### Instruction: Compute $\tan (-3645^\circ)$.### Response: To solve this problem, I need to recall the definition of the tangent function and the properties of negative angles.
I know that $\tan (\theta) = \frac{\sin (\theta)}{\cos (\theta)}$ for any angle $\theta$, and that $\sin (-\theta) = -\sin (\theta)$ and $\cos (-\theta) = \cos (\theta)$ for any angle $\theta$.
This means that $\tan (-\theta) = -\tan (\theta)$ for any angle $\theta$.
So, to find $\tan (-3645^\circ)$, I can first find $\tan (3645^\circ)$ and then multiply it by -1.
Now, I need to simplify the angle $3645^\circ$.
I notice that $3645^\circ$ is more than one full revolution of $360^\circ$, so I can subtract $360^\circ$ from it as many times as needed until I get an angle between $0^\circ$ and $360^\circ$.
I can divide $3645^\circ$ by $360^\circ$ to see how many full revolutions I have.
I get $3645^\circ \div 360^\circ = 10.125$, which means I have 10 full revolutions and a remainder of $0.125 \times 360^\circ = 45^\circ$.
So, $3645^\circ = 10 \times 360^\circ + 45^\circ$.
This means that $\tan (3645^\circ) = \tan (45^\circ)$, since the tangent function is periodic with a period of $360^\circ$.
I recall that $\tan (45^\circ) = 1$, since $\sin (45^\circ) = \cos (45^\circ) = \frac{\sqrt{2}}{2}$.
Therefore, $\tan (-3645^\circ) = -\tan (3645^\circ) = -\tan (45^\circ) = -1$.