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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