Sum of the series $\csc^{-1} \sqrt{10}+ \csc^{-1} \sqrt{50}+\csc^{-1}\sqrt{170}...$ 
Find the sum of the series:
   $\csc^{-1}\sqrt{10}+ \csc^{-1}\sqrt{50}+\csc^{-1}\sqrt{170}...\csc^{-1}\sqrt{(n^2+1)(n^2+2n+2)}$

I converted the series to $\sum^{n} _{i=0}\arcsin \dfrac{1}{\sqrt{(i^2+1)(i^2+2i+2)}}$ and then tried to use $\arcsin(x)+\arcsin(y)= \arcsin({x\sqrt{1+y^2}+y\sqrt{1+x^2}}) \forall x,y \ge 0 $ and $x^2+y^2\le1$. 
There was no clear pattern in the terms obtained. How to solve this question? And what is the general trick to solve inverse trigonometric summations?   

The given sum can be represented as $$\tan ^{-1} \frac {1}{3}+\tan ^{-1} \frac {1}{7}+\tan ^{-1} \frac {1}{13}\cdots \tan ^{-1} \frac {1}{n^2+n+1}$$
This can be represented as $$\sum_{k=1}^n \tan ^{-1} \left(\frac {1}{k^2+k+1}\right)$$
$$\sum_{k=1}^n \tan ^{-1} \left(\frac {(k+1)-(k) }{1+(k+1)(k)}\right) $$
 $$\sum_{k=1}^n \tan ^{-1} (k+1) -\tan ^{-1} (k)$$
Which telescopes to $$\tan^{-1} (n+1)- \frac {\pi}{4}$$
