<s>[INST] 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.
 [/INST] 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.
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