How do you find the exact value of #sec[arcsin(-sqrt3/2)] #?

1 Answer
Bdub
Oct 13, 2016

#sec[arcsin(-sqrt3/2)]=2#

Explanation:

#sec[arcsin(-sqrt3/2)]#

The restriction for the range of arcsin is #[-pi/2,pi/2]# since the argument is negative it means that our triangle is in quadrant four with the opposite side length of# -sqrt 3# , hypotenuse of 2, and therefore the adjacent is 1.

Then the ratio for secant of # theta# from our triangle is #sec theta=1/cos theta=1/(1/2) = 2#

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