Q. If $\tan^{- 1}{x - \tan^{- 1}{y = \tan^{- 1}{A,then\ A\ is\ equal\ to}}}$
A) $x - y$
B) $x + y$
C) $\frac{x - y}{1 + xy}$
D) $\frac{x + y}{1 - xy}$
Explanation
(c)
Given that, $\tan^{- 1}{x - \tan^{- 1}{y = \tan^{- 1}A}}$
$\Rightarrow \tan^{- 1}{\left( \frac{x - y}{1 + xy} \right) = \tan^{- 1}A}$
Hence, $A = \frac{x - y}{1 + xy}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.