MCQ
If ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}A,$ then $A$
- A$x - y$
- B$x + y$
- ✓$\frac{{x - y}}{{1 + xy}}$
- D$\frac{{x + y}}{{1 - xy}}$
$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}A$.
Hence $A = \frac{{x - y}}{{1 + xy}}$.
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$(I)$ The curve $y=f(x)$ intersects the $x$-axis exactly at one point
$(II)$ The curve $y=f(x)$ intersects the $x$-axis at $\mathrm{x}=\cos \frac{\pi}{12}$
Then