If x > 0, y > 0 and x > y, then ${\tan }^{-1} (x/y)+$${\tan }^{-1} \left[\right(x+y)/(x-y\left)\right]$ is equal to

Since $\frac{x}{y}\cdot \frac{x+y}{x-y}>1$ The given expression is equal to

$\begin{array}{l}\pi +{\tan }^{-1} \left[\frac{\frac{x}{y}+\frac{x+y}{x-y}}{1-\frac{x}{y}\times \frac{x+y}{x-y}}\right]\\ =\pi +{\tan }^1 \frac{x^2+y^2}{-\left(x^2+y^2\right)}=\pi +{\tan }^{-1} (-1)=3\pi /4.\end{array}$