We have, cotθ = 2k/k = base/perpendicular (For some value of k)

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

∴AB² = BC² + AC²

AB² = (k)² + (2k)²

AB² = k² + 4k²

AB² = 5k² = (k√5)²

→ AB = k√5

Hence, the trignometeric ratios for the given θ are:

sinθ = BC/AB = k/(k√5) = 1/√5

cosθ = AC/AB = (2k)/(k√5) = 2/√5

tanθ = BC/AC = sinθ /cosθ = k/(2k) = 1/2

cotθ = 2

cosecθ = AB/BC = 1/sinθ = √5

secθ = AB/AC = 1/cosθ = √5/2