Question

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

Answer 1

`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`