The law of tangents still has its uses! But most modern school syllabuses seem to have ditched it, solving non-right-angled triangles using the law of sines and law of cosines instead, so your question is a fair one. There's a misconception here:
I was thinking about it and anything I could think about using the law of tangents for could just be done using the law of sines much faster and easier.
If you know two sides and the included angle, i.e. you have "side-angle-side", the law of tangents immediately finds the other two angles (see below) but the law of sines is unhelpful as you do not know a side and its opposite angle. In fact the law of cosines would be the alternative to the law of tangents here. But here's some general advice: try to avoid using the law of sines to find angles, or at least think twice before attempting it. It's often people's go-to formula because it's nice and simple. But the $\sin^{-1}$ button on your calculator, or ASIN function on your spreadsheet or programming language, only ever return an acute result. You may need an obtuse angle, in which case you must subtract the given answer from two right angles. Sometimes the data given in the question is ambiguous and both acute and obtuse cases are possible. Other times you know your result needs to be acute because e.g. you already know there's another angle which is obtuse, and a triangle can't have two obtuse angles. The advantage of working with the law of cosines to find angles is that ACOS always returns the correct angle, whether acute or obtuse. This advantage also holds for the law of tangents: ATAN will always return the correct angle half-sum or half-difference. So this is already one win for the law of tangents! Yes, if all you know is one angle and the length of its opposite side, and the length of one other side, then use the law of sines to find the angle opposite the other known side: but you still need to think whether you want the acute or obtuse result, or even both.
And here's the kicker, the LHS can be reduced down to an alternative law of sines $\frac{\sin A-\sin B}{\sin A+\sin B}$ so how could the law of tangents be useful?
When would $\frac{\sin A-\sin B}{\sin A+\sin B}$ be more useful than $\frac{\tan\frac{1}{2}(A-B)}{\tan\frac{1}{2}(A+B)}$? If I knew both angles, I could substitute into either formula without difficulty. If I knew only one angle, I might prefer the formula with the sines as we could manipulate the equation to make the unknown sine the subject. If I knew neither angle, there are two unknowns, so neither formula seems helpful at first sight. But what if we knew either the sum or difference of the angles? Go back to the case where we'd been told angle $C$: then we would prefer the tangents to the sines, since we could easily work out $\frac{1}{2}(A+B)$ using the angle sum of the triangle, and treat $\frac{1}{2}(A-B)$ as our unknown. The ease of the law of tangents really shines through when we think about half-sums and half-differences as our variables, rather than $A$ and $B$ directly, due to the straightforward relationship between them.
Given the half-sum and half-difference of two quantities, the larger one is the half-sum plus the half-difference and the smaller one is the half-sum subtract the half-difference. Algebraically, assuming $A > B$, this is $A = \frac{1}{2}(A+B) + \frac{1}{2}(A-B)$ and $B = \frac{1}{2}(A+B) - \frac{1}{2}(A-B)$. Writing the half-sum (i.e. mean) angle as $M = \frac{1}{2}(A+B)$ and half-difference as $D = \frac{1}{2}(A-B)$, we have $A = M + D$ and $B = M - D$.
Some textbooks (including the one I learned from!) write the law of tangents with the lengths given as half-sum and half-difference too:
$$\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(A+B)} = \frac{\frac{1}{2}(a-b)}{\frac{1}{2}(a+b)}$$
Obviously the halves on the right-hand side can be cancelled, but they serve a conceptual purpose. If we similarly define the half-sum (mean) length as $m = \frac{1}{2}(a+b)$ and the half-difference length as $d = \frac{1}{2}(a-b)$, we have $a = m + d$ and $b = m - d$ (note that $A > B$ implies $a > b$) and we can write the law of tangents more succinctly as
$$\frac{\tan D}{\tan M} = \frac{d}{m} $$
Or if you have an incurable addiction to the "easy" law of sines, you may prefer:
$$\frac{\tan D}{d} = \frac{\tan M}{m}$$
Thinking about things this way should clarify four situations which can be dealt with efficiently by the law of tangents: if we know two lengths and either the sum or difference of their opposite angles, or know all the angles and either the sum or difference of two lengths.
