Question

Solution of the equation $\cot \left(\sum _{r=1}^4{\cot }^{-1}2r^2\right)=\frac{3x+4}{3x+2}$ is equal to

Answer 1

Answer: 2

Consider $\sum _{r=1}^4{\tan }^{-1}\frac{1}{2r^2}=\sum _{r=1}^4{\tan }^{-1}\left(\frac{2}{1+4r^2-1}\right)$
$\sum _{r=1}^4{\tan }^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)=\sum _{r=1}^4({\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1))$
$=\left({\tan }^{-1}3-{\tan }^{-1}1\right)+\left({\tan }^{-1}5-{\tan }^{-1}3\right)+\left({\tan }^{-1}7-{\tan }^{-1}5\right)$
$+\left({\tan }^{-1}9-{\tan }^{-1}7\right)$
$=\left({\tan }^{-1}9-{\tan }^{-1}1\right)$
Now cot $\left({\tan }^{-1}9-{\tan }^{-1}1\right)=\frac{1}{\tan \left({\tan }^{-1}9-{\tan }^{-1}1\right)}$
$=\frac{1+9\times 1}{9-1}=\frac{10}{8}=\frac{5}{4}$
$\therefore$ We have $\frac{5}{4}=\frac{3x+4}{3x+2}$
$\Rightarrow 15x+10=12x+16$
$\Rightarrow 3x=6$
$\therefore x=2$