What is the principal value of cot^-1(√(3) - 2)?

Inverse Trigonometric Functions · Question 63 of 120

Question 63

HardOriginal question

What is the principal value of $\cot^{-1}\left(\sqrt{3} - 2\right)$ ?

- \frac{5 \pi}{12}

Correct answerYour answer

\frac{5 \pi}{12}

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\frac{7 \pi}{12}

Correct answerYour answer

\frac{11 \pi}{12}

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Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$\sqrt{3} - 2 = -\left(2 - \sqrt{3}\right)$ , and $2 - \sqrt{3} = \tan \frac{\pi}{12} = \cot \frac{5\pi}{12}$ .
On $(0, \pi)$ , a negative cotangent needs the angle $\pi - \frac{5\pi}{12} = \frac{7\pi}{12}$ .

Worth noting

$-\frac{5\pi}{12}$ has the right cotangent but is negative; $\cot^{-1}$ never returns a negative angle.

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

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