<s>[INST] 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?
 [/INST] 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 $v = \pm \frac{1}{2}$, $2v^6-v^4+4v^2-1 = \frac{1}{32}$.
</s>