If angles A,B and C are in AP, then `(a+c)/(b)` is equal to

Options

A. `2 sin ((A-C)/(2))`
B. `2 cos ((A-C)/(2))`
C. `cos ((A-C)/(2))`
D. `sin ((A-C)/(2))`

Answer

To solve the problem, we need to show that if angles A, B, and C are in Arithmetic Progression (AP), then \((A + C)/B\) is equal to \(2 \cos\left(\frac{A - C}{2}\right)\).

### Step-by-Step Solution:

1. **Understanding the Condition of AP**:
Since angles A, B, and C are in AP, we can express this condition mathematically:
\[ 2B = A + C \]
This implies that:
\[ A + C = 2B \]

**Hint**: Remember that in an arithmetic progression, the middle term is the average of the other two terms.

2. **Using the Sine Rule**:
According to the sine rule in a triangle:
\[ \frac{A}{\sin A} = \frac{B}{\sin B} = \frac{C}{\sin C} = k \text{ (a constant)} \]
From this, we can express A, B, and C in terms of k:
\[ A = k \sin A, \quad B = k \sin B, \quad C = k \sin C \]

**Hint**: The sine rule relates the angles of a triangle to the lengths of its sides.

3. **Expressing \(A + C\)**:
We can substitute A and C into the expression \(A + C\):
\[ A + C = k \sin A + k \sin C = k(\sin A + \sin C) \]

**Hint**: Factor out the common term to simplify the expression.

4. **Substituting into \((A + C)/B\)**:
Now we substitute \(A + C\) and B into the expression:
\[ \frac{A + C}{B} = \frac{k(\sin A + \sin C)}{k \sin B} \]
The k terms cancel out:
\[ \frac{A + C}{B} = \frac{\sin A + \sin C}{\sin B} \]

**Hint**: Cancelling out common factors can simplify your calculations significantly.

5. **Using the Sine Addition Formula**:
We can use the sine addition formula:
\[ \sin A + \sin C = 2 \sin\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right) \]
Substituting this into our expression gives:
\[ \frac{A + C}{B} = \frac{2 \sin\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right)}{\sin B} \]

**Hint**: Familiarize yourself with trigonometric identities, as they can help simplify expressions.

6. **Substituting \(A + C = 2B\)**:
Since we know \(A + C = 2B\), we can substitute:
\[ \frac{A + C}{B} = \frac{2 \sin(B) \cos\left(\frac{A - C}{2}\right)}{\sin B} \]
This simplifies to:
\[ \frac{A + C}{B} = 2 \cos\left(\frac{A - C}{2}\right) \]

**Hint**: Always look for opportunities to substitute known values or relationships to simplify your expressions.

### Final Result:
Thus, we conclude that:
\[ \frac{A + C}{B} = 2 \cos\left(\frac{A - C}{2}\right) \]