Question

The sum of the infinite terms of the series ${\cot }^{-1}(1^2+\frac{3}{4})+{\cot }^{-1}(2^2+\frac{3}{4})+{\cot }^{-1}(3^2+\frac{3}{4})+..$ is equal to:

• A. ${\tan }^{-1}\left(1\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\tan }^{-1}\left(3\right)$
• D. ${\tan }^{-1}\left(4\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(1\right)$

${\cot }^{-1}(1^2+\frac{3}{4})+{\cot }^{-1}(2^2+\frac{3}{4})+{\cot }^{-1}(3^2+\frac{3}{4})$

$+\dots$

$\Rightarrow$ Take common terms,

$T_r={\cot }^{-1}(x^2+\frac{3}{4})$

$={\cot }^{-1}\left(\frac{4x^2+3}{4}\right)$

$={\tan }^{-1}\left(\frac{4}{4r^2+3}\right)$

$2{\tan }^{-1}\left(\frac{1}{r^2+\frac{3}{4}}\right)$

$={\tan }^{-1}\left(\frac{1}{1+r^2-\frac{1}{4}}\right)$

$={\tan }^{-1}\left(\frac{x+1/2-(r-\frac{1}{2})}{1+(r+\frac{1}{2}){(r-\frac{1}{2})}^2}\right)$

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

Now,

${\sum }_{r=1}^{\infty 0}$

${\tan }^{-1}(1+\frac{1}{2})-{\tan }^{-1}\left(\frac{1}{2}\right)$

$r=2;{\tan }^{-}\left(\frac{2+1}{2}\right)-{\tan }^{-1}(2-\frac{1}{2})$

$r=\infty$

$={\tan }^{-1}\infty$

$={\tan }^{-1}\infty -\tan \frac{-1}{2}$

$=\frac{\pi }{2}-{\tan }^{-1}\frac{1}{2}$

$(\because {\tan }^{-1}\infty =\frac{\pi }{2})$

$={\cot }^{-1}\frac{1}{2}$

$\therefore \tan xx+{\cot }^{-1}x=\frac{\pi }{2}$

or ${\tan }^{-1}2$

${\cot }^{-1}x=\frac{\pi }{2}-{\tan }^{-1}x$

$\text{option}B={\tan }^{-1}2$