How do I start solving this trigonometric system of equations? [closed]

I was passing by this forum for curiosity and stumbled upon your question, I was intrigued to solve and came with this, hope it can help.

Well, first things first, your system:

\begin{cases} \tan x + \tan y = 1\\ x - y = \fracπ4 \end{cases}

First thing you want to do is isolate x (or y) in the second line, I'll do with x just because:

$$x = \fracπ4 + y$$

Then, plug your "new" value of x in the first equation:

$$\tan (\fracπ4 + y) + \tan y = 1$$

So, this doesn't help much right, but there is a trigonometry formula that states:

$$\tan(u±v)=\frac{\tan(u) ± \tan(v)}{1 ∓ \tan(u)\cdot\tan(v)}$$

Font: https://www.varsitytutors.com/hotmath/hotmath_help/topics/trigonometric-identities

It's pretty helpful, it will separate the elements inside of tangent and we will be able to work with them.

So,

$$\tan(\frac{\pi}{4} + y) + \tan (y) = 1\\$$

Turns in:

$$\frac{\tan(\frac{\pi}{4}) + \tan(y)}{1 - \tan(\fracπ4)\cdot\tan(y)} + \tan (y) = 1\\$$

From here is just simple algebra, hope that helps :)

I finished the solution, that user rookie had offered. $$ \frac{\tan(\frac{\pi}{4})+\tan(y)}{1-\tan(\frac{\pi}{4})\cdot\tan(y)}+\tan(y) =1, \\ \left(\frac{1+\tan(y)}{1-\tan(y)}+\frac{\tan(y)-\tan^{2}(y)-1+\tan(y)}{1-\tan(y)}\right)=0, \implies \\ \implies \begin{cases} -\tan^{2}(y)+3\tan(y)=0,\\ {1-\tan(y)}\neq{0}. \end{cases} \iff \begin{cases} \tan(y)(\tan(y)-3)=0,\\ {\tan(y)}\neq{1}. \end{cases} \iff \\ \iff \bbox[lightgreen] { \begin{cases} y=\pm{\pi{n}},\quad\arctan(3)\pm{\pi{n}},\quad{n}\in{Z},\\ {y}\neq{\left(\frac{\pi}{4}\pm\pi{n}\right),\quad{n}\in{Z}}. \end{cases} } $$