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

$\cot^{-1}(\cot(-11)) + 10 \sin\left(2 \cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)\right) + 10 \sin(2 \tan^{-1}(2))$

is

For the first term, $\cot^{-1}(\cot(-11))$:
The principal value branch of $\cot^{-1}(x)$ is $(0, \pi)$.
Since the cotangent function is periodic with period $\pi$, $\cot(-11) = \cot(-11 + 4\pi)$.
Using $\pi \approx 3.14$, we have $4\pi \approx 12.56$, so $-11 + 4\pi \approx 1.56$, which lies in the interval $(0, \pi)$.
Thus, $\cot^{-1}(\cot(-11)) = 4\pi - 11$.

For the second term, $10 \sin\left(2 \cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)\right)$:
Since $\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$, we get:
$10 \sin\left(2 \times \dfrac{\pi}{4}\right) = 10 \sin\left(\dfrac{\pi}{2}\right) = 10(1) = 10$.

For the third term, $10 \sin(2 \tan^{-1}(2))$:
Let $\theta = \tan^{-1}(2) \Rightarrow \tan(\theta) = 2$.
Using the multiple angle formula $\sin(2\theta) = \dfrac{2\tan(\theta)}{1 + \tan^2(\theta)}$, we obtain:
$\sin(2\theta) = \dfrac{2(2)}{1 + 2^2} = \dfrac{4}{5}$.
Therefore, $10 \sin(2 \tan^{-1}(2)) = 10 \times \dfrac{4}{5} = 8$.

Adding the three evaluated terms together:
$(4\pi - 11) + 10 + 8 = 4\pi + 7$.

Answer: $4\pi + 7$