Question

If $S=\sum _{r=1}^{90}{\text{tan}}^{-1}\left(\frac{2r}{2+r^2+r^4}\right),$ then $8190\left(\cot S\right)$ is equal to

• A. $8192$
• B. $8190$
• C. $9108$
• D. $8109$

Answer 1

The correct option is A $8192$

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

$={\text{tan}}^{-1}\left(\frac{(r^2+1+r)-(r^2+1-r)}{1+(r^2+1+r)(r^2+1-r)}\right)$
$={\text{tan}}^{-1}(r^2+1+r)-{\text{tan}}^{-1}(r^2+1-r)$
$S=\sum _{r=1}^{90}[{\text{tan}}^{-1}(r^2+1+r)-{\text{tan}}^{-1}(r^2+1-r)]$
$={\tan }^{-1}\left(8191\right)-{\text{tan}}^{-1}\left(1\right)$
$={\tan }^{-1}\left(\frac{8190}{8192}\right)$
$\Rightarrow \tan S=\frac{8190}{8192}$
$\Rightarrow \cot S=\frac{8192}{8190}$
$\therefore 8190\cot S=8192$