Solveeit Logo
Login

Question

Mathematics Question on Trigonometric Functions

If sin⁡ 4A−cos⁡ 2A=cos⁡ 4A−sin⁡ 2A(0<A<π4)\sin \, 4A - \cos \, 2A = \cos \, 4A- \sin \, 2A \left( 0 < A < \frac{\pi}{4} \right) then the value of tan 4A =

A

1

B

13\frac{1}{\sqrt{3}}

C

3\sqrt{3}

D

3−13+1\frac{\sqrt{3} - 1}{\sqrt{3} + 1}

Answer

3\sqrt{3}

Explanation

Solution

Given sin⁡4A−cos⁡2A=cos⁡4A−sin⁡2A\sin4A -\cos 2A = \cos 4A - \sin2A ⇒sin⁡4A+sin⁡2A=cos⁡4A+cos⁡2A \Rightarrow \sin4A + \sin 2A = \cos 4A + \cos2A ⇒2sin⁡3Acos⁡A=2cos⁡3Acos⁡A \Rightarrow 2\sin 3A \cos A = 2 \cos 3A \cos A ⇒tan⁡3A=1 \Rightarrow \tan3A = 1    (∵cosA≠0)\,\,\,(\because cosA\ne0) ⇒3A=π4⇒A=π12⇒4A=π3\Rightarrow 3A = \frac{\pi}{4} \Rightarrow A = \frac{\pi}{12} \Rightarrow 4A = \frac{\pi}{3} ⇒tan⁡4A=tan⁡π3=3\Rightarrow \tan4A = \tan \frac{\pi}{3} = \sqrt{3}