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,