If `cos alpha=(2cos beta-1)/(2-cos beta)` then `tan (alpha/2)` is equal to

Options

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

Answer

To solve the problem where \( \cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta} \) and find \( \tan \left( \frac{\alpha}{2} \right) \), we can follow these steps:

### Step 1: Apply the Componendo and Dividendo Rule
We start with the given equation:
\[ \cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta} \]
Using the Componendo and Dividendo rule, we can rewrite the equation as:
\[ \frac{\cos \alpha - 1}{\cos \alpha + 1} = \frac{2 \cos \beta - 1}{2 - (2 \cos \beta - 1)} \]
This simplifies to:
\[ \frac{\cos \alpha - 1}{\cos \alpha + 1} = \frac{2 \cos \beta - 1}{3 - 2 \cos \beta} \]

### Step 2: Cross-Multiply
Cross-multiplying gives us:
\[ (\cos \alpha - 1)(3 - 2 \cos \beta) = (2 \cos \beta - 1)(\cos \alpha + 1) \]

### Step 3: Expand Both Sides
Expanding both sides:
\[ 3 \cos \alpha - 2 \cos \alpha \cos \beta - 3 + 2 \cos \beta = 2 \cos \beta \cos \alpha + 2 \cos \beta - \cos \alpha - 1 \]

### Step 4: Rearranging Terms
Rearranging the terms gives us:
\[ 3 \cos \alpha - 2 \cos \alpha \cos \beta + \cos \alpha + 1 = 2 \cos \beta \cos \alpha + 2 \cos \beta + 3 \]
Combining like terms leads to:
\[ 4 \cos \alpha - 2 \cos \alpha \cos \beta = 2 \cos \beta + 4 \]

### Step 5: Factor Out Common Terms
Factoring out \( \cos \alpha \):
\[ \cos \alpha (4 - 2 \cos \beta) = 2 \cos \beta + 4 \]

### Step 6: Solve for \( \tan \left( \frac{\alpha}{2} \right) \)
Using the half-angle identity:
\[ \tan \left( \frac{\alpha}{2} \right) = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} \]
Substituting \( \cos \alpha \):
\[ \tan \left( \frac{\alpha}{2} \right) = \sqrt{\frac{1 - \frac{2 \cos \beta - 1}{2 - \cos \beta}}{1 + \frac{2 \cos \beta - 1}{2 - \cos \beta}}} \]

### Step 7: Simplifying the Expression
This simplifies to:
\[ \tan \left( \frac{\alpha}{2} \right) = \sqrt{\frac{(2 - \cos \beta) - (2 \cos \beta - 1)}{(2 - \cos \beta) + (2 \cos \beta - 1)}} \]
This leads to:
\[ \tan \left( \frac{\alpha}{2} \right) = \sqrt{\frac{3 - \cos \beta}{1 + \cos \beta}} \]

### Final Result
Thus, we find:
\[ \tan \left( \frac{\alpha}{2} \right) = \sqrt{3} \tan \left( \frac{\beta}{2} \right) \]