If 9 secθ = 41 . Then find the value of 5cotθ .

9 secθ = 41

secθ = \(\frac{41}{9}\)  (where 41 → H and 9 → B)

P = \(\sqrt {(41)^2 - (9)^2}\)

P = \(\sqrt {1600}\) = 40

5cotθ = \(\frac{B}{P}\) =  5 × \(\frac{9}{40}\) =  \(\frac{9}{8}\)