Is there a way to write this expression differently: $\arctan 1/(1+n+n^2)$? $$ \arctan\left(\frac{1}{1+n+n^2}\right)$$
My professor wrote this as
$$\arctan(n+1) - \arctan(n)$$ 
I don't understand how this expression is right?

We have $$\tan^{-1}\left( \frac{1}{1+n+n^2}\right)$$ $$=\tan^{-1}\left( \frac{(n+1)-n}{1+n(n+1)}\right)$$ Now, let $\color{blue}{n+1=\tan\alpha}$ ($\implies \color{blue}{\alpha=\tan^{-1}(n+1)}$) & $\color{blue}{n=\tan\beta}$ ($\implies \color{blue}{\beta=\tan^{-1}(n)}$) $\forall \quad \alpha>\beta$ 
Thus by sustituitng the correspoding values in above expression, we get $$\tan^{-1}\left( \frac{\tan \alpha-\tan\beta}{1+\tan\alpha\tan \beta}\right)$$ $$=\tan^{-1}\left( \tan(\alpha-\beta)\right)$$$$=\alpha-\beta $$ $$=\color{green}{\tan^{-1}(n+1)-\tan^{-1}(n)}$$
