Find the Exact Value csc(arcsec(( square root of 6)/2))

Draw a triangle in the plane with vertices ${\displaystyle (1,\sqrt{{\left(\frac{\sqrt{6}}{2}\right)}^2-1^2})}$, ${\displaystyle (1,0)}$, and the origin. Then ${\displaystyle \mathrm{arcsec}\left(\frac{\sqrt{6}}{2}\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (1,\sqrt{{\left(\frac{\sqrt{6}}{2}\right)}^2-1^2})}$. Therefore, ${\displaystyle \csc \left(\mathrm{arcsec}\left(\frac{\sqrt{6}}{2}\right)\right)}$ is ${\displaystyle \sqrt{3}}$.