Question

If $\tan \left(\sum _{r=1}^{19}{\tan }^{-1}\frac{1}{1+r+r^2}\right)=\frac{a}{b}$, where, a and b are coprime, then find the value of $(a+b)$.

• A. $40$
• B. $50$
• C. $60$
• D. $70$

Answer 1

The correct option is A $40$

$tan\left(\sum _{r=1}^{19}ta{n}^{-1}\frac{1}{1+r+r^2}\right)=\frac{a}{b}$
$\Rightarrow tan\left(\sum _{r=1}^{19}{\tan }^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)\right)=\frac{a}{b}$
$\Rightarrow tan\left(\sum _{r=1}^{19}ta{n}^{-1}\right(r+1)-ta{n}^{-1}(r\left)\right)=\frac{a}{b}$
$\Rightarrow \tan (({\tan }^{-1}2-{\tan }^{-1}1)+({\tan }^{-1}3-{\tan }^{-1}2)+...({\tan }^{-1}20-{\tan }^{-1}19))=\frac{a}{b}\Rightarrow \tan ({\tan }^{-1}20-{\tan }^{-1}1)=\frac{a}{b}\Rightarrow \tan \left({\tan }^{-1}\left(\frac{20-1}{1+20.1}\right)\right)=\frac{a}{b}\Rightarrow \frac{a}{b}=\frac{19}{21}$
$\Rightarrow a+b=40$