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

(A) $\frac{\pi }{2}$

(B) $\frac{\pi }{3}$

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

(D) $\frac{{ - 3\pi }}{4}$

Option C is correct

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

$= {\tan ^{ - 1}}\left[ {\frac{{x(x + Y) - y(x - y)}}{{y(x + y) + x(x - y)}}} \right] = {\tan ^{ - 1}}\left( {\frac{{{x^2} + {y^2}}}{{{x^2} + {y^2}}}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4}.$