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?

The two given equations imply that
\begin{align}
&(1-\tan x)^2=\frac{\sin^2y}{1-\sin^2y},\tag{1}\\
&\cos x=\frac{1}{\sqrt{2}\sin y}.\tag{2}
\end{align}
Suppose both $x$ lies inside the first quadrant. From $(2)$ we see that $y$ must lie inside the first or the second quadrant. Also,
\begin{align}
\tan x&=\frac{\sqrt{1-\cos^2x}}{\cos x}=\sqrt{2\sin^2y-1},\\
\sin^2y&=\frac{1+\tan^2x}{2}.\\
\end{align}
Substitute the expression for $\sin^2y$ into $(1)$, we obtain
\begin{aligned}
&(1-t)^2=\frac{(1+t^2)/2}{1-(1+t^2)/2},\\
&(1-t^2)(1-t)^2=(1+t^2),\\
&(1+t^2)-(1-t^2)(1-t)^2=0,\\
&t\left[t(t-1)^2+2\right]=0.
\end{aligned}
Since $x$ lies inside the first quadrant, $t=\tan x$ is nonnegative and the equation above has a unique root $t=0$. Hence $x=0,\ \sin^2 y=\frac{1+t^2}{2}=\frac12$ and $y=\frac{\pi}{4}$.
I guess the other cases (where $x$ or $y$ or both lie inside other quadrants) can be handled in a similar manner.
