MCQ
${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right) = .....$
- ✓$\frac{\pi }{4}$
- B$\frac{\pi }{3}$
- C$\frac{\pi }{2}$
- D$\pi $
${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{\frac{x}{y} - \frac{{x - y}}{{x + y}}}}{{1 + \frac{x}{y}\left( {\frac{{x - y}}{{x + y}}} \right)}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{{x^2} + xy - xy + {y^2}}}{{xy + {y^2} + {x^2} - xy}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{{x^2} + {y^2}}}{{{x^2} + {y^2}}}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4}$
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