To solve the problem, we need to analyze the given expressions for \( \alpha \) and \( \beta \):
1. **Given:**
\[
\alpha = 3 \sin^{-1}\left(\frac{6}{11}\right)
\]
\[
\beta = 3 \cos^{-1}\left(\frac{4}{9}\right)
\]
2. **Step 1: Analyze \( \alpha \)**
- We know that \( \sin^{-1}(x) \) is defined for \( x \) in the range \([-1, 1]\). Here, \( \frac{6}{11} \) is valid.
- Since \( \frac{6}{11} > \frac{1}{2} \), we can compare \( \sin^{-1}\left(\frac{6}{11}\right) \) with \( \sin^{-1}\left(\frac{1}{2}\right) \).
- We know \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \).
- Thus, \( \sin^{-1}\left(\frac{6}{11}\right) > \frac{\pi}{6} \).
Therefore:
\[
3 \sin^{-1}\left(\frac{6}{11}\right) > 3 \cdot \frac{\pi}{6} = \frac{\pi}{2}
\]
Hence, \( \alpha > \frac{\pi}{2} \).
3. **Step 2: Analyze \( \beta \)**
- We know \( \cos^{-1}(x) \) is defined for \( x \) in the range \([-1, 1]\). Here, \( \frac{4}{9} \) is valid.
- Since \( \frac{4}{9} < \frac{1}{2} \), we can compare \( \cos^{-1}\left(\frac{4}{9}\right) \) with \( \cos^{-1}\left(\frac{1}{2}\right) \).
- We know \( \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \).
- Thus, \( \cos^{-1}\left(\frac{4}{9}\right) > \frac{\pi}{3} \).
Therefore:
\[
3 \cos^{-1}\left(\frac{4}{9}\right) > 3 \cdot \frac{\pi}{3} = \pi
\]
Hence, \( \beta > \pi \).
4. **Step 3: Combine \( \alpha \) and \( \beta \)**
- Now we have:
\[
\alpha > \frac{\pi}{2} \quad \text{and} \quad \beta > \pi
\]
Adding these inequalities:
\[
\alpha + \beta > \frac{\pi}{2} + \pi = \frac{3\pi}{2}
\]
5. **Step 4: Analyze the options**
- **Option 1:** \( \cos(\beta) > 0 \)
- Since \( \beta > \pi \), \( \cos(\beta) < 0 \). Thus, this option is **incorrect**.
- **Option 2:** \( \sin(\beta) < 0 \)
- Since \( \beta > \pi \), \( \sin(\beta) < 0 \). Thus, this option is **correct**.
- **Option 3:** \( \cos(\alpha + \beta) > 0 \)
- Since \( \alpha + \beta > \frac{3\pi}{2} \), \( \cos(\alpha + \beta) < 0 \). Thus, this option is **incorrect**.
- **Option 4:** \( \cos(\alpha) < 0 \)
- Since \( \alpha > \frac{\pi}{2} \), \( \cos(\alpha) < 0 \). Thus, this option is **correct**.
6. **Final Conclusion:**
The correct options are **2 and 4**.
