Solving $\tan^{-1}(2x)+\tan^{-1}(3x)=\frac\pi4$. Why not use other cases of the $\tan^{-1}x+\tan^{-1}y$ formula? A question is given in my book.

Solve for $x$:
$$\tan^{-1}(2x)+\tan^{-1}(3x)=\frac\pi4$$

I understood the solution of this question which is as follows:
We have,
$$\begin{align}
&\tan^{-1}(2x)+\tan^{-1}(3x) =\frac\pi4 \tag1\\[6pt]
\implies\quad &\tan^{−1}\frac{(2x)+(3x)}{1-(2x)(3x)} = 1, \text{if}\; (2x)(3x) <1 \tag2\\[6pt]
\implies\quad &\frac{5x}{1-6x^2}=1 \;\text{and}\; x^2<\frac16 \tag3\\[6pt]
\implies\quad &(6x-1)(x+1)=0 \;\text{and}\; -\frac{1}{\sqrt{6}}<x<\frac1{\sqrt{6}} \tag4\\[6pt]
\implies\quad &x=-1,\frac16 \;\text{and}\; -\frac{1}{\sqrt{6}}<x<\frac1{\sqrt{6}} \tag5\\[6pt]
\implies\quad &x=\frac16 \tag6
\end{align}$$
My doubt is that  other two Properties of tan; i.e.,
$$\tan^{−1}x + \tan^{−1}y = \begin{cases}
\phantom{-}\pi+\tan^{−1}\dfrac{x+y}{1−xy}, &\text{for}\; xy>1, x>0, y>0 \\[6pt] 
-\pi+\tan^{−1}\dfrac{x+y}{1−xy}, &\text{for}\; xy>1, x<0, y<0
\end{cases}$$

Why we have not used these two properties at step $(2)$?

As it may be possible that some value of $x$ satisfies the condition of these properties and we get one more solution of this equation.  Am I thinking wrong so? Please tell me. And if I am right, then in book why only one property has been used?
 A: For any real number $t$, whether $t = \frac{x+y}{1-xy}$ or
$t = \frac{2x+3x}{1-(2x)(3x)}$ or something else,
$\tan^{-1}(t)$ is always an angle within the interval
$\left(-\frac\pi2, \frac\pi2\right)$, that is,
$$ -\frac\pi2 < \tan^{-1}(t) < \frac\pi2.$$
It is therefore not possible that
$$ \pi + \tan^{-1}(t) = \frac\pi4.$$
That equation could only be satisfied if $\tan^{-1}(t) = -\frac{3\pi}{4},$
but the $-\frac{3\pi}{4}$ is not a possible value of $\tan^{-1}(t)$ for any real $t.$
It is also not possible that
$$ -\pi + \tan^{-1}(t) = \frac\pi4.$$
Therefore the only possible case is the one where
$$ \tan^{-1}(t) = \frac\pi4.$$
That is, the only possible solution is $t = 1.$
Of course you must then figure out if there is an $x$ that fits your problem such that $t = 1,$ but you seem to have worked that part out already.

But I would approach it this way:
Let $\alpha = \tan^{-1}(2x)$. Let $\beta = \tan^{-1}(3x)$.
Then $\frac\pi4 = \alpha + \beta $ and therefore
$$ \tan\left(\frac\pi4\right) = \tan(\alpha + \beta) =
     \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}. $$
But $\tan(\alpha) = \tan\left(\tan^{-1}(2x)\right) = 2x$, and so forth.
This way we never use the formula for adding two arc tangents, so we never even have to think about its three cases, much less explain why two of them do not apply.
Note that the order of function applications is important here:
$\tan\left(\tan^{-1}(t)\right) = t$ always,
but $\tan^{-1}\left(\tan(\theta)\right)$ is not always equal to $\theta$.
Do you see why?
