If `(sinA-sinC)/(cosC-cosA)=cotB`, then A, B, C are in

Options

A. A.P.
B. GP.
C. H.P.
D. none of these

Answer

To solve the problem, we need to show that if \((\sin A - \sin C) / (\cos C - \cos A) = \cot B\), then angles \(A\), \(B\), and \(C\) are in Arithmetic Progression (AP).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \frac{\sin A - \sin C}{\cos C - \cos A} = \cot B \]

2. **Use the sine and cosine difference identities:**
We can use the identities:
\[ \sin A - \sin C = 2 \cos\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right) \]
\[ \cos C - \cos A = -2 \sin\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right) \]

3. **Substituting these identities into the equation:**
\[ \frac{2 \cos\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right)}{-2 \sin\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right)} = \cot B \]

4. **Simplifying the equation:**
The \(2\) and \(\sin\left(\frac{A - C}{2}\right)\) cancel out (assuming \(\sin\left(\frac{A - C}{2}\right) \neq 0\)):
\[ -\frac{\cos\left(\frac{A + C}{2}\right)}{\sin\left(\frac{A + C}{2}\right)} = \cot B \]
This implies:
\[ \cot\left(\frac{A + C}{2}\right) = -\cot B \]

5. **Using the cotangent property:**
If \(\cot\left(\frac{A + C}{2}\right) = -\cot B\), then:
\[ \frac{A + C}{2} + B = 180^\circ \quad \text{(since cotangent is negative)} \]

6. **Rearranging the equation:**
\[ A + C + 2B = 360^\circ \]
This can be rearranged to:
\[ A + B + C = 180^\circ \quad \text{(which indicates that A, B, and C are in AP)} \]

### Conclusion:
Thus, we have shown that if \((\sin A - \sin C) / (\cos C - \cos A) = \cot B\), then angles \(A\), \(B\), and \(C\) are in Arithmetic Progression (AP).