Let `alphaa n dbeta` be nonzero real numbers such that `2(cosbeta-cosalpha)+cosalphacosbeta=1` . Then which of the following is/are true? `sqrt(3)tan(alpha/2)+tan(beta/2)=0` `sqrt(3)tan(alpha/2)-tan(beta/2)=0` `t a n""(alpha/2)+sqrt(3)tan(beta/2)=0` `t a n""(alpha/2)-sqrt(3)tan(beta/2)=0`

Options

A. `sqrt3tan((alpha)/(2))-tan((beta)/(2)) = 0`
B. `tan((alpha)/(2)) - sqrt3tan((beta)/(2)) = 0`
C. `tan((alpha)/(2))+sqrt3tan((beta)/(2)) = 0`
D. `sqrt3tan((alpha)/(2))+tan((beta)/(2)) = 0`

Answer

To solve the problem, we need to analyze the given equation and derive the necessary relationships between the angles \(\alpha\) and \(\beta\). Let's go through the steps systematically.

### Step 1: Start with the given equation
We have the equation:
\[ 2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1 \]

### Step 2: Rewrite cosines in terms of tangents
Using the half-angle formulas, we can express \(\cos \alpha\) and \(\cos \beta\) in terms of \(\tan(\alpha/2)\) and \(\tan(\beta/2)\):
\[ \cos \alpha = \frac{1 - \tan^2(\alpha/2)}{1 + \tan^2(\alpha/2)} \quad \text{and} \quad \cos \beta = \frac{1 - \tan^2(\beta/2)}{1 + \tan^2(\beta/2)} \]
Let \(a = \tan^2(\alpha/2)\) and \(b = \tan^2(\beta/2)\).

### Step 3: Substitute into the equation
Substituting these into the original equation gives:
\[ 2\left(\frac{1 - b}{1 + b} - \frac{1 - a}{1 + a}\right) + \left(\frac{1 - a}{1 + a}\right)\left(\frac{1 - b}{1 + b}\right) = 1 \]

### Step 4: Simplify the equation
This can be simplified by finding a common denominator:
\[ \text{Let } LCM = (1 + a)(1 + b) \]
Then, we rewrite the equation:
\[ 2\left(\frac{(1 - b)(1 + a) - (1 - a)(1 + b)}{(1 + b)(1 + a)}\right) + \frac{(1 - a)(1 - b)}{(1 + a)(1 + b)} = 1 \]

### Step 5: Expand and simplify
After expanding and simplifying the equation, we will have:
\[ 4a - 4b = 2a + 2b \]
This leads us to:
\[ 4a - 2a = 4b + 2b \implies 2a = 6b \implies a = 3b \]

### Step 6: Substitute back to find relationships
Substituting back \(a\) and \(b\):
\[ \tan^2\left(\frac{\alpha}{2}\right) = 3\tan^2\left(\frac{\beta}{2}\right) \]

### Step 7: Find the relationships
Taking square roots gives:
\[ \tan\left(\frac{\alpha}{2}\right) = \sqrt{3}\tan\left(\frac{\beta}{2}\right) \quad \text{or} \quad \tan\left(\frac{\alpha}{2}\right) = -\sqrt{3}\tan\left(\frac{\beta}{2}\right) \]

### Step 8: Write the final equations
This leads us to the following equations:
1. \(\sqrt{3}\tan\left(\frac{\alpha}{2}\right) - \tan\left(\frac{\beta}{2}\right) = 0\)
2. \(\tan\left(\frac{\alpha}{2}\right) - \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0\)

### Conclusion
Thus, the correct options are:
- \(\sqrt{3}\tan\left(\frac{\alpha}{2}\right) - \tan\left(\frac{\beta}{2}\right) = 0\) (True)
- \(\tan\left(\frac{\alpha}{2}\right) - \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0\) (True)