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
Step-by-Step Solution
Key Concept: The principal range of $\cot^{-1}$ is $(0,\pi)$. To evaluate $\cot^{-1}(\cot(x))$ for $x \notin (0,\pi)$, shift by the unique multiple of $\pi$ that places the argument in $(0,\pi)$.
**Step 1: Evaluate $\cot^{-1}(\cot(-11))$**
Principal range of $\cot^{-1}$ is $(0, \pi)$. We need to bring $-11$ into $(0,\pi)$ by adding multiples of $\pi$. $-11 + 4\pi \approx -11 + 12.566 = 1.566 \in (0,\pi)$. So $\cot^{-1}(\cot(-11)) = -11 + 4\pi$.
**Step 2: Evaluate the second term**
$\cos^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$. So $\sin\!\left(2 \cdot \dfrac{\pi}{4}\right) = \sin\dfrac{\pi}{2} = 1$. Second term $= 10 \times 1 = 10$.
**Step 3: Evaluate the third term**
Let $\theta = \tan^{-1}(2)$, so $\tan\theta = 2$. Using $\sin 2\theta = \dfrac{2\tan\theta}{1+\tan^2\theta} = \dfrac{4}{5}$. Third term $= 10 \times \dfrac{4}{5} = 8$.
**Step 4: Sum all three terms**
$(-11 + 4\pi) + 10 + 8 = 4\pi + 7$.
Correct Answer: C