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.
 A: The proof can be established starting from the angle addition identity $$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$  Then with $x = \frac{\pi}{2} - a$, $y = -b$, we get $$\cos(a+b) = \sin(\tfrac{\pi}{2} - a - b) = \cos a \cos b - \sin a \sin b.$$  We then find $$\tan(x+y) = \frac{\sin(x+y)}{\cos(x+y)} = \frac{(\sin x \cos y + \cos x \sin y)/(\cos x \cos y)}{(\cos x \cos y - \sin x \sin y)/(\cos x \cos y)} = \frac{\tan x + \tan y}{1 - \tan x \tan y}.$$  Now letting $u = \tan x$, $v = \tan y$, we get $$\tan(\tan^{-1} u + \tan^{-1}v) = \frac{u+v}{1-uv},$$ or $$\tan^{-1} u + \tan^{-1} v = \tan^{-1} \frac{u+v}{1-uv}.$$  Now letting $u = x$, $v= x^3$, we find $$\frac{u+v}{1-uv} = \frac{x+x^3}{1-x^4} = \frac{x(1+x^2)}{(1-x^2)(1+x^2)} = \frac{x}{1-x^2},$$ and the result immediately follows.
A: 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.
A: Part I)
Taking the tan of the RHS, we have a form $tan(A+B)$ where $A = tan^{-1}(x)$, $B = tan^{-1}(x^3)$.
$$
tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)*tan(B)}
= \frac{x + x^3}{1 - x^4}
= \frac{x(1 + x^2)}{(1-x^2)(1+x^2)}
= \frac{x}{1-x^2}
$$
Now the tan of LHS is also $\frac{x}{1-x^2}$.
Therefore the identity is true.
