Let $\alpha = 3\sin^{-1}\left(\dfrac{6}{11}\right)$ and $\beta = 3\cos^{-1}\left(\dfrac{4}{9}\right)$, 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:

We are given $\alpha = 3\sin^{-1}\left(\dfrac{6}{11}\right)$ and $\beta = 3\cos^{-1}\left(\dfrac{4}{9}\right)$.

First, we find the range of $\alpha$.
Since $\dfrac{1}{2} < \dfrac{6}{11} < \dfrac{1}{\sqrt{2}}$, we have $\sin^{-1}\left(\dfrac{1}{2}\right) < \sin^{-1}\left(\dfrac{6}{11}\right) < \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$.
$\Rightarrow \dfrac{\pi}{6} < \sin^{-1}\left(\dfrac{6}{11}\right) < \dfrac{\pi}{4}$
Multiplying by $3$, we get $\dfrac{\pi}{2} < \alpha < \dfrac{3\pi}{4}$.
Since $\alpha$ lies in the second quadrant, $\cos(\alpha) < 0$. Thus, Statement II is true.

Next, we find the range of $\beta$.
Since $0 < \dfrac{4}{9} < \dfrac{1}{2}$ and $\cos^{-1}(x)$ is a decreasing function, we have $\cos^{-1}\left(\dfrac{1}{2}\right) < \cos^{-1}\left(\dfrac{4}{9}\right) < \cos^{-1}(0)$.
$\Rightarrow \dfrac{\pi}{3} < \cos^{-1}\left(\dfrac{4}{9}\right) < \dfrac{\pi}{2}$
Multiplying by $3$, we get $\pi < \beta < \dfrac{3\pi}{2}$.
Thus, $\beta$ lies in the third quadrant.

Now, we determine the sign of $\cos(\alpha+\beta)$.
Adding the inequalities for $\alpha$ and $\beta$:
$\dfrac{\pi}{2} + \pi < \alpha + \beta < \dfrac{3\pi}{4} + \dfrac{3\pi}{2}$
$\Rightarrow \dfrac{3\pi}{2} < \alpha + \beta < \dfrac{9\pi}{4}$
The angle $(\alpha + \beta)$ lies in the interval $\left(\dfrac{3\pi}{2}, \dfrac{9\pi}{4}\right)$, which covers the fourth quadrant and a part of the first quadrant. In both of these regions, the cosine function is strictly positive.
Therefore, $\cos(\alpha+\beta) > 0$. Thus, Statement I is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true