# If cotθ = 2, find

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

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