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

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

The correct answer is Option (3) → \(\frac{\pi}{4}\)

$\tan^{-1}A-\tan^{-1}B=\tan^{-1}(\frac{A-B}{1+AB})$

$⇒\frac{x}{y}-\frac{x-y}{x+y}=\frac{x(x+y)-y(x-y)}{y(x+y)}$

$=\frac{x^2+y^2}{y(x+y)}$

and,

$1+\frac{x}{y}-\frac{x-y}{x+y}=\frac{x^2+y^2}{y(x+y)}$

$⇒\tan^{-1}\frac{x}{y}-\tan^{-1}(\frac{x-y}{x+y})=\tan^{-1}\left(\frac{x^2+y^2}{y(x+y)}×\frac{y(x+y)}{x^2+y^2}\right)$

$=\tan^{-1}1=\frac{\pi}{4}$