Show that $\displaystyle \forall x \in R$, that

$\displaystyle arctan(x+1) - arctanx = arctan\frac{1}{x^2 + x + 1}$

tan is a continuous function, so (can anyone tell me why that is important when doing these kinda proofs?)

$\displaystyle tan(arctan(x+1) - arctanx) = tan\left(arctan\frac{1}{x^2 + x + 1}\right)$

$\displaystyle u = arctan(x+1), v = arctanx \implies$

$\displaystyle tan(u-v) = \frac{tanu - tanv}{1 + tanu \cdot tanv} = tan\left(arctan\frac{1}{x^2 + x + 1}\right)$

Now all I need to prove is that

$\displaystyle \frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$

Suggestions?

(How do you write the symbol for real numbers in latex by the way?)