Let a = 6, b = 3 and `cos (A -B) = (4)/(5)`
Area (in sq. units) of the triangle is equal to

Options

A. 9
B. 12
C. 11
D. 10

Explanation

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

### Step 1: Use the Cosine of Angle Difference Formula
We know that:
\[ \cos(A - B) = \frac{4}{5} \]
This can be used to find \( \tan\left(\frac{A - B}{2}\) using the identity:
\[ \cos(A - B) = \frac{1 - \tan^2\left(\frac{A - B}{2}\right)}{1 + \tan^2\left(\frac{A - B}{2}\right)} \]

### Step 2: Set Up the Equation
Let \( x = \tan\left(\frac{A - B}{2}\right) \). Then we have:
\[ \frac{1 - x^2}{1 + x^2} = \frac{4}{5} \]

### Step 3: Cross-Multiply and Solve for \( x^2 \)
Cross-multiplying gives:
\[ 5(1 - x^2) = 4(1 + x^2) \]
Expanding both sides:
\[ 5 - 5x^2 = 4 + 4x^2 \]
Rearranging gives:
\[ 5 - 4 = 5x^2 + 4x^2 \]
\[ 1 = 9x^2 \]
Thus,
\[ x^2 = \frac{1}{9} \quad \Rightarrow \quad x = \frac{1}{3} \]

### Step 4: Use the Tangent Half-Angle Formula
We know that:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cdot \cos\left(\frac{C}{2}\right) \]
Substituting the values of \( a \) and \( b \):
\[ \frac{1}{3} = \frac{6 - 3}{6 + 3} \cdot \cos\left(\frac{C}{2}\right) \]
This simplifies to:
\[ \frac{1}{3} = \frac{3}{9} \cdot \cos\left(\frac{C}{2}\right) \quad \Rightarrow \quad \frac{1}{3} = \frac{1}{3} \cdot \cos\left(\frac{C}{2}\right) \]
Thus,
\[ \cos\left(\frac{C}{2}\right) = 1 \]

### Step 5: Find Angle \( C \)
Since \( \cos\left(\frac{C}{2}\right) = 1 \), we have:
\[ \frac{C}{2} = 0 \quad \Rightarrow \quad C = 0 \]
However, \( C \) cannot be zero in a triangle. Therefore, we conclude that \( C = 90^\circ \).

### Step 6: Calculate the Area of the Triangle
The area \( A \) of a triangle is given by:
\[ A = \frac{1}{2}ab \sin C \]
Substituting the known values:
\[ A = \frac{1}{2} \cdot 6 \cdot 3 \cdot \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \):
\[ A = \frac{1}{2} \cdot 6 \cdot 3 \cdot 1 = \frac{18}{2} = 9 \]

### Final Answer
The area of the triangle is \( \boxed{9} \) square units.

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