`cos 2 alpha =(3 cos 2 beta -1)/( 3-cos 2 beta),` then `tan alpha=`

Options

A. `tan beta`
B. `2 tan beta`
C. `sqrt2 tan beta`
D. None of these

Answer

To solve the equation \( \cos 2\alpha = \frac{3 \cos 2\beta - 1}{3 - \cos 2\beta} \) and find \( \tan \alpha \), we will follow these steps:

### Step 1: Start with the given equation
We have:
\[ \cos 2\alpha = \frac{3 \cos 2\beta - 1}{3 - \cos 2\beta} \]

### Step 2: Apply Componendo and Dividendo
Using the Componendo and Dividendo method, we can rewrite the equation as:
\[ \frac{\cos 2\alpha - 1}{\cos 2\alpha + 1} = \frac{(3 \cos 2\beta - 1) - (3 - \cos 2\beta)}{(3 \cos 2\beta - 1) + (3 - \cos 2\beta)} \]

### Step 3: Simplify the right-hand side
Now, simplify the numerator and denominator:
- Numerator:
\[ (3 \cos 2\beta - 1) - (3 - \cos 2\beta) = 4 \cos 2\beta - 4 = 4(\cos 2\beta - 1) \]
- Denominator:
\[ (3 \cos 2\beta - 1) + (3 - \cos 2\beta) = 2 \cos 2\beta + 2 = 2(1 + \cos 2\beta) \]

So we have:
\[ \frac{\cos 2\alpha - 1}{\cos 2\alpha + 1} = \frac{4(\cos 2\beta - 1)}{2(1 + \cos 2\beta)} = 2 \cdot \frac{\cos 2\beta - 1}{1 + \cos 2\beta} \]

### Step 4: Rewrite using trigonometric identities
Using the identities \( 1 - \cos 2\theta = 2 \sin^2 \theta \) and \( 1 + \cos 2\theta = 2 \cos^2 \theta \), we can rewrite the equation:
\[ \frac{1 - \cos 2\alpha}{1 + \cos 2\alpha} = 2 \cdot \frac{2 \sin^2 \beta}{2 \cos^2 \beta} = \frac{2 \sin^2 \beta}{\cos^2 \beta} \]

### Step 5: Substitute the identities
Thus, we have:
\[ \frac{2 \sin^2 \alpha}{2 \cos^2 \alpha} = \frac{2 \sin^2 \beta}{\cos^2 \beta} \]

### Step 6: Cancel the common factors
Cancel the 2's:
\[ \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{\sin^2 \beta}{\cos^2 \beta} \]

### Step 7: Relate the tangents
This gives us:
\[ \tan^2 \alpha = \tan^2 \beta \]

### Step 8: Find \( \tan \alpha \)
Taking the square root, we find:
\[ \tan \alpha = \sqrt{2} \tan \beta \]

### Conclusion
Thus, the final result is:
\[ \tan \alpha = \sqrt{2} \tan \beta \]

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