Know two sides and the included angle ("side-angle-side")
We could use any other information that lets you work out the sum of the two unknown angles $A$ and $B$ opposite our known sides $a$ and $b$, but knowing the included angle $C$ is the most common scenario. Just use the angle sum of a triangle to figure out $A+B$. Calculate $M$, $m$ and $d$. Find the angle half-difference by
$$D = \tan^{-1} \left( \frac{d}{m} \tan M \right)$$
then $A = M + D$ and $B = M - D$.
E.g. given $a = 26$, $b = 22$ and enclosed angle $C = 100^\circ$ we must have $A + B = 80^\circ$ and so the angle half-sum is $M = \frac{1}{2}(A + B) = 80/2 = 40^\circ$, the length half-sum is $m = \frac{1}{2}(a+b) = 24$ and the length half-difference is $d = \frac{1}{2}(26-22) = 2$. (We could also just use the length sum and difference in $\frac{a-b}{a+b}$ without halving, but my point here is to reconceptualise the lengths as $24 \pm 2$, and we'll shortly think about the angles in the same way.) The law of tangents gives the angle half-sum as $D = \tan^{-1} (2/24 \times \tan 40^\circ) \approx 4.00^\circ$ so we can think of our angles like $40^\circ \pm 4^\circ$: we have $A = M + D = 40^\circ + 4.00^\circ \approx 44.00^\circ$ and $B = M - D = 40^\circ - 4.00^\circ \approx 36.00^\circ$
There are other ways to solve this problem without using the law of tangents: see Formula for angle in triangle when two sides and the included angle are known. Since you only know one angle but not its opposite side, the law of sines isn't so helpful here. You could use the law of cosines to find the remaining side, then solve for the missing angles. Historically, though, this was one of the main use case of the law of tangents: it solves for the angles more "directly" and avoids having to take a square root, which was a pain before electronic calculators. Even today, for some applications it's worth avoiding the law of cosines due to numerical issues e.g. when $a$ and $b$ are similar and $C$ is small. There's some discussion of this on Wikipedia and in another answer I found quite noticeable numerical errors on a Casio calculator's answers using the law of cosines which were improved by using the law of tangents.
Know two sides and difference between their opposite angles
E.g. you are told $A$ is $40^\circ$ wider than $B$, and that $a = 7$ and $b = 3$. You can calculate $D = 20^\circ$, $m = 5$, $d = 2$, then use
$$M = \tan^{-1} \left( \frac{m}{d} \tan D \right)$$
to find $M \approx 42.30^\circ$. Then $A = M + D \approx 62.30^\circ$ and $B = M - D \approx 22.30^\circ$.
This can be done without the law of tangents, e.g. using trigonometric identities for an angle sum or difference, but it's not pleasant.
Know all angles and sum of two side lengths
This is a common practical situation, where you know e.g. the total length of a rod or string with a kink in it, so it now forms two sides of a triangle. When I say knowing "all" angles, we really only need to know two to find the third. (The two we'll actually use will be the ones opposite the sides whose total length we know, but these might not be the two which are given.) Let's say the total length of our two sides is $a + b = 8$, with opposite angles $A = 80^\circ$ and $B = 50^\circ$. Then $m = 4$, $M = 65^\circ$ and $D = 15^\circ$, and
$$d = m \frac{\tan D}{\tan M}$$
gives $d \approx 0.500$. Our two lengths are $a = m + d \approx 4.500$ and $b = m - d \approx 3.500$.
It's not hard to solve this using the law of sines either, we just write $b = 8 - a$ so
$$\frac{a}{\sin 80^\circ} = \frac{8 - a}{\sin 50^\circ}$$
which can be solved for $a$ e.g. by rearranging to $(\sin 50^\circ + \sin 80^\circ)a = 8 \sin 80^\circ$. The law of tangents perhaps spares us a little algebra.
Know all angles and difference of two side lengths
This is very similar to the previous scenario. Let's say we know $a$ is $6$ units longer than $b$, that $A = 65^\circ$ and $B = 45^\circ$. We calculate $d = 3$, $M = 55^\circ$ and $D = 10^\circ$, and
$$m = d \frac{\tan M}{\tan D}$$
gives $m \approx 24.30$. Our two lengths are $a = m + d \approx 27.30$ and $b = m - d \approx 21.30$.
If we wrote $a = b + 6$ we could have used the law of sines without too much difficulty,
$$\frac{b + 6}{\sin 65^\circ} = \frac{b}{\sin 45^\circ}$$
giving $(\sin 65^\circ - \sin 45^\circ)b = 6 \sin 45^\circ$.
