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 : $\quad \cos (\alpha+\beta)>0$.

Statement II : $\quad \cos (\alpha)<0$.

In the light of the above statements, choose the correct answer from the options given below :

If $\sin \left(\tan ^{-1}(x \sqrt{2})\right)=\cot \left(\sin ^{-1} \sqrt{1-x^2}\right), x \in(0,1)$, then the value of $x$ is:

Let $0<\alpha<1, \beta=\frac{1}{3 \alpha}$ and $\tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)=\frac{\pi}{4}$. Then $6(\alpha+\beta)$ is equal to:

Considering the principal values of inverse trigonometric functions, the value of the expression

$$\tan \left( 2 \sin^{-1}\left( \frac{2}{\sqrt{13}} \right) - 2 \cos^{-1}\left( \frac{3}{\sqrt{10}} \right) \right)$$

is equal to :