Question

Show that $(1+\sec 2\theta )(1+\sec 4\theta )(1+\sec 8\theta )=\frac{\tan 8\theta }{\tan \theta }$.

Answer 1

$(1+\sec 2\theta )(1+\sec 4\theta )(1+\sec 8\theta )$
$=(1+\frac{1}{\cos 2\theta })(1+\frac{1}{\cos 4\theta })(1+\frac{1}{\cos 8\theta })$
$=\left(\frac{\cos 2\theta +1}{\cos 2\theta }\right)\left(\frac{\cos 4\theta +1}{\cos 4\theta }\right)\left(\frac{\cos 8\theta +1}{\cos 8\theta }\right)$
$=\left(\frac{2{\cos }^2\theta -1+1}{\cos 2\theta }\right)\left(\frac{2{\cos }^{2}2\theta }{\cos 4\theta }\right)\left(\frac{{\cos }^{2}4\theta +1-1}{\cos 8\theta }\right)$
$=\frac{8{\cos }^2\theta }{\cos 8\theta }\cos 2\theta \cos 4\theta$
$=\frac{8\sin \theta \cos \theta \cos 2\theta \cos 4\theta \cos \theta }{\sin \theta \cos 8\theta }$
$=\frac{4\sin 2\theta \cos 2\theta \cos 4\theta \cos \theta }{\sin \theta \cos 8\theta }$
$=\frac{2\sin 4\theta \cos 4\theta }{\sin \theta \cos 8\theta }\cos \theta =\frac{\sin 8\theta \cos \theta }{\cos \theta \sin 8\theta }=\frac{\tan 8\theta }{\tan \theta }$