Question

Find the value of x which satisfy equation: ${\tan }^{-1}\frac{x-1}{x+2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$

• A. $x=+\sqrt{5/2}$
• B. $x=-\sqrt{5/2}$
• C. $x=\pm \sqrt{5/2}$
• D. None of these

Answer 1

The correct option is A $x=+\sqrt{5/2}$

Given, ${\tan }^{-1}\frac{x-1}{x+2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$
${\tan }^{-1}\frac{\frac{x-1}{x+2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x+2}\right)\left(\frac{x+1}{x+2}\right)}=\frac{\pi }{4}$
$\therefore \left(\frac{2x(x+2)}{x^2+4+4x-x^2+1}\right)=\tan \frac{\pi }{4}$
$\therefore \frac{2x(x+2)}{4x+5}=\tan \frac{\pi }{4}=1$

$2x^2+4x=4x+5$
$\therefore x=\pm \sqrt{\frac{5}{2}}$
But for $x=-\sqrt{5/2}$, L.H.S. is negative. Hence, $x=\sqrt{5/2}$