I have a function $f(x, y) = \csc(x) \csc(x + y) \sin(y) + \cot(y)$, and I know from graphing it that the function is symmetric in $x$ and $y$ such that $f(x, y) = f(y, x)$. However, that fact isn't obvious just by looking at the equation, since the two variables don't seem to be treated the same.
Can the expression be written in such a way that its symmetry is obvious from the expression itself, without needing to graph it?
I've tried using trig identities to replace the cosecants and cotangent with sines and cosines, using the angle addition identities to isolate one-variable-per-trig-function, and rearranging and regrouping terms, but without success.
Expanding my comment ...
We have
$$ f(x,y) = \frac{\sin y}{\sin x\sin(x+y)}+\frac{\cos y}{\sin y} = \frac{\sin^2y+\sin x\cos y\sin(x+y)}{\sin x\sin y\sin(x+y)} \tag1$$
The denominator is symmetric in $x$ and $y$, so let's focus on the numerator ...
$$\begin{align} &\sin^2y+\sin x\cos y(\sin x\cos y+\cos x\sin y) \tag2\\[4pt] \to\quad &(1-\cos^2y)+\sin^2x\cos^2y+\sin x\cos x\sin y\cos y \tag3\\[4pt] \to\quad & 1-(1-\sin^2x)\cos^2y+\sin x\cos x\sin y\cos y \tag4\\[4pt] \to\quad & 1 - \cos^2x\cos^2y+\sin x\cos x\sin y\cos y \tag5\\[4pt] \to\quad & 1 - \cos x\cos y(\cos x\cos y-\sin x\sin y) \tag6\\[4pt] \to\quad & 1 - \cos x\cos y \cos(x+y) \tag7 \end{align}$$
Therefore, $$f(x,y) \;=\; \frac{1-\cos x\cos y\cos(x+y)}{\sin x\sin y\sin(x+y)} \;=\; f(y,x) \tag{$\star$}$$