Question

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

Answer 1

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

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

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

             =\({\pi\over4}=tan^{-1}(1) \)

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

\(\therefore x=\pm\frac{1}{\sqrt2}\)