If earth is considered as a short magnet with its centre coinciding, with the centre of earth then the angle of dip $\delta$ at a place where the magnetic latitude $\lambda$ is given by

For $Msin\lambda$ , point is axial point. Hence $B_1=\frac{{\mu }_0}{4\pi }\frac{2M\sin \lambda }{R^3}$

 For $MCos\lambda$ , point is equatorial point. Hence $B_2=\frac{{\mu }_0}{4\pi }\frac{M\cos \lambda }{R^3}$

But $B_1,B_2$  represent $B_V,B_H$ at this point where we want the relation

$\therefore$Dip at this point is $Tan\delta =\frac{B_V}{B_H}=\frac{B_1}{B_2}=2\tan \lambda$

$\therefore \tan \delta =2\tan \lambda$