Explanation

(d) Let $\alpha$ be the angle which one of the planes make with the magnetic meridian. The other plane makes an angle $(90{^\circ} - \alpha)$ with it. The components of $H$ in these planes will be $H\cos\alpha\ $and $H\sin\alpha$ respectively. If $\phi_{1}$ and $\phi_{2}$ are the apparent dips in these two planes, then
$\tan{\phi_{1} = \frac{V}{H\cos\alpha}},\ i.e.,\cos\alpha = \frac{V}{H\tan\phi_{1}}\ \ \ldots(i)$
$\tan\phi_{2} = \frac{V}{H\sin\alpha},\ i.e.,\sin\alpha = \frac{V}{H\tan\phi_{2}}\ \ \ldots(ii)$
Squaring and adding (i) and (ii), we get
$\cos^{2}\alpha + \sin^{2}\alpha = \left( \frac{V}{H} \right)^{2}\left( \frac{1}{\tan^{2}\phi_{1}} + \frac{1}{\tan^{2}\phi_{2}} \right)$
$i.e.,\ 1 = \frac{V^{2}}{H^{2}}\left\lbrack \cot^{2}\phi_{1} + \cot^{2}\phi_{2} \right\rbrack$
or $\frac{H^{2}}{V^{2}} = \cot^{2}\phi_{1} + \cot^{2}\phi_{2},\ i.e.,\cot^{2}\phi = \cot^{2}\phi_{1} + \cot^{2}\phi_{2}$
This is the required result