`alpha and beta` are acute angles and `cos2alpha = (3cos2beta-1)/(3-cos2beta)` then `tan alpha cot beta =`

Options

A. `sqrt3`
B. `sqrt2`
C. 1
D. none of these

Answer

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

\[ \cos 2\alpha = \frac{3\cos 2\beta - 1}{3 - \cos 2\beta} \]

### Step 1: Apply Componendo and Dividendo
Using the Componendo and Dividendo theorem, we can rewrite the equation:

\[ \frac{a}{b} = \frac{c}{d} \implies \frac{a+b}{a-b} = \frac{c+d}{c-d} \]

Here, let \( a = \cos 2\alpha \), \( b = 1 \), \( c = 3\cos 2\beta - 1 \), and \( d = 3 - \cos 2\beta \).

Thus, we have:

\[ \frac{\cos 2\alpha + 1}{\cos 2\alpha - 1} = \frac{(3\cos 2\beta - 1) + (3 - \cos 2\beta)}{(3\cos 2\beta - 1) - (3 - \cos 2\beta)} \]

### Step 2: Simplify the Right Side
Now simplify the right-hand side:

\[ \text{Numerator: } (3\cos 2\beta - 1) + (3 - \cos 2\beta) = 2\cos 2\beta + 2 \]

\[ \text{Denominator: } (3\cos 2\beta - 1) - (3 - \cos 2\beta) = 2\cos 2\beta - 4 \]

So the equation becomes:

\[ \frac{\cos 2\alpha + 1}{\cos 2\alpha - 1} = \frac{2(\cos 2\beta + 1)}{2(\cos 2\beta - 2)} \]

### Step 3: Cancel Out the 2's
Dividing both sides by 2 gives:

\[ \frac{\cos 2\alpha + 1}{\cos 2\alpha - 1} = \frac{\cos 2\beta + 1}{\cos 2\beta - 2} \]

### Step 4: Express in Terms of Sine and Cosine
Using the double angle identities:

\[ \cos 2\theta = 2\cos^2\theta - 1 \quad \text{and} \quad \cos 2\theta = 1 - 2\sin^2\theta \]

We can express:

\[ \cos 2\alpha + 1 = 2\cos^2\alpha \quad \text{and} \quad \cos 2\alpha - 1 = -2\sin^2\alpha \]

Thus, we rewrite the left side:

\[ \frac{2\cos^2\alpha}{-2\sin^2\alpha} = -\frac{\cos^2\alpha}{\sin^2\alpha} = -\cot^2\alpha \]

For the right side, we have:

\[ \cos 2\beta + 1 = 2\cos^2\beta \quad \text{and} \quad \cos 2\beta - 2 = -2\sin^2\beta \]

So, the right side becomes:

\[ \frac{2\cos^2\beta}{-2\sin^2\beta} = -\cot^2\beta \]

### Step 5: Set the Two Sides Equal
Now we have:

\[ -\cot^2\alpha = -\cot^2\beta \]

Thus:

\[ \cot^2\alpha = \cot^2\beta \]

### Step 6: Find \( \tan \alpha \cot \beta \)
Taking square roots (since \(\alpha\) and \(\beta\) are acute angles):

\[ \cot\alpha = \cot\beta \implies \tan\alpha = \tan\beta \]

Thus, we find:

\[ \tan\alpha \cot\beta = \tan\beta \cdot \frac{1}{\tan\beta} = 1 \]

### Final Answer
Therefore, the value of \( \tan\alpha \cot\beta \) is:

\[ \sqrt{2} \]