Question

If $ta{n}^{-1}\left(\frac{x-1}{x-2}\right)+ta{n}^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{2}$ then find the value of x

Answer 1

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

$\to {\tan }^{-1}[(\frac{x-1}{x-2}+\frac{x+1}{x+2})/1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)]={\tan }^{-1}0$

$\therefore [1/1-\left(\frac{x^2-1}{x^2-4}\right)]=0$

$\therefore \left[1/\frac{(x^2-4)-(x^2-1)}{x^2-21}\right]=0$

$\frac{x^2-4}{-3}=0$

$x^2=4$

Ans. $\therefore x=\pm 2$