Get lost in $\cot$

Let $\displaystyle \theta_n = \cot^{-1} \left(1+\sum_{k=1}^n 2k\right) = \cot^{-1} \left(n^2+n+1\right)$ and $\displaystyle S_m = \sum_{n=1}^m \theta_n$ . Then we have:

\(\begin{array} {} m = 1 & \implies S_1 = \theta_1 = \cot^{-1} 3 & \implies S_{\color{blue}1} = \cot^{-1} \left(\dfrac {{\color{blue}1}+2}{\color{blue}1} \right) \\ m = 2 & \implies S_2 = \theta_1 + \theta_2 = \cot^{-1} 3 + \cot^{-1} 7 = \cot^{-1} \dfrac {3\times 7 - 1}{3 + 7} = \cot^{-1} 2 & \implies S_{\color{blue}2} = \cot^{-1} \left(\dfrac {{\color{blue}2}+2}{\color{blue}2} \right) \\ m = 3 & \implies S_3 = S_2 + \theta_3 = \cot^{-1} 2 + \cot^{-1} 13 = \cot^{-1} \dfrac {2\times 13 - 1}{2 + 13} = \cot^{-1} \dfrac 53 & \implies S_{\color{blue}3} = \cot^{-1} \left(\dfrac {{\color{blue}3}+2}{\color{blue}3} \right) \end{array} \)

This implies that $S_m = \cot^{-1} \dfrac {m+2}m$ , which can be proven by induction as follows:

$\begin{aligned} S_{m+1} & = S_m + \cot^{-1} \left((m+1)^2 + m+1+1\right) \\ & = \cot^{-1} \frac {m+2}m + \cot^{-1} \left(m^2+3m+3\right) \\ & = \cot^{-1} \frac {\frac {m+2}m(m^2+3m+3)-1}{\frac {m+2}m+m^2+3m+3} \\ & = \cot^{-1} \frac {m^3+5m^2+8m+6}{m^3+3m^2+4m+2} \\ & = \cot^{-1} \frac {(m+3)(m^2+2m^2+2)}{(m+1)(m^2+2m^2+2)} \\ & = \cot^{-1} \frac {m+3}{m+1} & \small \color{#3D99F6} \text{Proven} \end{aligned}$

Therefore, $\displaystyle X_m = \cot\left(\sum_{n=1}^m \cot^{-1} \left(1 + \sum_{k=1}^n 2k\right) \right) = \cot S_m = \cot \left(\cot^{-1} \frac {m+2}m \right) = \frac {m+2}m$ . For $m=4$ , $X_4 = \dfrac {4+2}4 = \dfrac 32$ . $\implies A-B = 3-2 = \boxed 1$ .

$\sum_{i=1}^n2i = 2(\frac {n(n+1)}{2})=n(n+1)$

$cot^{-1}(n^2+n+1)=tan^{-1}(\frac {1}{1+n(n+1)})$

= $tan^{-1}\frac {(n+1)-(n)}{1+n(n+1)}=tan^{-1}(n+1)-tan^{-1}n$

Summation of this upto n=4 will give you $tan^{-1}5-tan^{-1}1$

Ans:- $\frac {1}{tan(tan^{-1}5 - \frac {π}{4})}= \frac{1}{\frac {5-1}{1+5\cdot 1}}=\frac {6}{4}= \frac{3}{2}$

Thus, the answer = 3 - 2 = 1.