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

Inverse Trigonometric Functions · Question 47 of 120

Question 47

MediumOriginal question

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

- \frac{\pi}{6}

Correct answerYour answer

\frac{\pi}{6}

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\frac{2 \pi}{3}

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\frac{5 \pi}{6}

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Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

Find the angle $\theta$ in the branch $(0, \pi)$ with $\cot \theta = - \sqrt{3}$ .
That angle is $\frac{5 \pi}{6}$ .

Worth noting

$\cot^{-1}$ lives in $(0, \pi)$ , so a negative input gives an obtuse angle, $\pi - \frac{\pi}{6}$ , not $-\frac{\pi}{6}$ .

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

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