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 $\sin4A -\cos 2A = \cos 4A - \sin2A$
$ \Rightarrow \sin4A + \sin 2A = \cos 4A + \cos2A$
$ \Rightarrow 2\sin 3A \cos A = 2 \cos 3A \cos A$
$ \Rightarrow \tan3A = 1$$\,\,\,(\because cosA\ne0)$
$ \Rightarrow 3A = \frac{\pi}{4} \Rightarrow A = \frac{\pi}{12} \Rightarrow 4A = \frac{\pi}{3} $
$\Rightarrow \tan4A = \tan \frac{\pi}{3} = \sqrt{3} $