If : `sin (alpha + beta)=1 and sin (alpha-beta)=(1)/(2),`
`"then" : tan (alpha+2beta)*tan(2alpha+beta)=`

Options

A. 1
B. `-1`
C. zero
D. none of these

Answer

To solve the problem step by step, let's start from the given equations:

1. **Given Equations:**
\[ \sin(\alpha + \beta) = 1 \]
\[ \sin(\alpha - \beta) = \frac{1}{2} \]

2. **Analyzing the First Equation:**
The sine function equals 1 when its argument is \(90^\circ\) (or \(\frac{\pi}{2}\) radians). Therefore, we can write:
\[ \alpha + \beta = 90^\circ \]

3. **Analyzing the Second Equation:**
The sine function equals \(\frac{1}{2}\) when its argument is \(30^\circ\) (or \(\frac{\pi}{6}\) radians). Therefore, we can write:
\[ \alpha - \beta = 30^\circ \]

4. **Solving the System of Equations:**
Now we have a system of two equations:
\[ \alpha + \beta = 90^\circ \quad (1) \]
\[ \alpha - \beta = 30^\circ \quad (2) \]

We can add these two equations:
\[ (\alpha + \beta) + (\alpha - \beta) = 90^\circ + 30^\circ \]
This simplifies to:
\[ 2\alpha = 120^\circ \]
Dividing both sides by 2 gives:
\[ \alpha = 60^\circ \]

5. **Finding Beta:**
Now, substitute \(\alpha = 60^\circ\) back into equation (1):
\[ 60^\circ + \beta = 90^\circ \]
Solving for \(\beta\):
\[ \beta = 90^\circ - 60^\circ = 30^\circ \]

6. **Finding \( \tan(\alpha + 2\beta) \) and \( \tan(2\alpha + \beta) \):**
Now we need to calculate:
\[ \tan(\alpha + 2\beta) \quad \text{and} \quad \tan(2\alpha + \beta) \]

First, calculate \( \alpha + 2\beta \):
\[ \alpha + 2\beta = 60^\circ + 2 \times 30^\circ = 60^\circ + 60^\circ = 120^\circ \]
Now calculate \( \tan(120^\circ) \):
\[ \tan(120^\circ) = -\frac{1}{\sqrt{3}} \]

Next, calculate \( 2\alpha + \beta \):
\[ 2\alpha + \beta = 2 \times 60^\circ + 30^\circ = 120^\circ + 30^\circ = 150^\circ \]
Now calculate \( \tan(150^\circ) \):
\[ \tan(150^\circ) = -\frac{1}{\sqrt{3}} \]

7. **Final Calculation:**
Now we multiply the two tangent values:
\[ \tan(\alpha + 2\beta) \cdot \tan(2\alpha + \beta) = \left(-\frac{1}{\sqrt{3}}\right) \cdot \left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{3} \]

8. **Conclusion:**
Therefore, the final answer is:
\[ \tan(\alpha + 2\beta) \cdot \tan(2\alpha + \beta) = 1 \]