<s>[INST] Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$ 
$x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$ 

Substituting $y=x+45^{\circ}$, we get
$$3\tan{(y-60^{\circ})}=\tan{(y-30^{\circ})}$$
$$3\frac{\tan y - \sqrt3}{1+\sqrt3\tan y}=\frac{\tan y - 1/\sqrt3}{1+1/\sqrt3\cdot\tan y}$$ 
$$3(\tan ^2 y-3)=3\tan ^2-1$$ 
$$9=1$$ 
The solution provided by the book $x=n\pi  + \pi/4$ fits, so why did i get $9=1$?
 [/INST] You got $$0\cdot\tan^2{y}+9=1.$$ Id est, you got that $\tan{y}$ does not exist.
Thus, $$y=\frac{\pi}{2}+\pi n,$$ where $n\in\mathbb{Z},$ or
$$x+45^{\circ}=\frac{\pi}{2}+\pi n,$$ which gives $$x=\pi n+\frac{\pi}{4}.$$
We did not get a contradiction in Math!
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