Question 4

Step 1: Evaluate $\cot^{-1}(\cot(-11))$.

The range of $\cot^{-1}(x)$ is $(0, \pi)$. We know $\cot^{-1}(\cot x) = x + n\pi$.

For $x = -11$, we need $-11 + n\pi \in (0, \pi)$.

Taking $n = 4$, we get $4\pi - 11 \approx 4(3.14159) - 11 \approx 12.566 - 11 = 1.566$, which is in $(0, \pi)$.

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

Step 2: Evaluate $10 \sin(2 \cos^{-1}(1/\sqrt{2}))$.

$\cos^{-1}(1/\sqrt{2}) = \pi/4$.

$10 \sin(2 \cdot \pi/4) = 10 \sin(\pi/2) = 10(1) = 10$.

Step 3: Evaluate $10 \sin(2 \tan^{-1}(2))$.

Let $\theta = \tan^{-1}(2) \Rightarrow \tan \theta = 2$.

We know $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2(2)}{1+2^2} = \frac{4}{5}$.

So, $10 \sin(2\theta) = 10 \times \frac{4}{5} = 8$.

Step 4: Sum the values.

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