In `Delta ABC, (sin A (a - b cos C))/(sin C (c -b cos A))=`

Options

A. `-2`
B. `-1`
C. 0
D. 1

Explanation

To solve the problem, we need to simplify the expression:

\[ \frac{\sin A (a - b \cos C)}{\sin C (c - b \cos A)} \]

### Step-by-Step Solution:

1. **Start with the given expression:**
\[ \frac{\sin A (a - b \cos C)}{\sin C (c - b \cos A)} \]

2. **Use the Law of Cosines to express \(a\) and \(c\):**
- According to the Law of Cosines:
\[ a^2 = b^2 + c^2 - 2bc \cos A \]
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
- From these, we can express \(a\) and \(c\) in terms of \(b\) and the angles.

3. **Substitute \(a\) and \(c\) into the expression:**
- We can express \(a\) and \(c\) as:
\[ a = b \cos C + c \cos B \]
\[ c = a \cos B + b \cos A \]

4. **Substituting back into the expression:**
- Substitute these values into the original expression:
\[ \frac{\sin A \left(b \cos C + c \cos B - b \cos C\right)}{\sin C \left(a \cos B + b \cos A - b \cos A\right)} \]

5. **Simplify the expression:**
- The \(b \cos C\) terms cancel out in the numerator, and the \(b \cos A\) terms cancel out in the denominator:
\[ \frac{\sin A \cdot c \cos B}{\sin C \cdot a \cos B} \]

6. **Cancel \(\cos B\):**
- Assuming \(\cos B \neq 0\):
\[ \frac{\sin A \cdot c}{\sin C \cdot a} \]

7. **Using the sine rule:**
- By the sine rule, we know that:
\[ \frac{a}{\sin A} = \frac{c}{\sin C} \]
- Thus, we can express \(c\) in terms of \(a\) and the sine values:
\[ c = \frac{a \sin C}{\sin A} \]

8. **Final simplification:**
- Substitute \(c\) back into the expression:
\[ \frac{\sin A \cdot \frac{a \sin C}{\sin A}}{\sin C \cdot a} = 1 \]

### Conclusion:
The expression simplifies to:
\[ \frac{\sin A (a - b \cos C)}{\sin C (c - b \cos A)} = 1 \]