Q. If $\sin{4A - \cos{2A = \cos{4A - \sin{2A,\ \left( 0 < A < \frac{\pi}{4} \right)}}}}$, then the value of $\tan{4A}$ is
A) 1
B) $\frac{1}{\sqrt{3}}$
C) $\sqrt{3}$
D) $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Explanation
(c)
Given, $\sin{4A + \sin{2A = \cos{4A + \cos{2A}}}}$
$\Rightarrow \ \ 2\sin{3A\cos{A = 2\cos{3A\cos A}}}$
$\therefore\ \ \ \ \ \tan{3A = 1}$ and $\cos{A = 0}$
$\Rightarrow \ \ \ A = \frac{\pi}{12}$ and $A = \frac{\pi}{2} \notin \left( 0,\frac{\pi}{4} \right)$
$\therefore\ \ \tan{4A = \tan{\frac{\pi}{3} = \sqrt{3}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.