Let sin (A+B) = 1 and sin (A-B) = `1//2` where `AB epsilon[0,(pi)/(2)]`.
What is the value of tan (A+2B). tan(2A+B)?

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the equations
Given:
1. \( \sin(A + B) = 1 \)
2. \( \sin(A - B) = \frac{1}{2} \)

From the first equation, since \( \sin(A + B) = 1 \), we can conclude that:
\[ A + B = 90^\circ \]
This is because the sine function reaches its maximum value of 1 at \( 90^\circ \).

### Step 2: Solve for \( A - B \)
From the second equation, since \( \sin(A - B) = \frac{1}{2} \), we can conclude that:
\[ A - B = 30^\circ \]
This is because \( \sin(30^\circ) = \frac{1}{2} \).

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \( A + B = 90^\circ \)
2. \( A - B = 30^\circ \)

We can add these two equations:
\[ (A + B) + (A - B) = 90^\circ + 30^\circ \]
This simplifies to:
\[ 2A = 120^\circ \]
So, we find:
\[ A = 60^\circ \]

Next, we can substitute \( A \) back into the first equation to find \( B \):
\[ 60^\circ + B = 90^\circ \]
Thus:
\[ B = 30^\circ \]

### Step 4: Calculate \( \tan(A + 2B) \) and \( \tan(2A + B) \)
Now we need to find \( \tan(A + 2B) \) and \( \tan(2A + B) \).

1. Calculate \( A + 2B \):
\[ A + 2B = 60^\circ + 2 \times 30^\circ = 60^\circ + 60^\circ = 120^\circ \]
Thus:
\[ \tan(A + 2B) = \tan(120^\circ) \]

2. Calculate \( 2A + B \):
\[ 2A + B = 2 \times 60^\circ + 30^\circ = 120^\circ + 30^\circ = 150^\circ \]
Thus:
\[ \tan(2A + B) = \tan(150^\circ) \]

### Step 5: Use the tangent values
Now we can use the tangent values:
- \( \tan(120^\circ) = -\sqrt{3} \)
- \( \tan(150^\circ) = -\frac{1}{\sqrt{3}} \)

### Step 6: Multiply the tangents
Now we multiply the two results:
\[ \tan(A + 2B) \cdot \tan(2A + B) = \tan(120^\circ) \cdot \tan(150^\circ) = (-\sqrt{3}) \cdot \left(-\frac{1}{\sqrt{3}}\right) \]
This simplifies to:
\[ 1 \]

### Final Answer
Thus, the value of \( \tan(A + 2B) \cdot \tan(2A + B) \) is:
\[ \boxed{1} \]