The advantage of this particular definition is that differences in pseudorapidity are invariant under boosts along the $z$ axis. Specifically, consider a Lorentz transformation corresponding to a boost by velocity $\beta \hat{z}$. Since the tangent is basically a transverse distance over a longitudinal distance, it transforms under a Lorentz boost along the longitudinal axis with a factor $\gamma $:

$\mathrm{tan}\varphi \sim \frac{\mathrm{\Delta}{x}_{T}}{\mathrm{\Delta}{x}_{L}}\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}\mathrm{tan}{\varphi}^{\prime}=\frac{\mathrm{\Delta}{x}_{T}}{\mathrm{\Delta}{x}_{L}/\gamma}=\gamma \mathrm{tan}\varphi $

and if you put that in the pseudorapidity formula, you find

$\eta =-\mathrm{log}{\textstyle (}\mathrm{tan}\frac{\theta}{2}{\textstyle )}\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}{\eta}^{\prime}=-\mathrm{log}{\textstyle (}\mathrm{tan}\frac{\theta}{2}{\textstyle )}-\mathrm{log}\gamma $

The first term is a function of the particle's trajectory, but the second term is a function of the boost parameter only - it doesn't depend on the particle at all. So if you have two particles coming out of a collision with pseudorapidities ${\eta}_{1}$ and ${\eta}_{2}$, the $\mathrm{log}\gamma $ term is the same for both, and thus when you take the difference it cancels out:

${\eta}_{1}-{\eta}_{2}={\eta}_{1}^{\prime}-{\eta}_{2}^{\prime}$

The reason why it's so important to keep the pseudorapidity difference invariant is that in particle physics, people like to make plots called "lego plots" which show the directional distribution of the particles detected after a collision. When you do this, you could plot the particle detections vs. the polar and azimuthal angles $\theta $ and $\varphi $. But if you use pseudorapidity instead of the angle $\theta $, the fact that $\mathrm{\Delta}\eta $ is invariant means that you can perform a Lorentz boost on your data by just translating the whole graph along the $\eta $ axis.

This is useful because in hadron collisions, it's often the case that one of the individual quarks or gluons involved in the collision may have a lot more momentum than the other, so all the particles produced come out near one end of the detector. But by translating the graph appropriately, you can effectively shift to the center-of-mass frame of the colliding quarks or gluons, where the particles come out symmetrically distributed, and it's a lot easier to analyze.

By the way, the reason it's $\mathrm{tan}\frac{\theta}{2}$ rather than just $\mathrm{tan}\theta $ is that we'd like a jet which comes out of the collision at $\theta \approx \frac{\pi}{2}$, where the resolution is best, to have the same shape in the lego plot as it does in physical space. In particular, a circular jet should appear circular on the graph. This requires that the two coordinates be scaled the same way. If we were using $(\theta ,\varphi )$ as the directional coordinates, this wouldn't be a problem, since both are measured in radians, so we just need to choose a scaling for $\eta $ such that a small increment in $\theta $ near $\frac{\pi}{2}$ corresponds to the same numerical increment in $\eta $:

$|}\frac{\mathrm{d}\eta}{\mathrm{d}\theta}{\textstyle |}{\textstyle (}\theta =\frac{\pi}{2}{\textstyle )}=1$

The argument in the first part of the post (along with some common-sense conditions) basically requires that psuedorapidity be defined as

$\eta =-\mathrm{log}(\mathrm{tan}a\theta )$

for some constant $a$. Plugging into the derivative gives

$|}\frac{\mathrm{d}\eta}{\mathrm{d}\theta}{{\textstyle |}}_{\frac{\pi}{2}}={\textstyle |}\frac{a\phantom{\rule{thinmathspace}{0ex}}{\mathrm{sec}}^{2}(a\theta )}{\mathrm{tan}(a\theta )}{{\textstyle |}}_{\frac{\pi}{2}}={\textstyle |}\frac{a}{\mathrm{sin}(a\theta )\mathrm{cos}(a\theta )}{{\textstyle |}}_{\frac{\pi}{2}}=\frac{2a}{\mathrm{sin}(2a\pi /2)}=1$

This is a transcendental equation, so you can't solve it analytically, but with a little mathematical reasoning it's not hard to convince yourself that $a=\pm \frac{1}{2}$ are the only nonzero solutions. Having a negative argument to a logarithm brings in an extra imaginary term, though it would cancel out anyway, so we might as well choose the positive one.

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