Problem with trigonometric equations

Question

In a given question about a trigonometric proof I was trying to solve, I got in $$ \tan(x - π/4) = -\, 1 $$ then I solved it just like any other trig equation, but, after I managed to finish the demonstration, I get back to that equation and tried a different approach, by applying the inverse function in both sides of the equation: $$ \arctan(\tan(x - π/4)) = \arctan(-\, 1) = -\,π/4. $$ But, after analysing both sides, I realized I've got an absurd after considering that $$ \arctan(-\, 1) = -\,π/4 $$ since, to succeed in my proof, I had to consider that $$ \arctan(-\, 1) = 3π/4.$$ My question is, how to solve simple trigonometric equations, like the one I posted above, by applying the inverse function without getting an absurd? Just to mention, the $$ x $$ in the tangent argument is equal to $$ \arctan(2) + \arctan(3), $$hence it satisfies the arctan condition.

Answer 1

$$\tan\left(x-\dfrac\pi4\right)=-1\implies x-\dfrac\pi4=m\pi+\arctan(-1)$$

where $m$ is any integer

Answer 2

As there are infinitely many values of $x$ having the same value of $\tan x$, we usually take the principal value only when using the $\arctan$ function. That is, $\displaystyle -\frac{\pi}{2}<\arctan k<\frac{\pi}{2}$. So, $\displaystyle \arctan(-1)=-\frac{\pi}{4}$ and $\displaystyle \frac{3\pi}{4}=\arctan(-1)+\pi$.

For $\displaystyle \tan\left(x-\frac{\pi}{4}\right)=-1$, $\displaystyle x-\frac{\pi}{4}=n\pi+\arctan(-1)=n\pi-\frac{\pi}{4}$ and hence $x=n\pi$. The value of $x$ may be further restricted by the conditions given in your original problem.

For the problem $x=\arctan(2)+\arctan(3)$, let $u=\arctan(2)$ and $v=\arctan(3)$. Then $\displaystyle \frac{\pi}{4}<u<v<\frac{\pi}{2}$. Since $x=u+v$, $\displaystyle \frac{\pi}{2}<x<\pi$.

$$\tan x=\tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v}=\frac{2+3}{1-(2)(3)}=-1$$

$$x=n\pi+\arctan(-1)=n\pi-\frac{\pi}{4}$$

So, $\displaystyle x=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$.