# 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 $$\arctan(-2)+\arctan(-3)=-135^{\circ}<0$$ By the way, $\tan(-135^{\circ})=1.$

• but then isn't -1 a valid solution? – Raknos13 Dec 15 '17 at 8:17
• @mettled mike You can see that $-1$ is not root of the equation. We see that $\tan45^{\circ}=\tan(-135^{\circ})$, but $\arctan\tan45^{\circ}\neq-135^{\circ}.$ – Michael Rozenberg Dec 15 '17 at 8:19

$\tan(x)$ is negative in the interval $\left(-\frac{\pi}{2},0\right)$, so it's inverse on the interval $(-\infty,0)$ will give a value from the interval $\left(-\frac{\pi}{2},0\right)$.

$\arctan x + \arctan y = \arctan\left(\dfrac{x+y}{1-xy}\right), \quad\quad xy < 1$

So

$x = -1$ doesn't work as $2x \times 3x = 6x^2 = 1 \nless 1$.

$x=\frac16$ works as $6x^2 = \frac1{6} \lt 1$.

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 $\frac{\pi}{4}$.