If θ1 and θ2be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by

Let BH and  BV  be the horizonal and vertical components of earth's magnetic field β→. Since θ is the angle of dip∴tanθ=BVBH  or  cotθ=BHBV… (i) Suppose planes 1 and 2 are two mutually perpendicular planes and respectively make angles θ and 90°-θ with the magnetic meridian. The vertical components of earth's magnetic field remain same in the two planes but the effective horizontal components in the planes will be B1=BHcosθ  and B2=BHsinθ  The angles of dipθ1 and θ2 in the two planes are given by tanθ1=BVB1tanθ1=BVBHcosθ or cotθ1=BHcosθBV                              . . . (ii) Similarly, cotθ2=BHsinθBV                                     . . . .(iii)  From eqns. (ii) and (iii)  cot2θ1+cot2θ2=BH2BV2cos2θ+sin2θ=BH2BV2 ∴cot2θ1+cot2θ2=cot2θ                               [ from eqn. (i)]