MCQ
If ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \frac{\pi }{4}$ then
- A$x + y - xy = 1$
- ✓$x + y + xy = 1$
- C$x + y + xy + 1 = 0$
- D$x + y - xy + 1 = 0$
${\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right) = {\tan ^{ - 1}}1$
$\frac{{x + y}}{{1 - xy}} = 1$;
$x + y + xy = 1$.
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