Question

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

Answer 1

Answer: 5

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