Find the solutions of an equation with arctan?

Question

I have to show that $1$ and $\frac{-1}{\sqrt{3}}$ are (maybe not) solutions of the following equation: $arctan(x)+arctan(x\sqrt{3})= \frac{7\pi}{12}$.

How can I do that ?

Thank you in advance

Answer 1

Hint

Consider $$f(x)=\tan^{-1}(x)+\tan^{-1}(x\sqrt{3})$$ and now, examine the two cases $$f(1)=\tan^{-1}(1)+\tan^{-1}(\sqrt{3})=\frac{\pi}{4}+\frac{\pi}{3}=\frac{7\pi}{12}$$ $$f(-\frac{1}{\sqrt3})=\tan^{-1}(-\frac{1}{\sqrt3})+\tan^{-1}(-1)=-\frac{\pi}{6}-\frac{\pi}{4}=-\frac{5\pi}{12}$$

I am sure that you can take from here.

Answer 2

Plug in the values in the equation and check if it holds. ie;

$$\tan^{-1} (1)+\tan^{-1}\left (\sqrt3 \cdot 1\right )=\frac{\pi}{4}+\frac{\pi}{3}=\frac{7\pi}{12}$$ ergo $x=1$ is a solution.