Find $\tan \frac{9 \pi}{4}.$
Converting to degrees,
\[\frac{9 \pi}{4} = \frac{180^\circ}{\pi} \cdot \frac{9 \pi}{4} = 405^\circ.\]Since the tangent function has period $360^\circ,$ $\tan 405^\circ = \tan (405^\circ - 360^\circ) = \tan 45^\circ = \boxed{1}.$
1
Level 1
Precalculus
Problem: Find $\tan \frac{9 \pi}{4}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\tan \frac{9 \pi}{4}$, first simplify the angle. The tangent function has a period of $\pi$, meaning $\tan(\theta + n\pi) = \tan(\theta)$ for any integer $n$. 

Calculate how many full periods of $\pi$ fit into $\frac{9 \pi}{4}$. Since $\pi$ is equivalent to $\frac{4\pi}{4}$, we can express the angle as:

$$\frac{9 \pi}{4} = 2\pi + \frac{\pi}{4}.$$

This shows that $\frac{9 \pi}{4}$ is equivalent to $\frac{\pi}{4}$ after subtracting $2\pi$ (which is $8\pi/4$). 

Now, evaluate $\tan \frac{\pi}{4}$. The value of $\tan \frac{\pi}{4}$ is known to be $1$. 

Thus, we conclude that $\tan \frac{9 \pi}{4} = \tan \frac{\pi}{4} = 1$. 

The final answer is $1$.<|END_OF_LATENT|> The final answer is $1$.