# Find the principal values of the following : $\cot^{-1} (\sqrt 3)$

$\begin{array}{1 1} \frac{\pi}{3} \\ \frac{-\pi}{3} \\ \frac{\pi}{6} \\ \frac{-\pi}{6} \end{array}$

## 2 Answers

Toolbox:
• The range of the principle value of $\cot^{-1} x$ is $[0, \pi]$
Let $\cot^{-1 } \sqrt 3 = x \implies \cot(x) = \sqrt{3}$
We know that the range of the principal value of $\cot^{-1} x$ is $[0, \pi]$
Therefore, $\cot(x) = \sqrt{3} = \cot \frac{\pi}{6}$
$\implies x = \frac{\pi}{6},$ where $x \epsilon [0, \pi]$
$\therefore$ The principal value of $\cot^{-1} (\sqrt 3)$ is $\frac{\pi}{6}$
answered Feb 22, 2013
edited Dec 4, 2017

Toolbox:
• The range of the principal value of $\; cot^{-1}x$ is $\left [ 0,\pi \right ]$
Let $cot^{-1}\sqrt 3 = x \Rightarrow cot (x) = \sqrt 3$
We know that the range of the principal value of $\; cot^{-1}x$ is $\left [ 0,\pi \right ]$
Therefore, $cot(x) = \sqrt 3 = cot \frac{\pi}{6}$
$\Rightarrow x=\frac{\pi}{6}$, where $x \;\epsilon\;$ $\left [ 0,\pi \right ]$
Therefore, the principal value of $cot^{-1}(\sqrt 3) is \frac{\pi}{6}$
answered Mar 2, 2013

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