Question

If tan-1(x -1)/(x+2) + tan -1(x+1)/(x+2) =π/4 , then x is equal to

Answer 1

Use the identity,

$\tan ^{-1}a+\tan ^{-1}b=\tan ^{-1}(\frac{a+b}{1-ab})$.

Then the given equation is

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

$\frac{2x(x+2)}{(x+2)^2-(x^2-1)}=1\\ \frac{2x(x+2)}{4x+5}=1\\ 2x(x+2)=4x+5\\ 2x^2=5\\ x=\pm \sqrt{\frac{5}{2}}$

The only admissible value is $x= \sqrt{\frac{5}{2}}$