If `cos theta =(cos alpha - cos beta)/( 1- cos alpha cos beta),` then one of the values of `tan ((theta )/(2)) `is

Options

A. `cot "" beta/2 tan "" alpha /2`
B. `tan alpha tan "" beta/2`
C. `tan ""beta/2 cot "" alpha/2`
D. `tan ^(2) "" alpha/ 2 tan ^(2)"" beta/2`

Answer

To solve the problem, we need to find one of the values of \( \tan\left(\frac{\theta}{2}\right) \) given that

\[ \cos \theta = \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}. \]

### Step-by-step Solution:

1. **Recall the formula for \( \tan\left(\frac{\theta}{2}\right) \)**:
We know that:
\[ \tan^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{1 + \cos \theta}. \]

2. **Substitute \( \cos \theta \)**:
Substitute the given expression for \( \cos \theta \):
\[ \tan^2\left(\frac{\theta}{2}\right) = \frac{1 - \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}}{1 + \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}}. \]

3. **Simplify the numerator**:
The numerator becomes:
\[ 1 - \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta} = \frac{(1 - \cos \alpha \cos \beta) - (\cos \alpha - \cos \beta)}{1 - \cos \alpha \cos \beta}. \]
Simplifying the numerator:
\[ = \frac{1 - \cos \alpha \cos \beta - \cos \alpha + \cos \beta}{1 - \cos \alpha \cos \beta} = \frac{(1 - \cos \alpha) + (\cos \beta - \cos \alpha \cos \beta)}{1 - \cos \alpha \cos \beta}. \]

4. **Simplify the denominator**:
The denominator becomes:
\[ 1 + \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta} = \frac{(1 - \cos \alpha \cos \beta) + (\cos \alpha - \cos \beta)}{1 - \cos \alpha \cos \beta}. \]
Simplifying the denominator:
\[ = \frac{(1 - \cos \alpha \cos \beta) + \cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta} = \frac{(1 + \cos \alpha - \cos \beta - \cos \alpha \cos \beta)}{1 - \cos \alpha \cos \beta}. \]

5. **Combine the results**:
Now, we can write:
\[ \tan^2\left(\frac{\theta}{2}\right) = \frac{(1 - \cos \alpha) + (\cos \beta - \cos \alpha \cos \beta)}{(1 + \cos \alpha - \cos \beta - \cos \alpha \cos \beta)}. \]

6. **Using the half-angle identities**:
We can relate this to the tangent half-angle formulas:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right)}{1 - \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right)}. \]

7. **Final Result**:
Thus, we find that:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right)}{1 - \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right)}. \]

### Conclusion:
One of the values of \( \tan\left(\frac{\theta}{2}\right) \) is given by:
\[ \tan\left(\frac{\theta}{2}\right) = \tan\left(\frac{\alpha}{2}\right) \cdot \tan\left(\frac{\beta}{2}\right). \]