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?
 A: 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}
A: $$\cos x \sin y=\frac{\sqrt2}{2}$$
$$\cos x \sin y=\frac{1}{\sqrt2}$$
$\frac{1}{\sqrt2}$ can be written as $1*\frac{1}{\sqrt2}$.
This gives rise to two cases-->

*

*$\cos x=1$ and $\sin y=\frac{1}{\sqrt2}$. So, $x=0$ and $y=\frac{\pi}{4}$.

*$\cos x=\frac{1}{\sqrt2}$ and $\sin y=1$. So, $x=\frac{\pi}{4}$ and $y=\frac{\pi}{2}$.

Solving Equation 1 for both cases-->

*

*$$\tan x+\tan y$$
$$\tan 0+\tan\frac{\pi}{4}$$
$$0+1=1$$
2.$$\tan x+\tan y$$
$$\tan\frac{\pi}{4}+\tan\frac{\pi}{2}$$
$$1+\infty=\infty$$
We see that only Case 1 satisfies the given equation. So, $x=0$ and $y=\frac{\pi}{4}$.
A: 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.
A: Substitute $u=\cos x, \sqrt{1-u^2}=\sin x, v=\cos y, \sqrt{1-v^2}=\sin y$.
The second equation:
$$ u\sqrt{1-v^2} = \frac{1}{\sqrt2} $$
$$\Rightarrow u=\frac{1}{\sqrt{2-2v^2}}$$
The first equation:
$$ \frac{\sqrt{1-u^2}}{u} + \frac{\sqrt{1-v^2}}{v} = 1 $$
$$ \Rightarrow \sqrt{1-2v^2} + \frac{\sqrt{1-v^2}}{v} = 1 $$
$$\Rightarrow 4v^8-4v^6+9v^4-6v^2+1=0$$
$$\Rightarrow (2v^2-1)(2v^6-v^4+4v^2-1)=0$$
If $2v^2-1=0$,
$$(x,y)= (2\pi m, 2\pi n + \frac{\pi}{4}), (2\pi m \pm \pi, 2\pi n - \frac{3\pi}{4})$$
$2v^6-v^4+4v^2-1=0$ has two solutions (same absolute value, different signs), and they are not neat. Though we can approximate $v \approx \pm \frac{1}{2}$ since when $v = \pm \frac{1}{2}$, $2v^6-v^4+4v^2-1 = \frac{1}{32}$.
