Q. $\frac{1 - \tan^{2}{(45{^\circ} - A)}}{1 + \tan^{2}{(45{^\circ} - A)}}$ is equal to
A) $\sin{2A}$
B) $\cos{2A}$
C) $\tan{2A}$
D) $\cot{2A}$
Explanation
(a)
$\frac{1 - \tan^{2}{(45{^\circ} - A)}}{1 + \tan^{2}{(45{^\circ} - A)}}$
$= \frac{\cos^{2}{(45{^\circ} - A) - \sin^{2}{(45{^\circ} - A)}}}{\cos^{2}{(45{^\circ} - A) + \sin^{2}{(45{^\circ} - A)}}}$
$= \frac{\cos{2(45{^\circ} - A)}}{1}$
$= \sin{2A}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.