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

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