If 5 tan θ – 4 = 0, then the value of 
is
Given,
5 tan θ – 4 = 0
So,
5 tan θ = 4
tan θ = 4/5
In ∆ ABC,
AC2 = AB2 + BC2
AC2 = (4)2 + (5)2
AC2 = 16 + 25
AC2 = 41
AC = √41
Therefore,
sin θ = AB/AC = 4/√41
cos θ = BC/AC = 5/√41
Now,
3
If 5 tan θ – 4 = 0, then the value of is
Given,
5 tan θ – 4 = 0
So,
5 tan θ = 4
tan θ = 4/5
In ∆ ABC,
AC2 = AB2 + BC2
AC2 = (4)2 + (5)2
AC2 = 16 + 25
AC2 = 41
AC = √41
Therefore,
sin θ = AB/AC = 4/√41
cos θ = BC/AC = 5/√41
Now,