Solve for $x$: $\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$ $$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$$
$$\arctan \left(\frac{2x+3x}{1-2x*3x}\right)=\frac {\pi}{4}$$
$$\frac {5x}{1-6x^2}=\tan \frac{\pi}{4}=1$$
$$6x^2 + 5x -1 = 0$$
$$(6x-1)(x+1)=0$$
$$x=-1, \frac{1}{6}$$

The answer however rejects the solution $x=-1$ saying that it makes the L.H.S of the equation negative. I don't understand this, I don't see how $x=-1$ makes the L.H.S. negative.
Because
$$\arctan(-2)+\arctan(-3)=-135^{\circ}&lt;0$$
By the way, $\tan(-135^{\circ})=1.$
Concep: Let $f$ be a function, which means  $$x_1=x_2\Rightarrow f(x_1)=f(x_2)$$
But note that if $f(x_1)=f(x_2)$, then $x_1=x_2$ is one of the possibility, among perhaps many possibilities if the function is not injective

Now Given $\tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}$, taking $\tan$ of both sides(since $\tan$ is a function),we get 
 $$\frac{2x+3x}{1-2x.3x}=1$$

which after simplification gives $x=\frac16,-1$. Taking $x=\frac16$, we want to find what is $\tan^{-1}\frac13+\tan^{-1}\frac12$. For that let us take $\tan^{-1}\frac13=\alpha,\tan^{-1}\frac12=\beta$, which is equal to $\tan\alpha=\frac13,and \tan\beta=\frac12$, where $\alpha,\beta$ being principal values belongs to $(-\frac{\pi}2,\frac{\pi}2)$, which means $\alpha+\beta\in(-\pi,\pi)$. Now we know $$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$ which implies $$\tan(\alpha+\beta)=\frac{\frac13+\frac12}{1-\frac13\frac12}=1$$ Hence $$\alpha+\beta=\tan^{-1}1=\frac{\pi}4$$ or $$\tan^{-1}\frac13+tan^{-1}\frac12=\frac{\pi}4$$ and $x=\frac16$ becomes a root of the above equation

Now take $x=-1$, then the lhs of the equation becomes $$\tan^{-1}(-2)+\tan^{-1}(-3)$$ which is a negative angle $-\frac{3\pi}{4}$ , which can not be a positive number $\frac{\pi}{4}$.