Question

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

• A. $\tan (\frac{\pi }{2}-{\cot }^{-1}2)$
• B. $3$
• C. $2$
• D. $\tan ({\tan }^{-1}2$)

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\tan (\frac{\pi }{2}-{\cot }^{-1}2)$
C $2$
D $\tan ({\tan }^{-1}2$)

$\text{solution:}{\sum }_{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{r^2+\frac{3}{4}}\right)$

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

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

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

$=\tan x^{-1}\frac{3}{2}-{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{5}{2}-\tan x^{-1}\frac{3}{2}+.$

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

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

$=\frac{\pi }{2}-{\cot }^{-1}\left(2\right)$

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

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

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

$=2$

Answer: option (B)