If 3cos θ = 5 sin θ, then the value of is

Given,


3 cos θ = 5 sin θ




In ∆ABC,


AC2 = AB2 + BC2


AC2 = (3)2 + (5)2


AC2 = 9 + 25 = 34


AC = √34


Therefore,


sin θ = 3/√34


cos θ = 5/√34


sec θ = √34/5


Now,






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