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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} $

1 Answer

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 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 
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