Find $\tan \left( -\frac{3 \pi}{4} \right).$
Converting to degrees,
\[-\frac{3 \pi}{4} = \frac{180^\circ}{\pi} \cdot \left( -\frac{3 \pi}{4} \right) = -135^\circ.\]Since the tangent function has period $180^\circ,$ $\tan (-135^\circ) = \tan (-135^\circ + 180^\circ) = \tan 45^\circ = \boxed{1}.$
1
Level 1
Precalculus
Problem: Find $\tan \left( -\frac{3 \pi}{4} \right).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\tan \left( -\frac{3 \pi}{4} \right)$, first recognize that the tangent function is periodic with a period of $\pi$. This means that $\tan(x) = \tan(x + n\pi)$ for any integer $n$. 

Next, simplify the angle $-\frac{3 \pi}{4}$ by adding $\pi$ to it. This gives:

\[
-\frac{3 \pi}{4} + \pi = -\frac{3 \pi}{4} + \frac{4 \pi}{4} = \frac{\pi}{4}.
\]

Now, we can use the property of the tangent function:

\[
\tan \left( -\frac{3 \pi}{4} \right) = \tan \left( \frac{\pi}{4} \right).
\]

The value of $\tan \left( \frac{\pi}{4} \right)$ is known to be $1$. 

However, we must consider the sign of the tangent function in the relevant quadrant. The angle $-\frac{3 \pi}{4}$ is in the third quadrant, where the tangent function is positive. Therefore, we conclude:

\[
\tan \left( -\frac{3 \pi}{4} \right) = 1.
\]

Thus, the final answer is $1$.<|END_OF_LATENT|> The final answer is $1$.