I'll show a theoretical way to get at this answer without these Taylor/Maclaurin series featured in the other answers, though this would take a lot of time to do by hand.
Setup
Let $s=\sin x^\circ$ and note that for $x<\dfrac\pi2$, we have $\cos x^\circ=\sqrt{1-s^2}$. Then by using the sum formulas repeatedly, we find that:
$$\tag{1}\sin \left(9x^\circ\right)=256s^9-576s^7+432s^5-120s^3+9s$$
$$\tag{2}\tan \left(8x^\circ\right)=\dfrac{\left(-128s^7+192s^5-80s^3+8s\right)\sqrt{1-s^2}}{128s^8-256s^6+160s^4-32s^2+1}$$
To compare $\sin 9^\circ$ and $\tan 8^\circ$, we need a bound on $\sin 1^\circ$.
Estimating $\sin 1^\circ$
One known value is $\sin 15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}$. The sum formulas give us that $\sin \left(15x^\circ\right)$ is
$$-16384 s^{15}+61440 s^{13}-92160 s^{11}+70400 s^9-28800 s^7+6048 s^5-560 s^3+15 s$$
Evaluating that polynomial at $s=\dfrac{1}{20}$ yields $\dfrac{1363735274101499}{2000000000000000}>\dfrac12>\dfrac{\sqrt3-1}{2\sqrt2}$. Therefore, $0<\sin1^\circ<\dfrac{1}{20}$ for sure. (In reality, $\sin1^\circ$ is more like $1/50$.)
Conclusion
Evaluating the polynomials from (1) and (2) at $s=\sin x^\circ=\dfrac{1}{20}$ yield $\sin \left(9x^\circ\right)=\dfrac{870269101}{2000000000}$ and $\tan\left(8x^\circ\right)=\dfrac{3900599\sqrt{399}}{184199201}<\dfrac{3900599*20}{184199201}<\dfrac{870269101}{2000000000}$.
It remains to check why this really means $\sin 9^\circ$ is greater than $\tan8^\circ$. Since $\tan$ has an asymptote at $90^\circ$, $\tan{8x^\circ}$ must overtake $\sin{9x^\circ}$ sometime before $x=90/8$, so the expression has an asymptote at $s=\sin\left(\dfrac{90^\circ}{8}\right)=\dfrac12\sqrt{2-\sqrt{2+\sqrt2}}\approx\dfrac15$. By considering concavity, the $\tan$ function should overtake the $\sin$ function exactly once before this, and since it hasn't happened by $s=\dfrac{1}{20}>\sin 1^\circ$, $\boxed{\sin 9^\circ>\tan8^\circ}$ after all.
Pictures
Here is a graph of the two functions of $s$. $\tan$ overtakes $\sin$ before $s=.06$.
![tan vs sin](https://cdn.statically.io/img/i.sstatic.net/TTfrE.png)
If you had large enough drawing materials, you might be able to compare the values directly with a construction like this:
![circle close](https://cdn.statically.io/img/i.sstatic.net/e30oK.png)