]
This simplifies to:
\[ \frac{3}{4} = \frac{1}{3} \cot\left(\frac{C}{2}\right) \]
Cross-multiplying gives:
\[ 3 \cdot 3 = 4 \cot\left(\frac{C}{2}\right) \implies 9 = 4 \cot\left(\frac{C}{2}\right) \]
Therefore,
\[ \cot\left(\frac{C}{2}\right) = \frac{9}{4} \]

3. **Find \( \tan\left(\frac{C}{2}\right) \)**:
Since \( \cot\left(\frac{C}{2}\right) = \frac{1}{\tan\left(\frac{C}{2}\right)} \):
\[ \tan\left(\frac{C}{2}\right) = \frac{4}{9} \]

4. **Determine angle \( C \)**:
From \( \tan\left(\frac{C}{2}\right) = \frac{4}{9} \), we can find \( C \) using the double angle formula:
\[ \tan(C) = \frac{2 \tan\left(\frac{C}{2}\right)}{1 - \tan^2\left(\frac{C}{2}\right)} = \frac{2 \cdot \frac{4}{9}}{1 - \left(\frac{4}{9}\right)^2} \]
Calculating the denominator:
\[ 1 - \frac{16}{81} = \frac{65}{81} \]
Therefore,
\[ \tan(C) = \frac{\frac{8}{9}}{\frac{65}{81}} = \frac{8 \cdot 81}{9 \cdot 65} = \frac{72}{65} \]

5. **Determine the area of triangle ABC**:
Since we have a right triangle (as \( C = 90^\circ \)), the area can be calculated using:
\[ \text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times 6 \times 3 = 9 \]

### Final Answer:
The area of triangle ABC is \( 9 \) square units.