Question

Prove $\frac{sec8\theta -1}{sec4\theta -1}=\frac{tan8\theta }{tan2\theta }$

Answer 1

Given,

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

LHS:

$\frac{\sec 8\theta -1}{\sec 4\theta -1}$

$=\frac{\frac{1}{\cos 8\theta }-1}{\frac{1}{\cos 4\theta }-1}$

$=\frac{\frac{1-\cos 8\theta }{\cos 8\theta }}{\frac{1-\cos 4\theta }{\cos 4\theta }}$

$=\frac{\frac{2{\sin }^{2}4\theta }{\cos 8\theta }}{\frac{2{\sin }^{2}2\theta }{\cos 4\theta }}$ ........... $(\because$ $1-cos2x=2si{n}^{2}x)$

$=\frac{4{\cos }^{2}2\theta \cos 4\theta }{\cos 8\theta }$

RHS:

$\frac{\tan 8\theta }{\tan 2\theta }$

$=\frac{\frac{\sin 8\theta }{\cos 8\theta }}{\frac{\sin 2\theta }{\cos 2\theta }}$

$=\frac{2\sin 4\theta \cos 4\theta \times \cos 2\theta }{\cos 8\theta \times \sin 2\theta }$ ............ $[\because \sin 2x=2\sin x\cos x]$

$=\frac{2\times 2\sin 2\theta \cos 2\theta \times \cos 4\theta \times \cos 2\theta }{\cos 8\theta \times \sin 2\theta }$

$=\frac{4{\cos }^{2}2\theta \cos 4\theta }{\cos 8\theta }$

$\therefore LHS=RHS$