I have came across this system of trigonometry equations.
$$\tan x + \tan y=1-\tan x\tan y$$
$$ \sin2y-\sqrt 2 \sin x=1$$
I have tried with expressing $\tan$ as $\frac{\sin}{\cos}$ and using the formula for $\sin2\alpha$. But I haven't gone any further than the start.
Best Answer
HINT
The first equation can be rewritten as \begin{align*} \tan(x) + \tan(y) = 1 - \tan(x)\tan(y) & \Longleftrightarrow \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)} = 1\\\\ & \Longleftrightarrow \tan(x+y) = \tan\left(\frac{\pi}{4}\right)\\\\ & \Longleftrightarrow x + y = \frac{\pi}{4} + k\pi \end{align*} where $k\in\mathbb{Z}$. Substituting such relation into the second equation, one has that \begin{align*} \sin(2y) - \sqrt{2}\sin(x) = 1 & \Longleftrightarrow \sin\left(\frac{\pi}{2} - 2x\right) - \sqrt{2}\sin(x) = 1\\\\ & \Longleftrightarrow \cos(2x) - \sqrt{2}\sin(x) = 1\\\\ & \Longleftrightarrow 1 - 2\sin^{2}(x) - \sqrt{2}\sin(x) = 1\\\\ & \Longleftrightarrow \sin(x)(\sqrt{2}\sin(x) + 1) = 0 \end{align*}
Can you take it from here?
EDIT
Notice that $\tan(x)\tan(y) = 1$ iff \begin{align*} \tan(x)\tan(y) = 1 \Longleftrightarrow \begin{cases} \sin(x)\sin(y) = \cos(x)\cos(y)\\\\ \cos(x)\cos(y) \neq 0 \end{cases} & \Longleftrightarrow \begin{cases} \cos(x+y) \neq 0\\\\ \cos(x)\cos(y) \neq 0 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} \displaystyle x + y \neq \frac{\pi}{2} + m\pi\\\\ \displaystyle x \neq \frac{\pi}{2} + n\pi\\\\ \displaystyle y \neq \frac{\pi}{2} + p\pi \end{cases} \end{align*}
where $m,n,p\in\mathbb{Z}$. As it can be checked, the proposed solutions satisfy such restrictions.