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

\(\alpha=3 \sin ^{-1} \frac{6}{11}\) and \(\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)\)
Since \(\frac{6}{11}>\frac{6}{12}\)
Taking \(\sin ^{-1}\) on both sides, we get
\(
\sin ^{-1}\left(\frac{6}{11}\right)>\sin ^{-1}\left(\frac{6}{12}\right)
\)
\(\ldots\left[\because \sin ^{-1} x\right.\) is an increasing function \(]\)
\(\Rightarrow 3 \sin ^{-1}\left(\frac{6}{11}\right)>3 \sin ^{-1}\left(\frac{1}{2}\right)\)
\(\Rightarrow \alpha>3\left(\frac{\pi}{6}\right)\)
\(\Rightarrow \alpha>\frac{\pi}{2}\)
Now, \(\frac{4}{9} < \frac{4}{8}\)
Taking \(\cos ^{-1}\) on both sides, we get \(\cos ^{-1}\left(\frac{4}{9}\right)>\cos ^{-1}\left(\frac{4}{8}\right)\)
\(\ldots\left[\because \cos ^{-1} x\right.\) is a decreasing function \(]\)
\(\Rightarrow 3 \cos ^{-1}\left(\frac{4}{9}\right)>3 \cos ^{-1}\left(\frac{1}{2}\right)\)
\(\Rightarrow \beta>3\left(\frac{\pi}{3}\right)\)
\(\Rightarrow \beta>\pi\)
From (i) and (ii), we have \(\alpha\) lies in II \(^{\text {nd }}\) quadrant and \(\beta\) lies in III \(^{\text {rd }}\) quadrant.
\(
\therefore \quad \cos \alpha < 0, \cos \beta < 0 \text { and } \sin \beta < 0
\)

Also, \(\alpha+\beta>\frac{\pi}{2}+\pi\)
....[From (i) and (ii)]
\(
\therefore \quad \alpha+\beta>\frac{3 \pi}{2}
\)

Thus, \(\alpha+\beta\) lies in IV \(^{\text {th }}\) quadrant.
So, \(\cos (\alpha+\beta)>0\)
[Note: Options (B), (C) and (D) are correct.]