Let $$\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)$$ and $$\beta = 3\cos^{-1}\left(\frac{4}{9}\right)$$, where inverse trigonometric functions take only the principal values.
Given below are two statements :
Statement I : $$\cos(\alpha + \beta) > 0$$.
Statement II : $$\cos(\alpha) < 0$$.
In the light of the above statements, choose the correct answer from the options given below :

$$\sin^{-1}\left(\frac{1}{2}\right) < \sin^{-1}\left(\frac{6}{11}\right) < \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$

$$\frac{\pi}{6} < \sin^{-1}\left(\frac{6}{11}\right) < \frac{\pi}{3}$$

$$3 \left(\frac{\pi}{6}\right) < 3 \sin^{-1}\left(\frac{6}{11}\right) < 3 \left(\frac{\pi}{3}\right)$$

$$\frac{\pi}{2} < \alpha < \pi$$

$$\cos(\alpha) < 0$$,Therefore, Statement II is true.

$$\beta = 3 \cos^{-1} \left( \frac{4}{9} \right)$$

$$\cos^{-1}\left(\frac{1}{2}\right) < \cos^{-1}\left(\frac{4}{9}\right) < \cos^{-1}(0)$$

$$\frac{\pi}{3} < \cos^{-1}\left(\frac{4}{9}\right) < \frac{\pi}{2}$$

$$3 \left(\frac{\pi}{3}\right) < 3 \cos^{-1}\left(\frac{4}{9}\right) < 3 \left(\frac{\pi}{2}\right)$$

$$\pi < \beta < \frac{3\pi}{2}$$

$$\left( \frac{\pi}{2} + \pi \right) < \alpha + \beta < \left( \pi + \frac{3\pi}{2} \right)$$

$$\frac{3\pi}{2} < \alpha + \beta < \frac{5\pi}{2}$$

$$\cos(\alpha + \beta) > 0$$

Therefore, Statement I is true.