Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$

Question

$x$ 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 $9=1$?

Answer 1

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!

Answer 2

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

$$\dfrac{3+1}{3-1}=\dfrac{\sin2x}{\sin30^\circ}$$

Answer 3

Hint:

Set $\tan y=\dfrac1a$

$$\dfrac{3(1-\sqrt3a)}{a+\sqrt3}=\dfrac{\sqrt3-a}{\sqrt3a+1}$$

$$\iff3-a^2=3(1-3a^2)\iff a=0$$