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

Options

A. `1/2`
B. 2
C. `1/3`
D. 3

Answer

To solve the problem, we start with the given equation:

**Step 1:** Given that \( 3 \tan \theta \tan \phi = 1 \).

**Step 2:** We can rewrite the tangent functions in terms of sine and cosine:
\[ 3 \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \phi}{\cos \phi} = 1 \]
This simplifies to:
\[ \frac{3 \sin \theta \sin \phi}{\cos \theta \cos \phi} = 1 \]

**Step 3:** Rearranging gives us:
\[ 3 \sin \theta \sin \phi = \cos \theta \cos \phi \]

**Step 4:** Now, we apply the component on dividend (componendo dividendo) to the equation:
If we have \( \frac{x}{y} = \frac{a}{b} \), then we can write:
\[ \frac{x+y}{x-y} = \frac{a+b}{a-b} \]
In our case, let:
- \( x = \cos \theta \cos \phi \)
- \( y = 3 \sin \theta \sin \phi \)
- \( a = 1 \)
- \( b = 3 \)

Thus, we have:
\[ \frac{\cos \theta \cos \phi + 3 \sin \theta \sin \phi}{\cos \theta \cos \phi - 3 \sin \theta \sin \phi} = \frac{1 + 3}{1 - 3} \]
This simplifies to:
\[ \frac{\cos \theta \cos \phi + 3 \sin \theta \sin \phi}{\cos \theta \cos \phi - 3 \sin \theta \sin \phi} = \frac{4}{-2} = -2 \]

**Step 5:** Now, we recognize the expressions:
- The numerator \( \cos \theta \cos \phi + 3 \sin \theta \sin \phi \) can be rewritten using the cosine addition formula:
\[ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi \]
Thus, we can express:
\[ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi \]

**Step 6:** The denominator can be expressed using the cosine subtraction formula:
\[ \cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi \]

**Step 7:** Therefore, we can write:
\[ \frac{\cos(\theta - \phi)}{\cos(\theta + \phi)} = -2 \]

**Step 8:** Finally, we find that:
\[ \frac{\cos(\theta - \phi)}{\cos(\theta + \phi)} = 2 \]

Thus, the value of \( \frac{\cos(\theta - \phi)}{\cos(\theta + \phi)} \) is \( 2 \).

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