Question

If $\sum _{r=1}^{10}{\tan }^{-1}\left(\frac{3}{9r^2+3r-1}\right)={\cot }^{-1}\left(\frac{m}{n}\right)$ (where $m$ and $n$ are coprime), then find the value of $(m-n)$

Answer 1

Answer: 7

We have
$a_n={\tan }^{-1}\left(\frac{3}{9n^2+3n-1}\right)={\tan }^{-1}\left(\frac{3}{1+(3n+2)(3n-1)}\right)={\tan }^{-1}\left(\frac{(3n+2)-(3n-1)}{1+(3n+2)(3n-1)}\right)$
$={\tan }^{-1}(3n+2)-{\tan }^{-1}(3n-1)$
$\therefore \text{ Sum of first }10\text{ terms }=\sum _{r=1}^{10}{a}_r=\sum _{r=1}^{10}\left({\tan }^{-1}(3r+2)-{\tan }^{-1}(3r-1)\right)$
$=\left({\tan }^{-1}5-{\tan }^{-1}2\right)+\left({\tan }^{-1}8-{\tan }^{-1}5\right)+\dots \dots ..+\left({\tan }^{-1}32-{\tan }^{-1}29\right)$
$={\tan }^{-1}32-{\tan }^{-1}2={\tan }^{-1}\left(\frac{32-2}{1+32\cdot 2}\right)={\tan }^{-1}\frac{30}{65}={\tan }^{-1}\left(\frac{6}{13}\right)={\cot }^{-1}\left(\frac{13}{6}\right)={\cot }^{-1}\left(\frac{m}{n}\right)$
$\therefore m=13\text{ and }n=6$
$\text{ Hence }(m-n)=13-6=7$