If `sin (a+b) =1, sin (a- b) = ( 1)/( 2) , a, b in [ 0, pi //2]`, then `tan ( a+ 2 b ) tan ( 2 a + b) =`

Options

A. 1
B. 2
C. `-1`
D. none

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
- \( \sin(a + b) = 1 \)
- \( \sin(a - b) = \frac{1}{2} \)

2. **Analyzing the First Equation:**
- The sine function equals 1 at \( \frac{\pi}{2} \). Therefore, we can set:
\[ a + b = \frac{\pi}{2} \quad \text{(Equation 1)} \]

3. **Analyzing the Second Equation:**
- The sine function equals \( \frac{1}{2} \) at \( \frac{\pi}{6} \) (or 30 degrees). Thus, we have:
\[ a - b = \frac{\pi}{6} \quad \text{(Equation 2)} \]

4. **Adding the Two Equations:**
- Now, we add Equation 1 and Equation 2:
\[ (a + b) + (a - b) = \frac{\pi}{2} + \frac{\pi}{6} \]
- This simplifies to:
\[ 2a = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]
- Dividing by 2 gives:
\[ a = \frac{\pi}{3} \quad \text{(or 60 degrees)} \]

5. **Substituting to Find \( b \):**
- Now we substitute \( a \) back into Equation 1:
\[ \frac{\pi}{3} + b = \frac{\pi}{2} \]
- Rearranging gives:
\[ b = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{6} \quad \text{(or 30 degrees)} \]

6. **Finding \( \tan(a + 2b) \) and \( \tan(2a + b) \):**
- Now we calculate \( a + 2b \):
\[ a + 2b = \frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \]
- Next, we calculate \( 2a + b \):
\[ 2a + b = 2 \times \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{3} + \frac{\pi}{6} = \frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6} \]

7. **Calculating \( \tan(a + 2b) \) and \( \tan(2a + b) \):**
- We find:
\[ \tan\left(\frac{2\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3} \]
- And:
\[ \tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} \]

8. **Finding the Product:**
- Now we calculate:
\[ \tan(a + 2b) \tan(2a + b) = (-\sqrt{3}) \left(-\frac{1}{\sqrt{3}}\right) = 1 \]

9. **Final Answer:**
- Therefore, the final answer is:
\[ \tan(a + 2b) \tan(2a + b) = 1 \]