a
Let \(B_H\) and \(B_v\) be the horizontal and vertical components of earth's magnetic field \(\vec B\). Since \(\theta\) is the angle of dip

\(\tan \theta = \frac{B_V}{B_H}\)

 or \(cot\theta = \frac{B_H}{B_V}\)

Suppose planes \(1\) and \(2\) are two mutually perpendicular planes and respectively make angles \(\theta\) and \(90°- \theta\) 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

\(B_{1}=B_{H} \cos \theta \text { and } B_{2}=B_{H} \sin \theta\)

The angles of \(\operatorname{dip} \theta_{1}\) and \(\theta_{2}\) in the two planes are given by

\(\tan \theta_{1}=\frac{B_{V}}{B_{1}}\)

\(\tan \theta_{1}=\frac{B_{V}}{B_{H} \cos \theta}\)

or \(\quad \cot \theta_{1}=\frac{B_{H} \sin \theta}{B_{V}}.........(ii)\)

\(\text { Similarly, } \cot \theta_{2}=\frac{B_{H} \sin \theta}{B_{V}}.........(iii)\)

From eqns. \((ii)\) and \((iii)\)

\(\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\frac{B_{H}^{2}}{B_{V}^{2}}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{B_{H}^{2}}{B_{V}^{2}}\)

\(\therefore \quad \cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\cot ^{2} \theta \quad[\text { from eqn. (i) }]\)