Question

Solve for $x$:

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

Answer 1

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

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

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

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

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

$\Rightarrow 2x^2=7$

$\Rightarrow x^2=\frac{7}{2}$

$\Rightarrow x=\pm \sqrt{\frac{7}{2}}$.