If eland ${\theta }_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\theta$ is given by [JIPMER 2017]

Let the B_H and B_V be the horizontal and vertical components of earth's magnetic field B

 $\tan \theta =\frac{B_V}{B_H}\Rightarrow \cot \theta =\frac{B_H}{B_V}$        …(i)

Let plane's 1 and 2 be mutually perpendicular planes making angles a and (90° $-$ $\theta$) with magnetic meridian. The vertical component of earth's magnetic field remain same in two  planes but effective horizontal components in the two planes is given by

$B_1=B_H\cos \theta$                …(ii)

and  $B_2=B_H\sin \theta$             …(iii)

Then,  $\tan {\theta }_1=\frac{B_V}{B_1}=\frac{B_V}{B_H\cos \theta }$

$\cot {\theta }_1=\frac{B_H\cos \theta }{B_V}$              (iv)

Similarly,  $\tan {\theta }_2=\frac{B_V}{B_2}=\frac{B_V}{B_H\sin \theta }$

$\Rightarrow \cot {\theta }_2=\frac{B_H\sin \theta }{B_V}$        (v)

From Eqs. (iv) and (v), we get

$\Rightarrow {\cot }^2{\theta }_1+{\cot }^2{\theta }_2=\frac{B_H^2{\cos }^2\theta }{B_V^2}+\frac{B_H^2{\sin }^2\theta }{B_V^2}$

$\Rightarrow {\cot }^2{\theta }_1+{\cot }^2{\theta }_2=\frac{B_H^2}{B_V^2}\left({\cos }^2\theta +{\sin }^2\theta \right)$

$\Rightarrow {\cot }^2{\theta }_1+{\cot }^2{\theta }_2={\cot }^2\theta$   [from Eq. (i)]