Question

If $ta{n}^{-1}\left(\frac{x-1}{x-2}\right)+ta{n}^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$ then find the value of x

Answer 1

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

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

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

$\Rightarrow \frac{x^2+x-2+x^2-x-2}{-3}=\tan \frac{\pi }{4}$

$\Rightarrow 2x^2-4=-3$

$\Rightarrow 2x^2=-3+4=1$

$\Rightarrow x^2=\frac{1}{2}$

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