Question

If ${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$. Find the value of x.

Answer 1

Using the formula,${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

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

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

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

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

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

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

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

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