If `alpha and beta` are non-zero real number such that `2(cos beta-cos alpha)+cos alpha cos beta=1.` Then which of the following is treu?

Options

A. `tan ((alpha )/(2)) + sqrt3 tan ((beta)/(2)) =0`
B. `sqrt3 tan ((alpha)/(2)) + tan ((beta)/(2)) =0`
C. `tan ((alpha )/(2)) - sqrt3 tan ((beta)/(2)) =0`
D. `sqrt3 tan (( alpha )/(2)) - tan ((beta)/(2))= 0`

Explanation

To solve the equation \( 2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ 2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1 \]
This can be rewritten as:
\[ 2\cos \beta - 2\cos \alpha + \cos \alpha \cos \beta = 1 \]

### Step 2: Rearranging the equation
Rearranging gives:
\[ 2\cos \beta + \cos \alpha \cos \beta - 2\cos \alpha = 1 \]

### Step 3: Multiply through by 2
To eliminate fractions, multiply the entire equation by 2:
\[ 4\cos \beta + 2\cos \alpha \cos \beta - 4\cos \alpha = 2 \]

### Step 4: Group terms
Now, group the terms:
\[ (4\cos \beta + 2\cos \alpha \cos \beta) - 4\cos \alpha = 2 \]

### Step 5: Factor out common terms
Factor out \( \cos \beta \):
\[ \cos \beta(4 + 2\cos \alpha) - 4\cos \alpha = 2 \]

### Step 6: Isolate \( \cos \beta \)
Rearranging gives:
\[ \cos \beta(4 + 2\cos \alpha) = 2 + 4\cos \alpha \]
Thus,
\[ \cos \beta = \frac{2 + 4\cos \alpha}{4 + 2\cos \alpha} \]

### Step 7: Simplifying the expression
Now simplify the expression:
\[ \cos \beta = \frac{2(1 + 2\cos \alpha)}{2(2 + \cos \alpha)} = \frac{1 + 2\cos \alpha}{2 + \cos \alpha} \]

### Step 8: Use trigonometric identities
Using the half-angle identities:
\[ 1 - \cos \alpha = 2\sin^2\left(\frac{\alpha}{2}\right) \quad \text{and} \quad 1 - \cos \beta = 2\sin^2\left(\frac{\beta}{2}\right) \]
We can rewrite:
\[ \frac{1 - \cos \alpha}{1 + \cos \alpha} = \tan^2\left(\frac{\alpha}{2}\right) \]
and
\[ \frac{1 - \cos \beta}{1 + \cos \beta} = \tan^2\left(\frac{\beta}{2}\right) \]

### Step 9: Setting up the final equation
Thus, we have:
\[ \frac{1 - \cos \alpha}{1 + \cos \alpha} = 3 \cdot \frac{1 - \cos \beta}{1 + \cos \beta} \]
This leads to:
\[ \tan^2\left(\frac{\alpha}{2}\right) = 3 \tan^2\left(\frac{\beta}{2}\right) \]

### Step 10: Final results
From this, we can conclude:
\[ \tan\left(\frac{\alpha}{2}\right) = \pm \sqrt{3} \tan\left(\frac{\beta}{2}\right) \]

### Summary
The derived relationships indicate that:
1. \( \tan\left(\frac{\alpha}{2}\right) + \sqrt{3} \tan\left(\frac{\beta}{2}\right) = 0 \)
2. \( \tan\left(\frac{\alpha}{2}\right) - \sqrt{3} \tan\left(\frac{\beta}{2}\right) = 0 \)