In a triangle ABC, `(c cos(A-theta) + acos (C + theta))/(b costheta)` is equal to

Answer

To solve the problem, we need to simplify the expression \((c \cos(A - \theta) + a \cos(C + \theta)) / (b \cos \theta)\).

### Step-by-Step Solution:

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

2. **Use the cosine addition and subtraction formulas:**
- Recall that:
\[ \cos(A - \theta) = \cos A \cos \theta + \sin A \sin \theta \]
\[ \cos(C + \theta) = \cos C \cos \theta - \sin C \sin \theta \]

3. **Substituting these formulas into the expression:**
\[ c \cos(A - \theta) = c (\cos A \cos \theta + \sin A \sin \theta) = c \cos A \cos \theta + c \sin A \sin \theta \]
\[ a \cos(C + \theta) = a (\cos C \cos \theta - \sin C \sin \theta) = a \cos C \cos \theta - a \sin C \sin \theta \]

4. **Combine these results:**
\[ c \cos(A - \theta) + a \cos(C + \theta) = (c \cos A + a \cos C) \cos \theta + (c \sin A - a \sin C) \sin \theta \]

5. **Now substitute back into the original expression:**
\[ \frac{(c \cos A + a \cos C) \cos \theta + (c \sin A - a \sin C) \sin \theta}{b \cos \theta} \]

6. **Separate the terms:**
\[ = \frac{(c \cos A + a \cos C)}{b} + \frac{(c \sin A - a \sin C) \sin \theta}{b \cos \theta} \]

7. **Now apply the sine rule:**
- According to the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
- This gives us:
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

8. **Substituting these values into the expression:**
\[ = \frac{(k \sin C \cos A + k \sin A \cos C)}{k \sin B} + \frac{(k \sin C \sin A - k \sin A \sin C) \sin \theta}{k \sin B \cos \theta} \]

9. **Simplifying further:**
- The first term simplifies to:
\[ \frac{\sin C \cos A + \sin A \cos C}{\sin B} = \frac{\sin(A + C)}{\sin B} \]
- Since \(A + C = \pi - B\), we have:
\[ \sin(A + C) = \sin B \]

10. **Final result:**
\[ = \frac{\sin B}{\sin B} = 1 \]

### Conclusion:
Thus, the expression simplifies to:
\[ \frac{c \cos(A - \theta) + a \cos(C + \theta)}{b \cos \theta} = 1 \]