In a `triangleABC`, if a,c,b are in A.P. then the value of `(a cos B-b cos A)/(a-b)`, is

Options

A. 3
B. 2
C. 1
D. none of these

Answer

To solve the problem, we need to find the value of \((a \cos B - b \cos A) / (a - b)\) given that \(a\), \(b\), and \(c\) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
Since \(a\), \(b\), and \(c\) are in A.P., we can express this condition mathematically:
\[ 2c = a + b \]

2. **Using the Cosine Rule**:
We will use the cosine rule to express \(\cos A\) and \(\cos B\):
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

3. **Substituting \(\cos A\) and \(\cos B\)**:
Substitute these expressions into the original equation:
\[ a \cos B - b \cos A = a \left(\frac{a^2 + c^2 - b^2}{2ac}\right) - b \left(\frac{b^2 + c^2 - a^2}{2bc}\right) \]

4. **Simplifying the Expression**:
This simplifies to:
\[ = \frac{a(a^2 + c^2 - b^2)}{2ac} - \frac{b(b^2 + c^2 - a^2)}{2bc} \]
\[ = \frac{(a^2 + c^2 - b^2)}{2c} - \frac{(b^2 + c^2 - a^2)}{2c} \]

5. **Combining the Terms**:
Combine the two fractions:
\[ = \frac{(a^2 + c^2 - b^2) - (b^2 + c^2 - a^2)}{2c} \]
\[ = \frac{a^2 - b^2 + a^2 - b^2}{2c} \]
\[ = \frac{2(a^2 - b^2)}{2c} = \frac{a^2 - b^2}{c} \]

6. **Factoring the Difference of Squares**:
Recognizing that \(a^2 - b^2\) can be factored:
\[ = \frac{(a - b)(a + b)}{c} \]

7. **Final Expression**:
Now substituting this back into our original expression:
\[ \frac{a \cos B - b \cos A}{a - b} = \frac{(a - b)(a + b)}{c(a - b)} \]
Canceling \(a - b\) (assuming \(a \neq b\)):
\[ = \frac{a + b}{c} \]

8. **Using the A.P. Condition**:
From the A.P. condition, we know that \(a + b = 2c\):
\[ = \frac{2c}{c} = 2 \]

### Conclusion:
Thus, the value of \(\frac{a \cos B - b \cos A}{a - b}\) is \(2\).