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

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Alan P. Share
Jul 14, 2015

Answer:

#sec^(-1)(sqrt(2)) = pi/4# or #(5pi)/4# (radians)
#color(white)("XXXX")#if restricted to #[0,2pi)#

Explanation:

If #sec^(-1)(sqrt(2)) = theta#
then
#color(white)("XXXX")##sec(theta) = sqrt(2)#
#color(white)("XXXX")##color(white)("XXXX")#by definition of inverse trig function

#rArr cos(theta) = 1/sqrt(2)#

This is a standard trigonometric triangle as pictured below.
#color(white)("XXXX")#(#45^o = pi/4# radians)

enter image source here

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