Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Now $\sum _{n=1}^{50}{\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)=\sum _{n=1}^{50}{\tan }^{-1}\left(\frac{\left(n+1\right)-n}{1+n\left(n+1\right)}\right)$

$=\sum _{n=1}^{50}\left[{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}n\right]$

$={\tan }^{-1}51-{\tan }^{-1}1={\tan }^{-1}\left(\frac{51-1}{1+51}\right)$

$={\tan }^{-1}\left(\frac{50}{52}\right)={\cot }^{-1}\left(\frac{26}{25}\right)$

So $\cot \left(\sum _{n=1}^{50}{\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)\right)=\cot \left({\cot }^{-1}\frac{26}{25}\right)=\frac{26}{25}$