Given that `cos ((alpha -beta)/(2)) = 2 cos ((alpha+ beta)/(2)),` then ` tan ""(alpha)/(2) tan ""(beta)/(2)` is equal to

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `1/8`

Answer

To solve the problem, we start with the given equation:

\[ \cos\left(\frac{\alpha - \beta}{2}\right) = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \]

### Step 1: Rewrite the cosine terms using angle subtraction and addition formulas
Using the cosine subtraction and addition formulas, we can express the left and right sides:

\[ \cos\left(\frac{\alpha - \beta}{2}\right) = \cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) + \sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) \]

\[ 2 \cos\left(\frac{\alpha + \beta}{2}\right) = 2\left(\cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) - \sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right)\right) \]

### Step 2: Set the equations equal
Now we set the two expressions equal to each other:

\[ \cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) + \sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) = 2\left(\cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) - \sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right)\right) \]

### Step 3: Simplify the equation
Expanding the right side gives:

\[ \cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) + \sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) = 2\cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) - 2\sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) \]

Rearranging terms leads to:

\[ \sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) + 2\sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) = 2\cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) \]

This simplifies to:

\[ 3\sin\left(\frac{\alpha}{2}\right) \sin\left(\frac{\beta}{2}\right) = 2\cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) \]

### Step 4: Divide both sides by \(\cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right)\)
We can express this as:

\[ 3 \frac{\sin\left(\frac{\alpha}{2}\right)}{\cos\left(\frac{\alpha}{2}\right)} \cdot \frac{\sin\left(\frac{\beta}{2}\right)}{\cos\left(\frac{\beta}{2}\right)} = 2 \]

This leads to:

\[ 3 \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) = 2 \]

### Step 5: Solve for \( \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) \)
Dividing both sides by 3 gives:

\[ \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) = \frac{2}{3} \]

### Final Answer
Thus, the value of \( \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) \) is:

\[ \frac{2}{3} \]