$\begin{aligned} & \frac{1}{2}<\frac{6}{11}<\frac{1}{\sqrt{2}} \\ & \sin ^{-1}\left(\frac{1}{2}\right)<\sin ^{-1}\left(\frac{6}{11}\right)<\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ & \frac{\pi}{6}<3 \sin ^{-1}\left(\frac{6}{11}\right)<\frac{\pi}{4} \\ & \frac{\pi}{2}<\alpha<\frac{3 \pi}{4} \quad \therefore \cos \alpha<0 \\ & 0<\frac{4}{9}<\frac{1}{2} \\ & \frac{\pi}{3}<\cos ^{-1}\left(\frac{4}{9}\right)<\frac{\pi}{2} \\ & \pi<3 \cos ^{-1}\left(\frac{4}{9}\right)<\frac{3 \pi}{2} \\ & \pi<\beta<\frac{3 \pi}{2}\end{aligned}$
$\begin{aligned} & \text { Now } \frac{3 \pi}{2}<\alpha+\beta<\frac{9 \pi}{4} \\ & \therefore \cos (\alpha+\beta)>0\end{aligned}$