$x+y = \frac{\pi}{4}$ and $\tan x + \tan y = 1$ where $n$ is an integer

Question

$x+y = \frac{\pi}{4}$ and $\tan x + \tan y = 1$ where $n$ is an integer

Prove that when $ x = n\pi + \frac\pi4 $ then $ y =-n\pi $

Clearly above statement is only true if n = 0 and not all integers.

Now we know addition of X and y and value of X so solving those equation we get

$ n\pi + \frac\pi4 + y = \frac\pi4 $

Thus $ n\pi = y $

And $ - n\pi = y $ from prove that statement.

Thus n = -n

Which is only true for n = 0

So it prove that statement even true in the first place or not?

Ignore that $\tan$ part if you find it useless.

Answer 1

Hint:

Use $1=\tan(x+y)=\cdots$

So, at least one of $\tan x,\tan y$ must be $0$

So, if $\tan x=0,x=m\pi,y=\dfrac\pi4-x=?$

As $m$ is any integer, replace $m$ with $-n$

Answer 2

Hint:Use that $$\tan\left(\frac{\pi}{4}-x\right)=\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}$$

Answer 3

You are correct that if $x = n\pi + \frac\pi4$ then the equation $x+y=\frac\pi4$ implies that

$$ n\pi + \frac\pi4 + y = \frac\pi4. $$

You can cancel $\frac\pi4$ (or add $-\frac\pi4$ on both sides) to get

$$ n\pi + y = 0. $$

This then implies that $y = -n\pi.$ It is not at all the same thing as $y=n\pi.$