We have, = perpendicular/hypotenuse (For some value of k)

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

∴AB² = BC² + AC²

(2k)² = (k√3)² + AC²

4k² = 3k² + AC²

AC² = (4 - 3)k²

AC² = k²

→ AC = k, for some number k

Hence, the trigonometric ratios for the given θ are:

sinθ =

cosθ = AC/AB = k/(2k) = 1/2

tanθ = BC/AC = sinθ /cosθ = (k√3)/k = √k

cotθ = 1/tanθ = AC/BC = k/(k√3) = 1/√3

cosecθ = 1/sinθ = AB/BC = (2k)/(k√3) = 2/√3

secθ = 1/cosθ = AB/AC = (2k)/k = 2