Finding $\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$ 
Find the sum of
  $$\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$$

I tried to convert all into 
$\tan {}^{ - 1}(x) $
using 
$\cot {}^{ - 1} (x) = { \tan {}^{ - 1} ( \frac{1}{x} ) }$ 
and then tried to simplify them using property 
$ \tan {}^{ - 1} ( \frac{x - y}{1 - xy} )  = \tan {}^{ - 1} (x) - \tan {}^{- 1} (y)  $
so that some terms cancel each other.
But I do not manage to simplify the problem. What to do?

After the fix of sign in front of $\frac{5}{12}$ in the question body, the sum becomes pretty simple.
Let $a_{\pm} = 3(r\pm \frac12)$, notice
$$\frac{a_+ - a_-}{1 + a_+ a_-} = \frac{3}{1 + 9(r^2-\frac14)}
= \frac{12}{36r^2 - 5} = \left(3r^2 - \frac{5}{12}\right)^{-1}$$
We have
$$\begin{align}\cot^{-1}\left(3r^2 - \frac{5}{12}\right)
&= \tan^{-1}\left(\frac{a_+ - a_-}{1 + a_+a_-}\right)
= \tan^{-1}a_+ - \tan^{-1}a_-\\
&= \tan^{-1}(3(r + \frac12)) - \tan^{-1}(3(r - \frac12))
\end{align}
$$
The new sum is a telescoping one and
$$\sum_{r=1}^n
\cot^{-1}\left(3r^2 - \frac{5}{12}\right)
= \tan^{-1}(3(n + \frac12)) - \tan^{-1}(\frac32)$$
