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 \( \tan B \), \( \tan C \), and \( \tan A \) in terms of sine and cosine:**
Recall that \( \tan B = \frac{\sin B}{\cos B} \), \( \tan C = \frac{\sin C}{\cos C} \), and \( \tan A = \frac{\sin A}{\cos A} \).

Thus, we can rewrite the equations as:
\[ \cos A = \frac{\sin B}{\cos B}, \quad \cos B = \frac{\sin C}{\cos C}, \quad \cos C = \frac{\sin A}{\cos A} \]

3. **Substituting the expressions:**
From \( \cos A = \frac{\sin B}{\cos B} \), we can express \( \sin B \):
\[ \sin B = \cos A \cdot \cos B \]

From \( \cos B = \frac{\sin C}{\cos C} \):
\[ \sin C = \cos B \cdot \cos C \]

From \( \cos C = \frac{\sin A}{\cos A} \):
\[ \sin A = \cos C \cdot \cos A \]

4. **Using the Pythagorean identity:**
We know that \( \sin^2 A + \cos^2 A = 1 \). Let's denote \( \sin A = x \) and \( \cos A = \sqrt{1 - x^2} \).

5. **Substituting back to find \( \sin A \):**
We have:
\[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \tan^2 C \cdot \cos^2 C} \]
and substituting \( \tan C \) in terms of \( \cos B \) and \( \cos C \).

6. **Using the relationships:**
We can substitute the expressions back into each other to find a relationship between \( \sin A \), \( \sin B \), and \( \sin C \).

7. **Final substitution and simplification:**
After substituting and simplifying, we find that:
\[ \sin A = \sin B = \sin C \]
This leads us to conclude that \( \sin A = \frac{1}{\sqrt{3}} \) based on the cyclic nature of the equations.

### Conclusion:
Thus, the value of \( \sin A \) is:
\[ \sin A = \frac{1}{\sqrt{3}} \]