If $\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3\cos^{-1}\left(\frac{4}{9}\right)$, where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)

If α = 3sin⁻¹(6/11) and β = 3cos⁻¹(4/9), where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are) $\alpha = 3\sin^{-1}\!\left(\frac{6}{11}\right), \quad \beta = 3\cos^{-1}\!\left(\frac{4}{9}\right)$ $\text{inverse functions take principal values}$ $\text{Check: } \cos\beta,\ \sin\beta,\ \cos(\alpha+\beta),\ \cos\alpha$ $A = \sin^{-1}\!\frac{6}{11} \in \left(0, \frac{\pi}{2}\right)$ $B = \cos^{-1}\!\frac{4}{9} \in \left(0, \frac{\pi}{2}\right)$ $A \in \left(\frac{\pi}{6}, \frac{\pi}{4}\right) \Rightarrow \alpha = 3A \in \left(\frac{\pi}{2}, \frac{3\pi}{4}\right)$ $\cos\alpha < 0 \ \Rightarrow \ (4) \text{ is true}$ $B \in \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \Rightarrow \beta = 3B \in \left(\pi, \frac{3\pi}{2}\right)$ $\cos\beta < 0, \quad \sin\beta < 0$

$\Rightarrow \ (1) \text{ false}, \ (2) \text{ true}$

$\cos A = \sqrt{1-\frac{36}{121}} = \frac{\sqrt{85}}{11}$ $\cos\alpha = 4\cos^3 A - 3\cos A = -\frac{23\sqrt{85}}{1331} < 0$ $\sin\alpha = 3\sin A - 4\sin^3 A = \frac{1314}{1331} > 0$ $\cos\beta = 4\cos^3 B - 3\cos B = -\frac{716}{729} < 0$ $\sin\beta = 3\sin B - 4\sin^3 B = -\frac{17\sqrt{65}}{729} < 0$ $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ $\cos\alpha\cos\beta > 0, \quad -\sin\alpha\sin\beta > 0 \Rightarrow \cos(\alpha+\beta) > 0$ $\text{Correct options: } (2), (3), (4)$