If $\delta _{1}$ and $\delta _{2}$ are the angles of dip observed in two vertical planes at right angles to each other and $\delta $ is the true value of dip then :
- $tan^{2}\delta =tan^{2}\delta _{1}+tan^{2}\delta _{2}$
- $cot^{2}\delta =cot^{2}\delta _{1}+cot^{2}\delta _{2}$
- $tan^{2}\delta =\dfrac{tan^{2}\delta _{1}+tan^{2}\delta _{2}}{tan^{2}\delta _{1}tan^{2}\delta _{2}}$
- $cot^{2}\delta =1+cot^{2}\delta _{1}cos^{2}\delta _{2}$
The relation between the true dip delta and the apparent dips delta1 and delta2 in two mutually perpendicular vertical planes is given by cot^2(delta) = cot^2(delta1) + cot^2(delta2). This is a standard result in geomagnetism.
The horizontal component of earth's magnetic field is constant, so we can write H equals tan delta for the true dip and H equals tan delta one times cos theta for a plane at an angle theta to the meridian. For two planes at right angles to each other, this gives tan delta one equals tan delta times cos theta and tan delta two equals tan delta times sin theta. Squaring and adding these two equations yields tan squared delta one plus tan squared delta two equals tan squared delta times (cos squared theta plus sin squared theta), which simplifies to tan squared delta. Taking the reciprocal of both sides gives the final identity cot squared delta equals cot squared delta one plus cot squared delta two.