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What is the exact value of ${sec}^{- 1} (\sqrt{2})$ $sec^-1 (sqrt2)$ ?

1 Answer

Jul 14, 2015

${sec}^{- 1} (\sqrt{2}) = \frac{π}{4}$ $sec^(-1)(sqrt(2)) = pi/4$ or $\frac{5 π}{4}$ $(5pi)/4$ (radians)
$XXXX$ $color(white)("XXXX")$ if restricted to $[0, 2 π)$ $[0,2pi)$

Explanation:

If ${sec}^{- 1} (\sqrt{2}) = θ$ $sec^(-1)(sqrt(2)) = theta$
then
$XXXX$ $color(white)("XXXX")$ $sec (θ) = \sqrt{2}$ $sec(theta) = sqrt(2)$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ by definition of inverse trig function

$\Rightarrow cos (θ) = \frac{1}{\sqrt{2}}$ $rArr cos(theta) = 1/sqrt(2)$

This is a standard trigonometric triangle as pictured below.
$XXXX$ $color(white)("XXXX")$ ( $45^{o} = \frac{π}{4}$ $45^o = pi/4$ radians)

enter image source here

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