Question

Prove that:
${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$

Answer 1

$tan-1\frac{x-1}{x-2}+tan-1\frac{x+1}{x+2}=\frac{\pi }{4}$

$ta{n}^{-1}x+ta{n}^{-1}y$= $ta{n}^{-1}\frac{x+y}{1-xy}$

$ta{n}^{-1}\left(\frac{(\frac{x-1}{x+2}+\frac{x+1}{x+2})}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right)$ $\Rightarrow \left(\frac{\frac{(x-1)(x+2)(x+1)(x-2)}{x^2-4}}{\frac{x^2-4-x^2+1}{x^2-4}}\right)$

$\Rightarrow ta{n}^{-1}\left[\frac{x^2+3x-2-x^2-3x-2}{-3}\right]$

$\Rightarrow ta{n}^{-1}\left[\frac{2(x^2-2)}{-3}\right]$ =$\frac{\pi }{4}$

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

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

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

$[x=\frac{+-1}{\sqrt{2}}]$