Q. Let $\displaystyle\sum_{k=1}^{\infty} \cot ^{-1}\left(\frac{k^2}{8}\right)=\left(\frac{p}{q}\right) \pi$ where $\frac{p}{q}$ is rational in its lowest form. Find the value of $(p+q)$.
Inverse Trigonometric Functions
Solution:
$T _{ k }=\tan ^{-1}\left(\frac{8}{ k ^2}\right)=\tan ^{-1}\left(\frac{2}{\frac{ k ^2}{4}}\right)=\tan ^{-1}\left(\frac{2}{1+\left(\frac{ k ^2}{4}-1\right)}\right)$
$=\tan ^{-1}\left(\frac{2}{1+\left(\frac{ k }{2}+1\right)\left(\frac{ k }{2}-1\right)}\right)=\tan ^{-1}\left(\frac{\left(\frac{ k }{2}+1\right)-\left(\frac{ k }{2}-1\right)}{1+\left(\frac{ k }{2}+1\right)\left(\frac{ k }{2}-1\right)}\right)$
