First term: The principal range of ${\cot }^{-1}$ is $(0,\pi )$. We need $4\pi -11\in (0,\pi )$, which is true since $11\approx 3.5\pi$ lies between $3\pi$ and $4\pi$. Using $\cot (-11)=\cot (4\pi -11)$, we get ${\cot }^{-1}(\cot (-11))=4\pi -11$.

Second term: ${\cos }^{-1}(1/\sqrt{2})=\pi /4$, so $10\sin (2\cdot \pi /4)=10\sin (\pi /2)=10$.

Third term: Let $\theta ={\tan }^{-1}(2)$ with $\theta \in (0,\pi /2)$ and $\tan \theta =2$. Then

$\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{4}{5}$

So the third term equals $10\cdot \frac{4}{5}=8$.

Total: $(4\pi -11)+10+8=4\pi +7$.

$\therefore$ The correct option is (C).