We have, cosecθ = (k√10)/k = 1/sinθ (For some value of k)

→ sinθ = k/(k√10) = perpendicular/hypotenuse

By Pythagoras theorem, (hypotenuse)² = (perpendicular)² + (base)²

∴AB² = BC² + AC²

(√10k)² = (k)² + AC²

AC² = 10k² - k²

AC² = 9k² = (3k)²

→ AC = 3k

Hence, the trignometeric ratios for the given θ are:

sinθ = 1/√10

cosθ = AC/AB = (3k)/(k√10) = 3/√10

tanθ = BC/AC = sinθ /cosθ = 1/3

cotθ = AC/BC = 1/tanθ = 3

cosecθ = √10

secθ = AB/AC = 1/cosθ = √10/3