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

$\sec \left[\arcsin \left(- \frac{\sqrt{3}}{2}\right)\right] = 2$
$\sec \left[\arcsin \left(- \frac{\sqrt{3}}{2}\right)\right]$
The restriction for the range of arcsin is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ 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 $\sec \theta = \frac{1}{\cos} \theta = \frac{1}{\frac{1}{2}} = 2$