Let ABCDEFGHIJKL be a regular dodecagon. Then the value of `(AB)/(AF) + (AF)/(AB)` is equal to ____

Explanation

To solve the problem of finding the value of \(\frac{AB}{AF} + \frac{AF}{AB}\) for a regular dodecagon ABCDEFGHIJKL, we can follow these steps:

### Step 1: Understand the Geometry of the Dodecagon
A regular dodecagon has 12 equal sides and is inscribed in a circle. The center of the dodecagon can be denoted as point O. The vertices are labeled as A, B, C, D, E, F, G, H, I, J, K, L in a clockwise manner.

### Step 2: Determine the Lengths of AB and AF
1. **Finding AB**:
- The angle subtended by each side at the center (O) is \( \frac{360^\circ}{12} = 30^\circ \).
- The length \( AB \) can be calculated using the formula for the chord length in a circle:
\[ AB = 2R \sin\left(\frac{30^\circ}{2}\right) = 2R \sin(15^\circ) \]

2. **Finding AF**:
- The angle subtended by \( AF \) at the center is \( 150^\circ \) (since A to F skips 5 vertices).
- The length \( AF \) can be calculated similarly:
\[ AF = 2R \sin\left(\frac{150^\circ}{2}\right) = 2R \sin(75^\circ) \]

### Step 3: Substitute the Lengths into the Expression
Now we can substitute \( AB \) and \( AF \) into the expression:
\[ \frac{AB}{AF} + \frac{AF}{AB} = \frac{2R \sin(15^\circ)}{2R \sin(75^\circ)} + \frac{2R \sin(75^\circ)}{2R \sin(15^\circ)} \]
This simplifies to:
\[ \frac{\sin(15^\circ)}{\sin(75^\circ)} + \frac{\sin(75^\circ)}{\sin(15^\circ)} \]

### Step 4: Use the Identity for Sine
Recall that \( \sin(75^\circ) = \cos(15^\circ) \). Therefore, we can rewrite the expression as:
\[ \frac{\sin(15^\circ)}{\cos(15^\circ)} + \frac{\cos(15^\circ)}{\sin(15^\circ)} = \tan(15^\circ) + \cot(15^\circ) \]

### Step 5: Simplify the Expression
Using the identity \( \tan(x) + \cot(x) = \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)} = \frac{1}{\sin(x)\cos(x)} \):
\[ \tan(15^\circ) + \cot(15^\circ) = \frac{1}{\sin(15^\circ)\cos(15^\circ)} = \frac{2}{\sin(30^\circ)} = 4 \]

### Final Answer
Thus, the value of \( \frac{AB}{AF} + \frac{AF}{AB} \) is equal to **4**.

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