Are known the trig identities
\begin{align}
&\sin(a\pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b),\tag1\\[4pt]
&\cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1 = 1-2\sin^2(a),\tag2\\[4pt]
&\cos(a) - \cos(b) = 2\sin\dfrac{a+b}2\sin\dfrac{b-a}2.\tag3\\
\end{align}
Using $(1) - (3),$ one can reduce the arguments of the LHS of the issue condition,
\begin{align}
&\sin(x-y)+\cos(x+2y)\sin (y) = \sin(x)\cos(y)+\cos(x+2y)\sin (y)-\cos(x)\sin(y) = \\[4pt]
&\sin(x)\cos(y)-2\sin(x+y)\sin^2(y) = (1-2\sin^2(y))\sin(x+y)- \sin(y)\cos(x) =\\[4pt]
&\cos(2y)\sin(x+y)- \sin(y)\cos(x).\\[4pt]
\end{align}
Then
\begin{align}
&(\cos(2y)\sin(x+y)- \sin(y)\cos(x))^2 = 4\sin(y)\cos(x)\sin(x+y),\\
&\cos^2(2y)\sin^2(x+y)-2(\cos(2y)+2)\sin(y)\cos(x)\sin(x+y)+\sin^2(y)\cos^2(x) = 0.\tag4\\
\end{align}
Due to the area of admissible values of the goal equality, $\cos(x)\not=0$ and $\cos(y)\not=0.$ So it is correct to divide the equation $(4)$ by $\cos^2(x)\cos^2(y).$
Taking in account $(1)-(2),$ this leads to the square equation in the form of
\begin{align}
&\cos^2(2y)(\tan(x)+\tan(y))^2-2(\cos(2y)+2)\tan(y)(\tan(x)+\tan(y))^2+\tan^2(y) = 0,\\[4pt]
&D = (\cos(2y)+2)^2 - \cos^2(2y) = 4(\cos(2y)+1) = 8\cos^2(y),\\[4pt]
&\tan(x)+\tan(y) = \dfrac{\cos(2y)+2\pm\sqrt8\ \cos(y)}{\cos^2(2y)}\tan(y),\\[4pt]
&\dfrac{\cos(2y)+2\pm2\sqrt2\ \cos(y)}{\cos^2(2y)}
= \dfrac{2\cos^2(y)\pm2\sqrt2\ \cos(y) +1}{(2\cos^2(y)-1)^2}
= \dfrac{\cos(2y)+2\pm\sqrt8\ \cos(y)}{(2\cos^2(y)-1)^2}\\[4pt]
&= \dfrac{\left(\sqrt2\cos(y)\pm1\right)^2}{\left(\sqrt2\cos(y)+1\right)^2\left(\sqrt2\cos(y)-1\right)^2}
= \dfrac1{\left(\sqrt2\cos(y)\pm1\right)^2},\\[4pt]
&\tan(x)+\tan(y) = \dfrac{\tan(y)}{\left(\sqrt2\cos(y)\pm1\right)^2}.\tag5\\[4pt]
\end{align}
In accordance with the formula $(5),$ the initial problem setting looks incorrect, with a minimal discrepancy in the answers.
Maybe something is missing in the conditions?