proving $\tan^{-1} \left( \frac{x}{1-x^2} \right) = \tan^{-1}x + \tan^{-1}(x^3)$ Two related questions, one easy, one just a bit harder:
1) Prove the identity
$$
\tan^{-1} \left( \frac{x}{1-x^2} \right) =
\tan^{-1}x + \tan^{-1}(x^3)
$$
2) Now try to find a geometric or trigonometric proof of that same geometry, without resorting to calculus.
I'll post an answer to both questions in a couple of days if nobody has one yet.

Part I: Consider $f(x) = tan^{-1}\left(\dfrac{x}{1-x^2}\right) - tan^{-1}x - tan^{-1}(x^3)$, taking derivative of $f$: 
$f'(x) = \dfrac{1+x^2}{1-x^2+x^4} - \dfrac{1}{1+x^2} - \dfrac{3x^2}{1+x^6} = \dfrac{(1+x^2)^2 - (1-x^2+x^4) - 3x^2}{1+x^6} = \dfrac{1+ 2x^2 + x^4 - 1 + x^2 - x^4 - 3x^2}{1+x^6} = 0$. Thus $f$ is constant on $\mathbb{R}$, and $f(x) = f(0) = 0$, and the identity follows.
