If 3 tan `theta tan phi=1, ` then `(cos (theta-phi))/(cos (theta+phi))` is

Answer

To solve the problem, we need to find the value of \(\frac{\cos(\theta - \phi)}{\cos(\theta + \phi)}\) given that \(3 \tan \theta \tan \phi = 1\).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ 3 \tan \theta \tan \phi = 1 \]
This implies:
\[ \tan \theta \tan \phi = \frac{1}{3} \]

2. **Express \(\tan \theta\) and \(\tan \phi\) in terms of sine and cosine:**
\[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \tan \phi = \frac{\sin \phi}{\cos \phi} \]
Therefore,
\[ \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \phi}{\cos \phi} = \frac{1}{3} \]

3. **Cross-multiply to get a new equation:**
\[ \sin \theta \sin \phi = \frac{1}{3} \cos \theta \cos \phi \]

4. **Take the reciprocal of both sides:**
\[ \frac{\cos \theta \cos \phi}{\sin \theta \sin \phi} = 3 \]

5. **Use the identity for cotangent:**
\[ \frac{\cos \theta \cos \phi}{\sin \theta \sin \phi} = \cot \theta \cot \phi \]
Thus,
\[ \cot \theta \cot \phi = 3 \]

6. **Apply the component and dividend rule:**
If \(\frac{a}{b} = \frac{c}{d}\), then:
\[ \frac{a + b}{a - b} = \frac{c + d}{c - d} \]
Here, let \(a = \cos \theta \cos \phi\) and \(b = \sin \theta \sin \phi\). Since we have \(\cot \theta \cot \phi = 3\), we can set:
\[ \frac{\cos \theta \cos \phi + \sin \theta \sin \phi}{\cos \theta \cos \phi - \sin \theta \sin \phi} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2 \]

7. **Use the cosine addition and subtraction formulas:**
Recall that:
\[ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi \]
\[ \cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi \]
Therefore, we have:
\[ \frac{\cos(\theta - \phi)}{\cos(\theta + \phi)} = 2 \]

8. **Conclusion:**
Thus, we conclude that:
\[ \frac{\cos(\theta - \phi)}{\cos(\theta + \phi)} = 2 \]

### Final Answer:
\[ \frac{\cos(\theta - \phi)}{\cos(\theta + \phi)} = 2 \]