Given $\cos (\alpha + \beta) + \sin (\alpha - \beta) = 0$ and $\tan \beta = \frac{1}{2000},$ find $\tan \alpha.$
I notice that the given equation involves both sine and cosine of sums and differences of angles, which can be simplified using some trigonometric identities. Specifically, I recall that $\cos (A + B) = \cos A \cos B - \sin A \sin B$ and $\sin (A - B) = \sin A \cos B - \cos A \sin B.$ Substituting these identities into the equation, I get $(\cos \alpha \cos \beta - \sin \alpha \sin \beta) + (\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 0.$ Simplifying, I get $\cos \alpha \cos \beta + \sin \alpha \cos \beta = \sin \alpha \sin \beta + \cos \alpha \sin \beta.$ Dividing both sides by $\cos \alpha \cos \beta,$ I get $1 + \tan \alpha = \tan \beta + \tan \alpha \tan \beta.$ Rearranging, I get $\tan \alpha (1 - \tan \beta) = \tan \beta - 1.$ Dividing both sides by $(1 - \tan \beta),$ I get $\tan \alpha = \frac{\tan \beta - 1}{1 - \tan \beta}.$ Now, I can plug in the given value of $\tan \beta = \frac{1}{2000}$ and simplify. I get $\tan \alpha = \frac{\frac{1}{2000} - 1}{1 - \frac{1}{2000}} = \frac{-\frac{1999}{2000}}{\frac{1999}{2000}} = -1.$

