Answer :

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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