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
$$
\sum_{n=1}^\infty \tan^{-1} \frac{1}{2n^2} = \frac{\pi}{4}
$$
 A: The RHS sign is wrong, it should be:

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

The Rule is simply 

$$ \tan^{-1} \frac{u -v}{1+u v}  = \tan^{-1} u - \tan^{-1} v $$ 

A: Hint: Apply the inverse function of $\tan^{-1}(x)$ which is $\tan(x)$ to both sides of the equation and then you can use the double angle formula.
A: Observe that $\tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}$.  Then, letting $x=\arctan(\frac{1}{2n+1})$ and $y=\arctan(\frac{1}{2n-1})$, we see that 
$$\begin{align}
\tan(x-y)&=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\\
&=\frac{\frac{1}{2n+1}-\frac{1}{2n-1}}{1+\frac{1}{2n+1}\frac{1}{2n-1}}\\
&=\frac{(2n-1)-(2n+1)}{1+(2n-1)(2n+1)}
\end{align}$$
Taking the inverse tangent of both sides of this last expression reveals
$$\begin{align}
x-y&=\arctan\left(\frac{1}{2n+1}\right)-\arctan\left(\frac{1}{2n-1}\right)\\
&=\arctan\left(\frac{(2n-1)-(2n+1)}{1+(2n-1)(2n+1)}\right)
\end{align}$$
A: $$
\tan^{-1} \frac1{2n-1} = \arg (2n-1 + i) \\
\tan^{-1} \frac1{2n+1} = \arg (2n+1 + i)
$$
hence
$$
\tan^{-1} \frac1{2n-1} - \tan^{-1} \frac1{2n+1} = \arg \frac{2n-1+i}{2n+1+i} \\
= \arg(2n-1+i)(2n+1-i) = \arg (4n^2 +2i) =\tan^{-1}\frac1{2n^2} \\
=\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n-1)(2n+1)}
$$
