$$ \arctan\left(\frac{1}{1+n+n^2}\right)$$
My professor wrote this as
$$\arctan(n+1) - \arctan(n)$$
I don't understand how this expression is right?
$\endgroup$ 14 Answers
$\begingroup$Use the formula
$$\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$$
with $\alpha=\arctan(n+1)$ and $\beta=\arctan n$.
$\endgroup$ $\begingroup$See this diagram (drawn for $n=3$, but the principle is general):
The red line has slope $\dfrac{1}{1+n+n^2}$ and makes an angle of $\tan^{-1}(n+1)-\tan^{-1}(n)$ with the horizontal axis.
$\endgroup$ $\begingroup$$$ \arctan x + \arctan y = \arctan \frac{x+y}{1-xy} $$ Applying this with $x=n+1$ and $y=-n$ we get $$ \frac{x+y}{1-xy} = \frac{(n+1)-n}{1+(n+1)n} = \frac 1 {n^2+n+1} $$
$\endgroup$ $\begingroup$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 $$\tan^{-1}\left( \frac{\tan \alpha-\tan\beta}{1+\tan\alpha\tan \beta}\right)$$ $$=\tan^{-1}\left( \tan(\alpha-\beta)\right)$$$$=\alpha-\beta $$ $$=\color{green}{\tan^{-1}(n+1)-\tan^{-1}(n)}$$
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