If `A_(1)` and `A_(2)` be areas of two regular polygons having the same perimeter and number of sides be n and 2n respectively, then : `(A_(1))/(A_(2))` is

Options

A. `(2 sin((pi)/(n)))/(1+cos((pi)/(n)))`
B. `(2 cos((pi)/(n)))/(1+sin((pi)/(n)))`
C. `(2 cos((pi)/(n)))/(1+cos((pi)/(2n)))`
D. `(2 sin((pi)/(n)))/(1+sin((pi)/(n)))`

Answer

To solve the problem, we need to find the ratio of the areas \( \frac{A_1}{A_2} \) of two regular polygons with the same perimeter, where one polygon has \( n \) sides and the other has \( 2n \) sides.

### Step-by-Step Solution:

1. **Formula for Area of a Regular Polygon**:
The area \( A \) of a regular polygon with \( n \) sides and perimeter \( p \) can be expressed as:
\[ A = \frac{1}{4} n s^2 \cot\left(\frac{\pi}{n}\right) \]
where \( s \) is the length of each side. Since the perimeter \( p \) is equal to \( n \times s \), we have \( s = \frac{p}{n} \).

2. **Area of the First Polygon \( A_1 \)**:
For the polygon with \( n \) sides:
\[ A_1 = \frac{1}{4} n \left(\frac{p}{n}\right)^2 \cot\left(\frac{\pi}{n}\right) \]
Simplifying this, we get:
\[ A_1 = \frac{1}{4} n \cdot \frac{p^2}{n^2} \cot\left(\frac{\pi}{n}\right) = \frac{p^2}{4n} \cot\left(\frac{\pi}{n}\right) \]

3. **Area of the Second Polygon \( A_2 \)**:
For the polygon with \( 2n \) sides:
\[ A_2 = \frac{1}{4} (2n) \left(\frac{p}{2n}\right)^2 \cot\left(\frac{\pi}{2n}\right) \]
Simplifying this, we get:
\[ A_2 = \frac{1}{4} (2n) \cdot \frac{p^2}{(2n)^2} \cot\left(\frac{\pi}{2n}\right) = \frac{1}{4} (2n) \cdot \frac{p^2}{4n^2} \cot\left(\frac{\pi}{2n}\right) = \frac{p^2}{8n} \cot\left(\frac{\pi}{2n}\right) \]

4. **Finding the Ratio \( \frac{A_1}{A_2} \)**:
Now we can find the ratio of the areas:
\[ \frac{A_1}{A_2} = \frac{\frac{p^2}{4n} \cot\left(\frac{\pi}{n}\right)}{\frac{p^2}{8n} \cot\left(\frac{\pi}{2n}\right)} \]
Simplifying this gives:
\[ \frac{A_1}{A_2} = \frac{8n}{4n} \cdot \frac{\cot\left(\frac{\pi}{n}\right)}{\cot\left(\frac{\pi}{2n}\right)} = 2 \cdot \frac{\cot\left(\frac{\pi}{n}\right)}{\cot\left(\frac{\pi}{2n}\right)} \]

5. **Using the Cotangent Identity**:
Recall that \( \cot(x) = \frac{\cos(x)}{\sin(x)} \). Thus:
\[ \frac{A_1}{A_2} = 2 \cdot \frac{\cos\left(\frac{\pi}{n}\right) \sin\left(\frac{\pi}{2n}\right)}{\sin\left(\frac{\pi}{n}\right) \cos\left(\frac{\pi}{2n}\right)} \]

6. **Final Expression**:
This can be further simplified using the double angle identity for cosine:
\[ \frac{A_1}{A_2} = \frac{2 \cos\left(\frac{\pi}{n}\right)}{2 \cos^2\left(\frac{\pi}{2n}\right)} = \frac{\cos\left(\frac{\pi}{n}\right)}{\cos^2\left(\frac{\pi}{2n}\right)} \]
Thus, the final result is:
\[ \frac{A_1}{A_2} = 2 \cdot \frac{\cos\left(\frac{\pi}{n}\right)}{1 + \cos\left(\frac{\pi}{n}\right)} \]