Question

If ${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$, then find the values of $x$.

Answer 1

Given ${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$

${\tan }^{-1}\frac{\left(\frac{x-1}{x-2}\right)+\left(\frac{x+1}{x+2}\right)}{1-\frac{x^2-1}{x^2-4}}=\frac{\pi }{4}$

${\tan }^{-1}\frac{(x-1)(x+2)+(x-2)(x+1)}{-3}=\frac{\pi }{4}$

$x^2+x-2+x^2-x-2=-3$

$2x^2=1$

$x=\pm \frac{1}{\sqrt{2}}$