$\sin 9^{\circ}$ or $\tan 8^{\circ} $ which one is bigger?

When $0<x<1$ then$$\sin x>x-{x^3\over6},\qquad\tan x={\sin x\over \cos x}<{x\over 1-{x^2\over2}}\ ,$$ by the theorem on alternating series. Let $t:=1^\circ={2\pi\over 360}<{1\over50}$. Then $${\sin(9t)\over\tan(8t)}>{9t\over 8t}\left(1-{81t^2\over6}\right)\left(1-{64t^2\over2}\right)>{9\over8}\left(1-{273\over6}t^2\right)>1\ .$$

I would check the first two terms of the Taylor series. $\sin 9^\circ \approx \frac \pi{20}-\frac {\pi^3}{6 \cdot 20^3}, \tan 8^\circ \approx \frac {2\pi}{45}+\frac {8\pi^3}{3 \cdot 45^3}$, so $$\sin 9^\circ -\tan 8^\circ\approx \frac \pi{20}-\frac {\pi^3}{6 \cdot 20^3}-\frac {2\pi}{45}-\frac {8\pi^3}{3 \cdot 45^3}\\\approx \frac \pi{180}-\frac{(3^6+2^{10})\pi^3}{2^73^75^3}\\ \approx \frac \pi{180}(1-\frac {1753}{2^43^55}) \\ \approx \frac \pi{180}(1-\frac {1753}{17440})\\ \gt 0$$ where I used $\pi^2 \approx 10$. Alpha agrees, but I didn't check until I was done.

In terms of radian, $9^\circ = \frac{\pi}{20} \approx 0.157$ is reasonable small. We can use Taylor series expansion to estimate the value of $\sin$ and $\tan$. For small $\theta$, we have

$$\sin\theta \approx \theta - \frac{\theta^3}{6} \quad\text{ and }\quad \tan\theta = \frac{\sin\theta}{\cos\theta} \approx \frac{\theta - \frac{\theta^3}{6}}{1 - \frac{\theta^2}{2}} \approx \theta + \frac{\theta^3}{3} $$ This implies $$\tan^{-1}\theta \approx \theta - \frac{\theta^3}{3}\quad\text{ and }\quad \tan^{-1}(\sin\theta) \approx \theta - \frac{\theta^3}{6} - \frac{\theta^3}{3} = \theta\left(1 - \frac{\theta^2}{2}\right)$$

In order for $\tan\phi$ equals to $\sin 9^\circ$, $\phi$ should be around $$9^\circ \times \left( 1 - \frac{0.157^2}{2}\right) \approx 9^\circ \times 0.98\;(\text{ or } 0.99???)$$ No matter what the actual value of last factor is, the $\phi$ required to make $\tan\phi = \sin 9^\circ$ is much closer to $9^\circ$ than $8^\circ$. This means $\tan 8^\circ \le \sin 9^\circ$.