If `x , y , z` are in A.P., then `(sinx-sinz)/(cosz-cosx)` is equal to (a)`tany` (b) `coty` (c) `siny` (d) `cosy`

Options

A. (a) `tan y`
B. (b) `cot y`
C. (c) `sin y`
D. (d) `cos y`

Explanation

To solve the problem, we need to evaluate the expression \((\sin x - \sin z) / (\cos z - \cos x)\) given that \(x\), \(y\), and \(z\) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding A.P.**: Since \(x\), \(y\), and \(z\) are in A.P., we have:
\[ y - x = z - y \]
This can be rearranged to:
\[ 2y = x + z \quad \text{or} \quad y = \frac{x + z}{2} \]

2. **Using Trigonometric Identities**: We will use the sine and cosine difference formulas:
- The formula for \(\sin A - \sin B\) is:
\[ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
- The formula for \(\cos A - \cos B\) is:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]

3. **Applying the Formulas**: Let \(A = x\) and \(B = z\):
- For \(\sin x - \sin z\):
\[ \sin x - \sin z = 2 \cos\left(\frac{x + z}{2}\right) \sin\left(\frac{x - z}{2}\right) \]
- For \(\cos z - \cos x\):
\[ \cos z - \cos x = -2 \sin\left(\frac{x + z}{2}\right) \sin\left(\frac{x - z}{2}\right) \]

4. **Substituting into the Expression**:
Now substituting these into our original expression:
\[ \frac{\sin x - \sin z}{\cos z - \cos x} = \frac{2 \cos\left(\frac{x + z}{2}\right) \sin\left(\frac{x - z}{2}\right)}{-2 \sin\left(\frac{x + z}{2}\right) \sin\left(\frac{x - z}{2}\right)} \]
The \(2\) and \(\sin\left(\frac{x - z}{2}\right)\) cancel out:
\[ = -\frac{\cos\left(\frac{x + z}{2}\right)}{\sin\left(\frac{x + z}{2}\right)} \]

5. **Using the Value of \(y\)**: Since we have established that:
\[ \frac{x + z}{2} = y \]
Therefore:
\[ -\frac{\cos\left(\frac{x + z}{2}\right)}{\sin\left(\frac{x + z}{2}\right)} = -\cot(y) \]

6. **Final Result**: Thus, we find:
\[ \frac{\sin x - \sin z}{\cos z - \cos x} = \cot y \]
Hence, the answer is:
\[ \text{(b) } \cot y \]