Question

Find all 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

We know that, ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-1}\left(\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{(x-1)(x+2)+(x+1)(x-2)}{(x^2-4)-(x^2-1)}=\tan \frac{\pi }{4}$
$\Rightarrow \frac{2x^2-4}{-3}=1$
$\Rightarrow 2x^2=1$
$\Rightarrow x=\pm \frac{1}{\sqrt{2}}$