How do you find the value of $sec^-1[(2sqrt 3)/3]$ ?

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Answer 1 · 2015-11-29T22:49:28

Find arcsec ((2sqrt3)/3)

Ans: $pi/6 and (5pi)/6$

sec x = 1/(cos x). First, find cos x
$cos x = 3/(2sqrt3) = sqrt3/2$ --> arc $x = +-pi/6$
Ans: $x = pi/6$ and $x = (5pi)/6$

How do you find the value of sec^-1[(2sqrt 3)/3]sec^-1[(2sqrt 3)/3]?