I came to this identity while doing some indefinite integrals.
$\arctan(x) + \arctan(x^3) = \arctan(2x+\sqrt{3}) + \arctan(2x-\sqrt{3})$
Seems weird to me, no idea why it's correct but it is.
I wonder if there's some geometric or trigonometric reasoning/insight behind it,
say something which can be useful to high-school students for solving some problems.
Any ideas?
Hint: \begin{eqnarray*} \tan^{-1}(A)+\tan^{-1}(B)=\tan^{-1} \left( \frac{A+B}{1-AB} \right). \end{eqnarray*}
$$\tan^{-1}(x)+\tan^{-1}(x^3)=\tan^{-1} \left( \frac{x(1+x^2)}{1-x^4}\right)=\tan^{-1} \left( \frac{x}{1-x^2} \right).$$
\begin{eqnarray*}\tan^{-1}(2x+\sqrt{3})+\tan^{-1}(2x-\sqrt{3})&=&\tan^{-1} \left( \frac{4x}{1-(4x^2-3)} \right)\\ &=&\tan^{-1} \left( \frac{x}{1-x^2} \right).\end{eqnarray*}
Taking the tangent of both members,
$$\frac{x+x^3}{1-x\,x^3}=\frac{4x}{1-(4x^2-3)}$$ is an identity. (Simplify $1+x^2$.)
An alternative will use complex number
$$ \begin{aligned} \arg{\left((1+ix)(1+ix^{3})\right)}=\arg{\left((1+i(2x+\sqrt{3}))(1+i(2x-\sqrt{3}))\right)} \end{aligned} $$
