If `A+C=2B, then (cosC-cosA)/(sinA-sinC)=`

Options

A. `cot B`
B. `cot 2B`
C. `tan2B`
D. `tanB`

Answer

To solve the problem, we need to evaluate the expression \(\frac{\cos C - \cos A}{\sin A - \sin C}\) given that \(A + C = 2B\).

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We start with the expression:
\[ \frac{\cos C - \cos A}{\sin A - \sin C} \]
We can factor out a negative sign from the numerator:
\[ = \frac{- (\cos A - \cos C)}{\sin A - \sin C} \]

2. **Use the Cosine Difference Formula**:
The formula for the difference of cosines is:
\[ \cos A - \cos C = -2 \sin\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right) \]
So, substituting this into our expression gives:
\[ = \frac{2 \sin\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right)}{\sin A - \sin C} \]

3. **Use the Sine Difference Formula**:
The formula for the difference of sines is:
\[ \sin A - \sin C = 2 \cos\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right) \]
Substituting this into our expression gives:
\[ = \frac{2 \sin\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right)}{2 \cos\left(\frac{A + C}{2}\right) \sin\left(\frac{A - C}{2}\right)} \]

4. **Cancel Common Terms**:
The \(\sin\left(\frac{A - C}{2}\right)\) terms cancel out (assuming \(\sin\left(\frac{A - C}{2}\right) \neq 0\)):
\[ = \frac{\sin\left(\frac{A + C}{2}\right)}{\cos\left(\frac{A + C}{2}\right)} \]

5. **Convert to Tangent**:
The expression simplifies to:
\[ = \tan\left(\frac{A + C}{2}\right) \]

6. **Substitute the Given Relation**:
We know from the problem statement that \(A + C = 2B\). Therefore:
\[ \tan\left(\frac{A + C}{2}\right) = \tan(B) \]

### Final Result:
Thus, the value of the expression \(\frac{\cos C - \cos A}{\sin A - \sin C}\) is:
\[ \boxed{\tan B} \]