Browse Questions

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

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

Ans : $sec^{-1}sec\frac{\pi}{6}=\frac{\pi}{6}$

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 $\frac{\pi}{6}$
edited Apr 19, 2016 by meena.p