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# The principal value of $\tan^{-1}\sqrt 3$ is _______________.

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

Can you answer this question?

Toolbox:
• The range of the principal value of $\; tan^{-1}x$ is $\left (- \large\frac{\pi}{2},\large\frac{\pi}{2} \right )$
Ans : $tan^{-1}tan\large\frac{\pi}{3}=\large\frac{\pi}{3}$

Let $tan^{-1}\sqrt 3 = x \Rightarrow tan (x) = \sqrt 3$

We know that the range of the principal value of $\; tan^{-1}x$ is $\left (- \large\frac{\pi}{2},\large\frac{\pi}{2} \right)$
$\therefore$ $tan(x) = \sqrt 3 = tan \large\frac{\pi}{3}$

$\Rightarrow x=\large\frac{\pi}{3}$, where$x\:\in\: \left (- \large\frac{\pi}{2},\large\frac{\pi}{2} \right )$
Therefore, the principal value of $tan^{-1}(\sqrt 3) is \large\frac{\pi}{3}$

answered Feb 18, 2013
edited Mar 16, 2013