Question

Prove :

$\frac{\sec 8\theta -1}{\sec 4\theta -1}=\frac{\tan 8\theta }{\tan 2\theta }$

Answer 1

$Given,theequationis\frac{\sec 8\theta -1}{\sec 4\theta -1}=\frac{\tan 8\theta }{\tan 2\theta }L.H.S:\frac{\sec 8\theta -1}{\sec 4\theta -1}=\frac{\frac{1}{\cos 8\theta }-1}{\frac{1}{\cos 4\theta }-1}=\frac{(1-\cos 8\theta )\cos 4\theta }{(1-\cos 4\theta )\cos 8\theta }=\frac{2{\sin }^{2}4\theta \cos 4\theta }{2{\sin }^{2}2\theta \cos 8\theta }=\frac{2\sin 4\theta \cos 4\theta \sin 4\theta }{2\sin 2\theta \cos 8\theta \sin 2\theta }=\frac{\sin 8\theta 2\sin 2\theta \cos 2\theta }{2\sin 2\theta \cos 8\theta \sin 2\theta }=\frac{\tan 8\theta }{\tan 2\theta }=R.H.Shenceproved.$