\(\tan^{-1}(\frac{x}{y})-\tan^{-1}(\frac{x-y}{x+y})\) is equal to

\(\frac{\pi}{2}\) \(\frac{\pi}{3}\) \(\frac{\pi}{4}\) \(-\frac{5\pi}{4}\)

\(\frac{\pi}{4}\)

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