Question

The value of $\tan \left(\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{4}{4r^2+3}\right)\right)$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

${\tan }^{-1}\left(\frac{4}{4r^2+3}\right)={\tan }^{-1}\left(\frac{1}{r^2+\frac{3}{4}}\right)={\tan }^{-1}\left(\frac{1}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right)={\tan }^{-1}\left(r+\frac{1}{2}\right)-{\tan }^{-1}\left(r-\frac{1}{2}\right)$
$\therefore \sum _{r=1}^n{\tan }^{-1}\left(\frac{4}{4r^2+3}\right)=\left({\tan }^{-1}\left(n+\frac{1}{2}\right)-{\tan }^{-1}\frac{1}{2}\right)$
$\Rightarrow \sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{4}{4r^2+3}\right)={\cot }^{-1}\frac{1}{2}=\left({\tan }^{-1}2\right)$
$\therefore \tan \left({\tan }^{-1}2\right)=2$