Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is not that brought me in a right way.
Do you have any idea how to solve this one?

Let $ t = \tan  \left(y\right)$ and $ u = \tan  \left(x\right)$. One has $\sin  \left(y\right) = \pm  \frac{t}{\sqrt{1+{t}^{2}}}$ and $\cos  \left(x\right) = \pm  \frac{1}{\sqrt{1+{u}^{2}}}$. The second equation then implies
\begin{equation}2 {t}^{2} = \left(1+{u}^{2}\right) \left(1+{t}^{2}\right)\end{equation}
but according to the first equation, we have $ u = 1-t$, hence the quadric equation
\begin{equation}\renewcommand{\arraystretch}{1.5}  \begin{array}{cc}&2 {t}^{2} = \left(1+{\left(1-t\right)}^{2}\right) \left(1+{t}^{2}\right)\\
\Longleftrightarrow &\left(t-1\right) \left({t}^{3}-{t}^{2}-2\right) = 0
\end{array}\end{equation}
This equation has two real solutions $ {t}_{0} = 1$ and $ {t}_{1} = \frac{1}{3} \left(d+1+\frac{1}{d}\right)$ with $ d = \sqrt[3]{3 \sqrt{87}+28}$.
The first solution gives
\begin{equation}x = k {\pi} + 2 n \pi, \quad  y = \frac{{\pi}}{4}+ k {\pi}\end{equation}
The second solution gives
\begin{equation}y = \arctan  \left({t}_{1}\right)+k {\pi} , \quad  x = \arctan  \left(1-{t}_{1}\right)+k {\pi}+2 n {\pi}\end{equation}
