<s>[INST] 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?   
 [/INST] 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 $$\tan^{-1} (n+1)- \frac {\pi}{4}$$
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