Q. If$\ \tan{A - \tan{B = x\ and\cot{B - \cot{A = y,\ then\cot(A - B)}}}}$ is equal to
A) $\frac{1}{x} + y$
B) $\frac{1}{xy}$
C) $\frac{1}{x} - \frac{1}{y}$
D) $\frac{1}{x} + \frac{1}{y}$
Explanation
(d)
Given that,
$\tan{A - \tan{B = x\ \ \ \ldots(i)}}$
and${\ cot}{B - \cot{A = y}}\ \ldots(ii)$
Now$,\cot{(A - B)} = \frac{1}{\tan{(A - B)}}$
$= \frac{1 + \tan{A\tan B}}{\tan{A - \tan B}}$
$= \frac{1}{\tan{A - \tan B}} + \frac{\tan{A\tan B}}{\tan{A - \tan B}}$
$= \frac{1}{x} + \frac{1}{y}\ \ \$[from Eqs.(i)and(ii)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.