Considering only the principal values of the inverse trigonometric functions, the value of $\cot^{-1}(\cot(-11)) + 10\sin\left(2\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) + 10\sin\left(2\tan^{-1}(2)\right)$ is

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

$\text{To find } E \text{ using principal values}$

$\text{choose the correct option}$

$1.\ \cot^{-1}(\cot(-11)) \text{ to its principal value}$

$2.\ 10\sin\!\left(2\cos^{-1}\tfrac{1}{\sqrt2}\right)$

$3.\ 10\sin\!\left(2\tan^{-1}2\right)$

$\cot^{-1}: \text{range } (0,\pi)$

$\cot(\theta+\pi)=\cot\theta \Rightarrow \cot(-11)=\cot(-11+k\pi)$

$-11+4\pi \approx 1.566 \in (0,\pi) \quad ✓$

$\cot^{-1}(\cot(-11)) = 4\pi - 11$

$\cos^{-1}\tfrac{1}{\sqrt2}=\tfrac{\pi}{4}\ \Rightarrow\ 2\cos^{-1}\tfrac{1}{\sqrt2}=\tfrac{\pi}{2}$

$10\sin\tfrac{\pi}{2} = 10\cdot 1 = 10$

$\theta=\tan^{-1}2 \Rightarrow \tan\theta=2$

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

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

$10\sin(2\tan^{-1}2) = 10\cdot\tfrac{4}{5} = 8$

$E = (4\pi - 11) + 10 + 8 = 4\pi + 7$

$\boxed{E = 4\pi+7 \ \Rightarrow\ \text{option (3)}}$