Question

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. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• E. $\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Answer 1

Answer: (C)

Let $\tan 4A=\sqrt{3}=\tan \frac{\pi }{3}$
$\Rightarrow$ $A=\frac{\pi }{12}$
$\therefore$ $sin4A-cos2A=sin\frac{\pi }{3}-cos\frac{\pi }{6}$
$=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=0$ and $\cos 4A-\sin 2A=\sin \frac{\pi }{3}-\sin \frac{\pi }{6}$
$=\frac{1}{2}-\frac{1}{2}=0$
$\therefore$ $sin4A-cos2A=cos4A-sin2A$
Hence, our assumption is true.
$\therefore$ $\tan 4A=\sqrt{3}$