How do you find the exact value of #sec^-1(sqrt2)#?

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Answer 1 · 2016-09-01T18:55:17

45degrees or #pi/4#

Draw a. Right angled isosceles triangle . The ratio of the sides 1:1:#sqrt2#
Sec is hypotenuse over adjacent

Answer 2 · 2016-09-01T18:57:34

#sec^-1 sqrt2=pi/4#.

The Defn. of #sec^-1# is :

#sec^-1 x=theta , |x|.=1 iff sec theta=x, theta in [0,pi]-{pi/2}#.

Knowing that, #sec (pi/4)= sqrt2, pi/4 in [0,2pi]-{pi/2}, sec^-1 sqrt2=pi/4#.