### Instruction: Compute $\tan 240^\circ$.### Response: To compute $\tan 240^\circ$, I need to recall the definition of the tangent function in terms of sine and cosine.
I know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ for any angle $\theta$.
So, to find $\tan 240^\circ$, I just need to plug in $\theta = 240^\circ$ and simplify.
I get $\tan 240^\circ = \frac{\sin 240^\circ}{\cos 240^\circ}$.
Now, I need to find the values of $\sin 240^\circ$ and $\cos 240^\circ$.
I can use the unit circle and the reference angles to help me.
I know that $240^\circ$ is in the third quadrant, and its reference angle is $60^\circ$.
That means that $\sin 240^\circ = -\sin 60^\circ$ and $\cos 240^\circ = -\cos 60^\circ$, since both sine and cosine are negative in the third quadrant.
I also know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$ by using the 30-60-90 triangle.
Therefore, I can substitute these values into the expression for $\tan 240^\circ$.
I get $\tan 240^\circ = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{1} = \sqrt{3}$.