Question

Find the sum of the values of $x$ satisfying the equation $\tan ^{-1}\left(\frac{2 x-1}{10}\right)+\tan ^{-1}\left(\frac{1}{2 x}\right)=\displaystyle\sum_{n=2}^{3} \cot ^{-1}(n) .$

Answer 1

$\displaystyle\sum_{n=2}^{3} \cot ^{-1}(x)=\frac{\pi}{4}=\tan ^{-1} 1$
$\therefore \tan ^{-1} \frac{2 x-1}{10}=\tan ^{-1} 1-\tan ^{-1} \frac{1}{2 x} $
$\frac{2 x-1}{10}=\frac{1-\frac{1}{2 x}}{1+\frac{1}{2 x}}$
$ \Rightarrow(2 x-1)(2 x-9)=0 $
$\Rightarrow x=\frac{1}{2}$
$x=\frac{9}{2}$
$\therefore $ Sum of all $x=5$