Question

sec 8A-1 / sec 4A-1 = prove tan 8A/ tan 2A​

Answer 1

Given : $\frac{\sec 8A-1}{\sec 4A-1}$

To prove : $\frac{\tan 8A}{\tan 2A}$

Step-by-step explanation:

$\frac{\sec 8A-1}{\sec 4A-1}\\\\\frac{\frac{1}{\cos 8 A}-1}{\frac{1}{\cos 4 A}-1}\\\\\frac{(1-\cos 8A)\cos 4A}{(1-\cos 4A)\cos 8A}\\\\\frac{2\sin^24A \cos 4A}{(2\sin^22A)\cos 8A}\\\\\frac{2\sin 4A\cos 4A\sin 4A}{(2\sin^22A)\cos 8A}\\\\\frac{\sin 8A 2\sin 2A\cos 2A}{(2\sin^22A)\cos 8A}\\\\\frac{\tan 8 A\cos 2A}{\sin 2A}\\\\\frac{\tan 8A}{\tan 2A}$

hence proved