Question

If $\sin 4A-\cos 2A=\cos 4A-\sin 2A\left(0<A<\frac{\pi }{4}\right)$ then the value of tan 4A =

• A. 1
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Answer 1

Answer: (C)

Given $\sin 4A-\cos 2A=\cos 4A-\sin 2A$
$\Rightarrow \sin 4A+\sin 2A=\cos 4A+\cos 2A$
$\Rightarrow 2\sin 3A\cos A=2\cos 3A\cos A$
$\Rightarrow \tan 3A=1$$(\because cosA\ne 0)$
$\Rightarrow 3A=\frac{\pi }{4}\Rightarrow A=\frac{\pi }{12}\Rightarrow 4A=\frac{\pi }{3}$
$\Rightarrow \tan 4A=\tan \frac{\pi }{3}=\sqrt{3}$