Question

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

Answer 1

Consider $\displaystyle\sum^{4}_{r = 1} \tan ^{-1} \frac{1}{2 r^{2}}=\displaystyle\sum^{4}_{r=1} \tan ^{-1}\left(\frac{2}{1+4r^2-1}\right)$
$\displaystyle\sum_{r=1}^{4} \tan ^{-1}\left(\frac{(2 r+1)-(2 r-1)}{1+(2 r+1)(2 r-1)}\right)=\displaystyle\sum_{r=1}^{4}\left(\tan ^{-1}(2 r+1)-\tan ^{-1}\right.(2 r-1))$
$=\left(\tan ^{-1} 3-\tan ^{-1} 1\right)+\left(\tan ^{-1} 5-\tan ^{-1} 3\right)+\left(\tan ^{-1} 7-\tan ^{-1} 5\right)$
$+\left(\tan ^{-1} 9-\tan ^{-1} 7\right)$
$=\left(\tan ^{-1} 9-\tan ^{-1} 1\right)$
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 $\frac{5}{4}=\frac{3 x+4}{3 x+2}$
$\Rightarrow 15 x+10=12 x+16$
$\Rightarrow 3 x=6$
$\therefore x=2$