Question

the possible values of x,which satisfy the trigonometric equation
${tan}^{-1}\left(\frac{x-1}{x-2}\right)+{tan}^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$

Answer 1

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

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

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

$\Rightarrow \frac{x^2+x-2+x^2-x-2}{-3}=1$

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

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

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