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Find the principal values of the following: \[ sec^{-1} \bigg( \frac {2} {\sqrt 3} \bigg) \]

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

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Ans : \( sec^{-1}sec\frac{\pi}{6}=\frac{\pi}{6} \)
answered Feb 22, 2013 by thanvigandhi_1
 
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Toolbox:
  • The range of the principal value of $\; sec^{-1}x$ is [ 0,$\pi$ ]-{$\frac{\pi}{2}$}
Let $sec^{-1}\frac{2}{\sqrt 3} = x \Rightarrow sec (x) = \frac{2}{\sqrt 3}$
We know that the range of the principal value of $\; sec^{-1}x$ is [ 0,$\pi$ ]-{$\frac{\pi}{2}$}
Therefore, $sec(x) = \frac{2}{\sqrt 3} = sec \frac{\pi}{6}$
$\Rightarrow x=\frac{\pi}{6}$, where $x \;\epsilon\;$ [ 0,$\pi$ ]-{$\frac{\pi}{2}$}
Hence the principal value of $\; sec^{-1}\frac{2}{\sqrt 3}$ is $\frac{\pi}{6}$
answered Mar 2, 2013 by balaji.thirumalai
edited Apr 19 by meena.p
 
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