If `costheta=cos alphacosbeta, then tan((theta+alpha)/(2))tan((theta-alpha)/(2))` is equal to

Options

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

Answer

To solve the problem, we need to find the value of \( \tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) \) given that \( \cos \theta = \cos \alpha \cos \beta \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \cos \theta = \cos \alpha \cos \beta \]

2. **Express the tangent in terms of sine and cosine**:
\[ \tan\left(\frac{\theta + \alpha}{2}\right) = \frac{\sin\left(\frac{\theta + \alpha}{2}\right)}{\cos\left(\frac{\theta + \alpha}{2}\right)} \]
\[ \tan\left(\frac{\theta - \alpha}{2}\right) = \frac{\sin\left(\frac{\theta - \alpha}{2}\right)}{\cos\left(\frac{\theta - \alpha}{2}\right)} \]

3. **Combine the two tangents**:
\[ \tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) = \frac{\sin\left(\frac{\theta + \alpha}{2}\right) \sin\left(\frac{\theta - \alpha}{2}\right)}{\cos\left(\frac{\theta + \alpha}{2}\right) \cos\left(\frac{\theta - \alpha}{2}\right)} \]

4. **Use the product-to-sum identities**:
\[ \sin A \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right] \]
Here, let \( A = \frac{\theta + \alpha}{2} \) and \( B = \frac{\theta - \alpha}{2} \):
\[ \sin\left(\frac{\theta + \alpha}{2}\right) \sin\left(\frac{\theta - \alpha}{2}\right) = \frac{1}{2} \left[ \cos\left(\frac{\theta - \alpha}{2} - \frac{\theta + \alpha}{2}\right) - \cos\left(\frac{\theta - \alpha}{2} + \frac{\theta + \alpha}{2}\right) \right] \]
This simplifies to:
\[ = \frac{1}{2} \left[ \cos(-\alpha) - \cos(\theta) \right] = \frac{1}{2} \left[ \cos \alpha - \cos \theta \right] \]

5. **Substituting \( \cos \theta \)**:
Substitute \( \cos \theta = \cos \alpha \cos \beta \):
\[ = \frac{1}{2} \left[ \cos \alpha - \cos \alpha \cos \beta \right] = \frac{1}{2} \cos \alpha (1 - \cos \beta) \]

6. **Now for the denominator**:
Using the identity for cosine:
\[ \cos\left(\frac{\theta + \alpha}{2}\right) \cos\left(\frac{\theta - \alpha}{2}\right) = \frac{1}{2} \left[ \cos(\theta) + \cos(\alpha) \right] \]
Substitute \( \cos \theta = \cos \alpha \cos \beta \):
\[ = \frac{1}{2} \left[ \cos \alpha \cos \beta + \cos \alpha \right] = \frac{1}{2} \cos \alpha (1 + \cos \beta) \]

7. **Final expression**:
Now, we can simplify:
\[ \tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) = \frac{\frac{1}{2} \cos \alpha (1 - \cos \beta)}{\frac{1}{2} \cos \alpha (1 + \cos \beta)} = \frac{1 - \cos \beta}{1 + \cos \beta} \]

### Conclusion:
Thus, the final result is:
\[ \tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) = \frac{1 - \cos \beta}{1 + \cos \beta} \]