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 $9=1$?

Another way to avoid confusion:
$$\dfrac31=\dfrac{\sin(x+15^\circ)\cos(x-15^\circ)}{\cos(x+15^\circ)\sin(x-15^\circ)}$$
Apply Componendo et Dividendo
$$\dfrac{3+1}{3-1}=\dfrac{\sin2x}{\sin30^\circ}$$
