SOLUTION : L.H.S. $=\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right) \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$
$=\tan ^{-1}\left[\frac{\frac{x}{y}-\left[\frac{x-y}{x+y}\right]}{1+\frac{x}{y}\left(\frac{x-y}{x+y}\right)}\right]=\tan ^{-1}\left[\frac{x^2+x y-x y+y^2 /(x+y) y}{x y+y^2+x^2-x y /(x+y) y}\right]=\tan ^{-1}\left[\frac{x^2+y^2}{x^2+y^2}\right]=\tan ^{-1} 1=\tan ^{-1} \cdot \tan \frac{\pi}{4}=\frac{\pi}{4}$
