why $A = 180n - B$ is NOT a solution of $\tan (A) = \tan (B)$?

Question

Solving: $\tan (A) = \tan (B)$, gives: $A = 180^\circ n + B$ , where $n$ is an integer.

But why is: $A = 180^\circ n - B$, not considered to be a solution of the equation?

Answer 1

If you use that definition for A, you get $\tan(A)=-\tan(B)$ as $\tan(x)$ is an odd function.

Answer 2

If you use the formula $$\tan(x+y)=\frac {\tan(x)+\tan(y)} {1-\tan(x)~\tan(y)}$$ first with $x=n\pi$ and $y=B$ and second with $x=n\pi$ and $y=-B$, you will notice the same as what Silynn wrote.