allery$\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) = $
A$\tfrac{\pi}{3}$
B$-\tfrac{\pi}{3}$
C$\tfrac{2\pi}{3}$
D$\tfrac{5\pi}{6}$
Answer & Solution
Correct answer: C. $\tfrac{2\pi}{3}$
$\cot^{-1}$ has principal range $(0, \pi)$.
At $\theta = \tfrac{2\pi}{3}$, $\cot\theta = \tfrac{\cos(2\pi/3)}{\sin(2\pi/3)} = \tfrac{-1/2}{\sqrt{3}/2} = -\tfrac{1}{\sqrt{3}}$.
**Option D ($5\pi/6$) yields $\cot = -\sqrt{3}$, not $-1/\sqrt{3}$ — common trap.**
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