If `cosA=tanB, cos B=tanC, cosC=tanA,` then sin A is equal to

Options

A. `sin 180^(@)`
B. `2sin18^(@)`
C. `2cos18^(@)`
D. `2 cos36^(@)`

Answer

To solve the problem where \( \cos A = \tan B \), \( \cos B = \tan C \), and \( \cos C = \tan A \), we need to find the value of \( \sin A \).

### Step-by-Step Solution:

1. **Start with the given equations**:
\[ \cos A = \tan B, \quad \cos B = \tan C, \quad \cos C = \tan A \]

2. **Express tangent in terms of sine and cosine**:
Recall that \( \tan B = \frac{\sin B}{\cos B} \). Therefore, we can rewrite the first equation:
\[ \cos A = \frac{\sin B}{\cos B} \]
Cross-multiplying gives:
\[ \cos A \cdot \cos B = \sin B \]

3. **Square both sides**:
\[ \cos^2 A \cdot \cos^2 B = \sin^2 B \]

4. **Use the Pythagorean identity**:
We know that \( \sin^2 B + \cos^2 B = 1 \). Thus, we can express \( \cos^2 B \) as:
\[ \cos^2 B = 1 - \sin^2 B \]
Substituting this into our squared equation gives:
\[ \cos^2 A \cdot (1 - \sin^2 B) = \sin^2 B \]

5. **Let \( \sin A = x \), \( \sin B = y \), and \( \sin C = z \)**:
We can rewrite the equation as:
\[ \cos^2 A \cdot (1 - y^2) = y^2 \]

6. **Express \( \cos^2 A \) in terms of \( x \)**:
Using \( \cos^2 A = 1 - x^2 \), we substitute:
\[ (1 - x^2)(1 - y^2) = y^2 \]

7. **Expand and rearrange**:
Expanding gives:
\[ 1 - y^2 - x^2 + x^2y^2 = y^2 \]
Rearranging leads to:
\[ x^2y^2 - 2y^2 + 1 - x^2 = 0 \]
This can be rewritten as:
\[ x^2y^2 - 2y^2 + 1 = x^2 \]

8. **Repeat the process for \( \cos B \) and \( \cos C \)**:
We can derive similar equations for \( y^2 \) and \( z^2 \) using the relationships \( \cos B = \tan C \) and \( \cos C = \tan A \).

9. **Substituting and solving**:
After deriving the equations, we will have a system of equations involving \( x^2, y^2, z^2 \).

10. **Final substitution**:
After solving the system, we find:
\[ \sin^2 A = \frac{3 - \sqrt{5}}{2} \]
Therefore, taking the positive square root (since sine is positive in the first quadrant):
\[ \sin A = \sqrt{\frac{3 - \sqrt{5}}{2}} \]

### Final Answer:
\[ \sin A = \frac{\sqrt{5} - 1}{2} \]