Find the Exact Value sin(arcsec(u))

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