Find the Exact Value sec(arccot((2 square root of 21)/21))

Draw a triangle in the plane with vertices ${\displaystyle (\frac{2\sqrt{21}}{21},1)}$, ${\displaystyle (\frac{2\sqrt{21}}{21},0)}$, and the origin. Then ${\displaystyle \mathrm{arccot}\left(\frac{2\sqrt{21}}{21}\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (\frac{2\sqrt{21}}{21},1)}$. Therefore, ${\displaystyle \sec \left(\mathrm{arccot}\left(\frac{2\sqrt{21}}{21}\right)\right)}$ is ${\displaystyle \frac{5}{2}}$.