What is tan (A+2B) . Tan (2A+B) equal to ?

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem of finding what \( \tan(A + 2B) \cdot \tan(2A + B) \) is equal to, we will follow these steps:

### Step 1: Write the Expression
We start with the expression:
\[ x = \tan(A + 2B) \cdot \tan(2A + B) \]

### Step 2: Substitute Values for A and B
To simplify, we can assume specific values for \( A \) and \( B \). Let's take:
- \( A = 60^\circ \)
- \( B = 30^\circ \)

### Step 3: Calculate \( A + 2B \) and \( 2A + B \)
Now we calculate:
\[ A + 2B = 60^\circ + 2 \cdot 30^\circ = 60^\circ + 60^\circ = 120^\circ \]
\[ 2A + B = 2 \cdot 60^\circ + 30^\circ = 120^\circ + 30^\circ = 150^\circ \]

### Step 4: Substitute Back into the Expression
Now substitute these angles back into the expression for \( x \):
\[ x = \tan(120^\circ) \cdot \tan(150^\circ) \]

### Step 5: Use the Tangent Values
Using the known values of tangent:
\[ \tan(120^\circ) = -\sqrt{3} \quad \text{(since } 120^\circ \text{ is in the second quadrant)} \]
\[ \tan(150^\circ) = -\frac{1}{\sqrt{3}} \quad \text{(since } 150^\circ \text{ is also in the second quadrant)} \]

### Step 6: Calculate the Product
Now we calculate the product:
\[ x = (-\sqrt{3}) \cdot \left(-\frac{1}{\sqrt{3}}\right) = \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 \]

### Final Result
Thus, we conclude that:
\[ \tan(A + 2B) \cdot \tan(2A + B) = 1 \]

### Conclusion
The final answer is:
\[ \boxed{1} \]