What is the principal value of cot^-1(√(3))?

Inverse Trigonometric Functions · Question 40 of 120

Question 40

EasyOriginal question

What is the principal value of $\cot^{-1}\left(\sqrt{3}\right)$ ?

- \frac{\pi}{6}

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\frac{\pi}{6}

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\frac{\pi}{3}

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\frac{7 \pi}{6}

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Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Find the angle $\theta$ in the branch $(0, \pi)$ with $\cot \theta = \sqrt{3}$ .
That angle is $\frac{\pi}{6}$ .

Worth noting

$\frac{\pi}{3}$ has cotangent $\frac{1}{\sqrt{3}}$ ; $\sqrt{3}$ belongs to $\frac{\pi}{6}$ .

Ten questions from Inverse Trigonometric Functions, at your own pace.

Ten questions from Inverse Trigonometric Functions, at your own pace.