As given that,
$$
\begin{aligned}
\mathrm{B}_{\mathrm{V}} &=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m} \cos \theta}{\mathrm{r}^3} \\
\mathrm{~B}_{\mathrm{H}} &=\frac{\mu_0}{4 \pi} \frac{\sin \theta \mathrm{m}}{\mathrm{r}^3}
\end{aligned}
$$


Squaring both the equations and adding, we get
$$
\begin{gathered}
\mathrm{B}^2=\mathrm{B}_{\mathrm{V}}^2+\mathrm{B}_{\mathrm{H}}^2 \\
=\left(\frac{\mu_0}{4 \pi}\right) \frac{\mathrm{m}^2}{\mathrm{r}^6}\left[4 \cos ^2 \theta+\sin ^2 \theta\right] \\
\mathrm{B}=\sqrt{\mathrm{B}_{\mathrm{V}}^2+\mathrm{B}_{\mathrm{H}}^2}=\frac{\mu_0}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^3}\left[3 \cos ^2 \theta+1\right]^{1 / 2} .
\end{gathered}
$$
From Eq. (iii), the value of B is minimum, if $\cos \theta=\frac{\pi}{2}$
$\theta=\frac{\pi}{2}$. So, the magnetic equator is the locus.
(b) For angle of dip $\delta$ is
$$
\begin{aligned}
&\tan \delta=\frac{B_{\mathrm{V}}}{B_H}=\frac{\frac{\mu_0}{4 \pi} \cdot \frac{2 m \cos \theta}{r^3}}{\frac{\mu_0}{4 \pi} \cdot \frac{\sin \theta \cdot m}{r^3}} \\
&=2 \cot \theta \\
&\tan \delta=2 \cot \theta \\
&
\end{aligned}
$$
For dip angle is zero, so, that $\delta=0$,
$$
\cot \theta=0, \theta=\frac{\pi}{2}
$$
So locus is magnetic equator.
(c) $\quad \tan \delta=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}}}$
If, $\left(\delta=\pm 45^{\circ}\right)$
or, $\frac{B_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}}}=\tan \left(\pm 45^{\circ}\right), \frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}}}=\tan 45^{\circ}$
$$
\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}}}=1, \mathrm{~B}_{\mathrm{V}}=\mathrm{B}_{\mathrm{H}}
$$
$2 \cot \theta=1$
From Eq. (iv)
$\cot \theta=\frac{1}{2}$
$\tan \theta=2$
So, $\quad \theta=\tan ^{-1}(2)$
Hence, $\theta=\tan ^{-1}(2)$ is the locus