Login

prove that : tan5A+tan3A / tan5A-tan3A = 4cos2Acos4A

$L . H . S . = \frac{\tan 5 A + \tan 3 A}{\tan 5 A - \tan 3 A} = \frac{\frac{\sin 5 A}{\cos 5 A} + \frac{\sin 3 A}{\cos 3 A}}{\frac{\sin 5 A}{\cos 5 A} - \frac{\sin 3 A}{\cos 3 A}} = \frac{\sin 5 A \cdot \cos 3 A + \sin 3 A \cdot \cos 5 A}{\sin 5 A \cdot \cos 3 A - \sin 3 A \cdot \cos 5 A} = \frac{\sin (5 A+ 3 A)}{\sin (5 A - 3 A)} = \frac{\sin 8A}{\sin 2 A} = \frac{\sin 2 (4 A)}{\sin 2 A} = \frac{2 \sin 4 A \cdot \cos 4 A}{\sin 2 A} = \frac{2 \sin 2 (2A) \cdot \cos 4 A}{\sin 2 A} = \frac{(2) (2) \sin 2A \cdot \cos 2 A \cdot \cos 4 A}{\sin 2 A} = 4 \cos 2 A \cdot \cos 4 A = R.H.S.$

33

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?