Given,

Velocity of boy, $\overrightarrow{v_b}=4 \mathrm{m} {\mathrm{s}}^{-1}$ and

Velocity of rain, $\overrightarrow{v_r}=8 \mathrm{m} {\mathrm{s}}^{-1}$.

Velocity of rain with respect to boy is $\overrightarrow{v_{rb}}=\overrightarrow{v_r}-\overrightarrow{v_b}$..

When the boy runs from east to west:

$\Rightarrow \tan \left(\theta \right)=\frac{v_r\sin \left(45°\right)+v_b}{v_r\cos \left(45°\right)}$

$\Rightarrow \tan \left(\theta \right)=\frac{\left({\displaystyle \frac{8}{\sqrt{2}}}+4\right)}{\left(\frac{8}{\sqrt{2}}\right)}=\frac{\sqrt{2}+1}{\sqrt{2}} ...\left(\mathrm{i}\right)$

When the boy runs from east to west:

$\Rightarrow \tan \left(\alpha \right)=\frac{v_r\sin \left(45°\right)-v_b}{v_r\cos \left(45°\right)}$

$\Rightarrow \tan \left(\alpha \right)=\frac{\left({\displaystyle \frac{8}{\sqrt{2}}-4}\right)}{\left(\frac{8}{\sqrt{2}}\right)}=\frac{\sqrt{2}-1}{\sqrt{2}} ...\left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$,

$\Rightarrow \frac{\tan \left(\theta \right)}{\tan \left(\alpha \right)}=\frac{{\displaystyle \sqrt{2}+1}}{\sqrt{2}{\displaystyle -}1}=\frac{{\displaystyle {\left(\sqrt{2}+1\right)}^2}}{{\displaystyle {\left(\sqrt{2}\right)}^2}{\displaystyle -}{\displaystyle {\left(1\right)}^2}}$,

$\therefore \frac{\tan \left(\theta \right)}{\tan \left(\alpha \right)}={\left(\sqrt{2}+1\right)}^2$.