Q. If $A + C = 2B$, then $\frac{\cos{C - \cos A}}{\sin{A - \sin C}}$ is equal to
A) $\cot B$
B) $\cot{2B}$
C) $\tan{2B}$
D) $\tan B$
Explanation
(d)
$\frac{\cos{C - \cos A}}{\sin{A - \sin C}} = \frac{2\sin{\left( \frac{A + C}{2} \right)\sin\left( \frac{A - C}{2} \right)}}{2\cos{\left( \frac{A + C}{2} \right)\sin\left( \frac{A - C}{2} \right)}}$
$= \frac{2\sin B}{2\cos B} = \tan B\ \ \ \ \lbrack\because A + C = 2B,\ given\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.