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 start with the given equation:

**Given**:
\[ \cos \theta = \cos \alpha \cos \beta \]

We need to find the value of:
\[ \tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) \]

### Step 1: Rewrite the equation
From the given equation, we can express it as:
\[ \frac{\cos \theta}{\cos \alpha} = \cos \beta \]

### Step 2: Apply the component and dividend method
Using the component and dividend method, we can rewrite the left-hand side:
\[ \frac{\cos \theta + \cos \alpha}{\cos \theta - \cos \alpha} = \frac{\cos \beta + 1}{\cos \beta - 1} \]

### Step 3: Use cosine addition and subtraction formulas
Now, we can apply the cosine addition and subtraction formulas:
1. \(\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)\)
2. \(\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)\)

Applying these to our equation:
- For the numerator:
\[ \cos \theta + \cos \alpha = 2 \cos\left(\frac{\theta + \alpha}{2}\right) \cos\left(\frac{\theta - \alpha}{2}\right) \]
- For the denominator:
\[ \cos \theta - \cos \alpha = -2 \sin\left(\frac{\theta + \alpha}{2}\right) \sin\left(\frac{\theta - \alpha}{2}\right) \]

### Step 4: Substitute into the equation
Substituting these into our equation gives:
\[ \frac{2 \cos\left(\frac{\theta + \alpha}{2}\right) \cos\left(\frac{\theta - \alpha}{2}\right)}{-2 \sin\left(\frac{\theta + \alpha}{2}\right) \sin\left(\frac{\theta - \alpha}{2}\right)} = \frac{\cos \beta + 1}{\cos \beta - 1} \]

### Step 5: Simplify the equation
The 2's cancel out, leading to:
\[ -\frac{\cos\left(\frac{\theta + \alpha}{2}\right) \cos\left(\frac{\theta - \alpha}{2}\right)}{\sin\left(\frac{\theta + \alpha}{2}\right) \sin\left(\frac{\theta - \alpha}{2}\right)} = \frac{\cos \beta + 1}{\cos \beta - 1} \]

### Step 6: Use the tangent identity
Using the identity \(\tan A = \frac{\sin A}{\cos A}\), we can rewrite the left-hand side:
\[ -\tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) = \frac{\cos \beta + 1}{\cos \beta - 1} \]

### Step 7: Find the value of the expression
Thus, we have:
\[ \tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) = -\frac{\cos \beta + 1}{\cos \beta - 1} \]

### Step 8: Recognize the relationship
From the trigonometric identities, we know:
\[ \frac{\sin^2\left(\frac{\beta}{2}\right)}{\cos^2\left(\frac{\beta}{2}\right)} = \tan^2\left(\frac{\beta}{2}\right) \]

### Conclusion
Therefore, we conclude that:
\[ \tan\left(\frac{\theta + \alpha}{2}\right) \tan\left(\frac{\theta - \alpha}{2}\right) = \tan^2\left(\frac{\beta}{2}\right) \]

Thus, the final answer is:
\[ \tan^2\left(\frac{\beta}{2}\right) \]