If `3(cos 2phi-cos2theta)=1-cos2phi cos2theta`, then `tan theta= k tan phi`, where `theta, phiin (0,(pi)/(2))`, where k =

Options

A. `sqrt(2)`
B. `(1)/(sqrt(2))`
C. `sqrt(3)`
D. `(1)/(sqrt(3))`

Explanation

To solve the equation \(3(\cos 2\phi - \cos 2\theta) = 1 - \cos 2\phi \cos 2\theta\) and find the value of \(k\) such that \(\tan \theta = k \tan \phi\), we can follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 3(\cos 2\phi - \cos 2\theta) = 1 - \cos 2\phi \cos 2\theta \]
Rearranging gives:
\[ 3\cos 2\phi - 3\cos 2\theta = 1 - \cos 2\phi \cos 2\theta \]
This can be rewritten as:
\[ 3\cos 2\phi + \cos 2\phi \cos 2\theta = 1 + 3\cos 2\theta \]

### Step 2: Factoring Out \(\cos 2\phi\)
Now, factor out \(\cos 2\phi\):
\[ \cos 2\phi(3 + \cos 2\theta) = 1 + 3\cos 2\theta \]
From this, we can express \(\cos 2\phi\) as:
\[ \cos 2\phi = \frac{1 + 3\cos 2\theta}{3 + \cos 2\theta} \]

### Step 3: Applying Componendo and Dividendo
Next, we apply the method of componendo and dividendo:
\[ \frac{-\cos 2\phi}{1 + \cos 2\phi} = \frac{3\cos 2\theta - 1}{3 + \cos 2\theta + 1} \]
This simplifies to:
\[ \frac{1 - \cos 2\phi}{1 + \cos 2\phi} = \frac{2 - 2\cos 2\theta}{4 + 4\cos 2\theta} \]

### Step 4: Simplifying the Ratio
This can be simplified further:
\[ \frac{1 - \cos 2\phi}{1 + \cos 2\phi} = \frac{1 - \cos 2\theta}{2(1 + \cos 2\theta)} \]

### Step 5: Using the Trigonometric Identity
Using the identity:
\[ \tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta} \]
We can write:
\[ \frac{1 - \cos 2\phi}{1 + \cos 2\phi} = \frac{1}{2} \tan^2 \theta \]
Thus,
\[ \tan^2 \phi = \frac{1}{2} \tan^2 \theta \]

### Step 6: Finding the Relationship Between \(\tan \theta\) and \(\tan \phi\)
Rearranging gives:
\[ \tan^2 \theta = 2 \tan^2 \phi \]
Taking the square root:
\[ \tan \theta = \sqrt{2} \tan \phi \]
Thus, we have:
\[ k = \sqrt{2} \]

### Final Answer
The value of \(k\) is:
\[ \boxed{\sqrt{2}} \]