Question

Sum infinite terms of the series ${\cot }^{-1}(1^2+\frac{3}{4})+{\cot }^{-1}(2^2+\frac{3}{4})+{\cot }^{-1}(3^2+\frac{3}{4})+\dots$ is

• A. $\pi /4$
• B. ${\tan }^{-1}2$
• C. ${\tan }^{-1}3$
• D. $Noneofthese$

Answer 1

The correct option is A $\pi /4$

Cot ${}^{-1}(1^2+\frac{3}{4})+{\cot }^{-1}(2^2+\frac{3}{4})+\dots \dots \dots ...\infty$

$\begin{array}{cc}{T}_n& ={\cot }^{-1}(n^2+\frac{3}{4})={\cot }^{-1}\left(\frac{4n^2+3}{4}\right)\\ & ={\tan }^{-1}\left(\frac{4}{4n^2+3}\right)={\tan }^{-1}\left(\frac{1}{n^2+\frac{3}{4}}\right)\\ & ={\tan }^{-1}\left(\frac{1}{n^2+1-1+\frac{3}{4}}\right)\\ & ={\tan }^{-1}\left(\frac{1}{1+n^2-\frac{1}{4}}\right)\\ & ={\tan }^{-1}\left(\frac{1}{1+(n-\frac{1}{2})(n+\frac{1}{2})}\right)\\ & ={\tan }^{-1}\left(\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n-\frac{1}{2})(n+\frac{1}{2})}\right)\end{array}$

$\begin{array}{cc}{T}_n& ={\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(n-\frac{1}{2})\\ \sum _{n=1}^n{T}_n=& {\tan }^{-1}\left(\frac{3}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{5}{2}\right)-{\tan }^{-1}\left(\frac{3}{2}\right)\\ & +\dots .....+{\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(K-\frac{1}{2})\\ & ={\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}\left(\frac{1}{2}\right)\end{array}$

$\begin{array}{cc}\text{Required sum}& =\underset{n\to \infty }{lim}\sum _{n=1}^n{T}_n\\ & =\underset{n\to \infty }{lim}\{{\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}\left(\frac{1}{2}\right)\}\\ & =\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{2}\right)\\ & ={\cot }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\left(2\right)\\ \therefore \text{Answer: option}& \text{(B).}\end{array}$