# What is the exact value of sec^-1 (sqrt2)?

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#### Explanation

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Alan P. Share
Jul 14, 2015

${\sec}^{- 1} \left(\sqrt{2}\right) = \frac{\pi}{4}$ or $\frac{5 \pi}{4}$ (radians)
$\textcolor{w h i t e}{\text{XXXX}}$if restricted to $\left[0 , 2 \pi\right)$

#### Explanation:

If ${\sec}^{- 1} \left(\sqrt{2}\right) = \theta$
then
$\textcolor{w h i t e}{\text{XXXX}}$$\sec \left(\theta\right) = \sqrt{2}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$by definition of inverse trig function

$\Rightarrow \cos \left(\theta\right) = \frac{1}{\sqrt{2}}$

This is a standard trigonometric triangle as pictured below.
$\textcolor{w h i t e}{\text{XXXX}}$(${45}^{o} = \frac{\pi}{4}$ radians)

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