Question

If tan^-1$\frac{x-1}{x+2}$ + tan^-1$\frac{x+1}{x+2}$ = $\frac{\pi }{4}$, then find the value of x.

Answer 1

L.H.S. =  tan^-1$\frac{x-1}{x+2}$ + tan^-1$\frac{x+1}{x+2}$ 

            =${\tan }^{-1}[\frac{\frac{x-1}{x+2}+\frac{x+1}{x+2}}{1-(\frac{x-1}{x+2})(\frac{x+1}{x+2})}]$

             =${\tan }^{-1}[\frac{2x^2-4}{-3}]$

             =$\frac{\pi }{4}=ta{n}^{-1}(1)$

$\therefore \frac{2x^2-4}{-3}=1$

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