If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression `(a)/(c) sin 2C + (c)/(a) sin 2A` is

Options

A. `(1)/(2)`
B. `(sqrt3)/(2)`
C. 1
D. `sqrt3`

Explanation

To solve the problem, we need to find the value of the expression \(\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A\) given that the angles \(A\), \(B\), and \(C\) of a triangle are in arithmetic progression.

### Step 1: Understand the relationship between angles
Since \(A\), \(B\), and \(C\) are in arithmetic progression, we can express them as:
\[ A = B - d, \quad B = B, \quad C = B + d \]
where \(d\) is the common difference.

### Step 2: Use the sum of angles in a triangle
The sum of angles in a triangle is \(180^\circ\):
\[ A + B + C = 180^\circ \]
Substituting the expressions for \(A\) and \(C\):
\[ (B - d) + B + (B + d) = 180^\circ \]
This simplifies to:
\[ 3B = 180^\circ \implies B = 60^\circ \]
Thus, we have:
\[ A = 60^\circ - d, \quad B = 60^\circ, \quad C = 60^\circ + d \]

### Step 3: Find the values of angles \(A\) and \(C\)
Since \(A + C = 120^\circ\) (because \(B = 60^\circ\)), we can express \(A\) and \(C\) as:
\[ A = 60^\circ - d, \quad C = 60^\circ + d \]

### Step 4: Use the sine double angle formula
We need to evaluate:
\[ \frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A \]
Using the sine double angle formula:
\[ \sin 2C = 2 \sin C \cos C, \quad \sin 2A = 2 \sin A \cos A \]
Substituting these into our expression gives:
\[ \frac{a}{c} (2 \sin C \cos C) + \frac{c}{a} (2 \sin A \cos A) \]
This simplifies to:
\[ 2 \left( \frac{a}{c} \sin C \cos C + \frac{c}{a} \sin A \cos A \right) \]

### Step 5: Express sides in terms of angles
Using the Law of Sines:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
Thus:
\[ a = 2R \sin A, \quad c = 2R \sin C \]
Substituting these into our expression:
\[ \frac{2R \sin A}{2R \sin C} \sin C \cos C + \frac{2R \sin C}{2R \sin A} \sin A \cos A \]
This simplifies to:
\[ \frac{\sin A}{\sin C} \sin C \cos C + \frac{\sin C}{\sin A} \sin A \cos A \]
Cancelling out terms gives:
\[ \sin A \cos C + \sin C \cos A \]

### Step 6: Use the angle sum identity
Using the identity \(\sin A \cos C + \sin C \cos A = \sin(A + C)\):
\[ \sin(A + C) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \]

### Step 7: Final expression
Thus, we have:
\[ 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

### Conclusion
The value of the expression \(\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A\) is \(\sqrt{3}\).