Simplifying An Inverse Tan Function I would like to know how this equality holds.

$$
\tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1} \frac{1}{2n-1} - \tan^{-1} \frac{1}{2n+1}.$$

I was told to use the double angle formula for $\tan \theta$ but I can't seem to show this.
Thank you. 
For some context, I was asked to show
$$
\sum_{n=1}^\infty \tan^{-1} \frac{1}{2n^2} = \frac{\pi}{4}
$$

$$
\tan^{-1} \frac1{2n-1} = \arg (2n-1 + i) \\
\tan^{-1} \frac1{2n+1} = \arg (2n+1 + i)
$$
hence
$$
\tan^{-1} \frac1{2n-1} - \tan^{-1} \frac1{2n+1} = \arg \frac{2n-1+i}{2n+1+i} \\
= \arg(2n-1+i)(2n+1-i) = \arg (4n^2 +2i) =\tan^{-1}\frac1{2n^2} \\
=\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n-1)(2n+1)}
$$
