Let $\alpha\in\mathbb{R}$ be so that $\cos\alpha\neq 1$. Simplify as much as possible $\dfrac{\sin(2022) + \sin(2022+\alpha)}{\cos(2022)-\cos(2022+\alpha)}$. To clarify, if for a finite number of values of $\alpha,$ the expression takes a numerical value, find that numeric value for each of those values of $\alpha$. If not, find a function of $\alpha$ that is as simplified as possible.
Clearly $\alpha$ is not an integer multiple of $2\pi.$ Let $f(\alpha)$ denote the expression that needs to be simplified. We have that $f(\alpha + 2k\pi) = f(\alpha)$ for any integer $k$, so we may assume WLOG that $0 < \alpha < 2\pi$. It might be easier to generalize and replace $2022$ with $m$. Perhaps we only need the fact that $2022$ is an integer? Using the addition law, we get that $f(\alpha) = \dfrac{\sin(m)(1+ \cos\alpha) + \sin(\alpha)\cos m}{\cos(m)(1 - \cos(\alpha)) + \sin(m)\sin(\alpha)}$. I'm not sure if Euler's formula will help simplify $f(\alpha)$. Multiply the denominator and numerator of $f(\alpha)$ by $\cos m(1+\cos \alpha) - \sin m \sin \alpha$. The denominator becomes $\cos^2 m \sin^2 \alpha + 2\sin m \cos m \sin \alpha \cos \alpha - \sin^2m\sin^2\alpha = \sin\alpha( \cos(2m) \sin \alpha + \sin(2m)\cos \alpha) = \sin \alpha \sin(2m+\alpha)$. The numerator becomes $\sin m \cos m (1+\cos \alpha)^2 -\sin^2 m \sin \alpha(1+\cos \alpha) + \cos^2 m \sin \alpha(1+\cos \alpha) - \sin m \cos m \sin^2 \alpha = \cos(2m) \sin (\alpha) ( 1+\cos \alpha) + \sin m \cos m(2\cos \alpha + 2\cos^2 \alpha) = (1+\cos \alpha)(\cos (2m) \sin (\alpha) + \sin(2m)\cos (\alpha)) = (1+\cos\alpha)\sin(2m+\alpha).$ Hence $f(\alpha) = \dfrac{1+\cos \alpha}{\sin \alpha}.$ Note that $m$ could've actually been any real number by the same proof as above.
