If earth is considered as a short magnet with its centre coinciding with the centre of earth. Then the angle of  dip 𝛅 at a place where the magnetic latitude  $\text{'}\mathrm{\lambda }\text{'}$ is given by:
 

For  $\mathrm{Msin\lambda }$, point P is at axial point 
Hence,  ${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{Msin\lambda }}{{\mathrm{R}}^3}$
For  $\mathrm{Mcos\lambda }$ point P is equatorial point
        ${\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{Mcos\lambda }}{{\mathrm{R}}^3}$
Here  $B_1, B_2$  represents  $B_V,B_H$ at the point ‘P’
Dip at the point ‘P’
$\mathrm{tan\delta }=\frac{{\mathrm{B}}_{\mathrm{V}}}{\mathrm{B}{}_{\mathrm{H}} }=\frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}=2\mathrm{tan\lambda }$
 $\Rightarrow \mathrm{tan\delta }=2\mathrm{tan\lambda }$