Derive the relation ..
tan delta= 2tan lamda, where delta is the angle of dip and lamda is the magnetic latitude
The derivation is quite involved and complex so we will attempt to make it short and meaningful...
the potential due to a dipole is given as
U = (M / 4πr2)cosθ
here M is the magnetic dipole moment and θ is the colatitude angle
now, the magnetic field induction will be given as the gradient of U (three dimensional derivative)
B = -μ0ΔU
this will give us the two components of the field B
Br = -Z = (μ0M / 2πr3)cosθm
Bθ = -H = (μ0M / 4πr3)sinθm
now, if B0 = (μ0M / 4πr3)
Br = 2B0cosθm
Bθ = B0sinθm
we define an angle of inclination which is the angle made between the magnetic field (vertical) and the dip or horizontal component.
tani = Br / Bθ
now, for a dipole
tani = cotθm
tani = 2tanλ