Question

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

Answer 1

L.H.S. $=\frac{\sec 8\theta -1}{\sec 4\theta -1}=\frac{\frac{1}{\cos 8\theta }-1}{\frac{1}{\cos 4\theta }-1}=\frac{1-\cos 8\theta }{\cos 8\theta }\times \frac{\cos 4\theta }{1-\cos 4\theta }$
$=\frac{2{\sin }^{2}4\theta }{\cos 8\theta }\times \frac{\cos 4\theta }{2{\sin }^{2}2\theta }$ $[\because 1-\cos 8\theta =2{\sin }^2\frac{8\theta }{2}=2{\sin }^{2}4\theta ]$ and $[\because 1-\cos 4\theta =2{\sin }^2\frac{4\theta }{2}=2{\sin }^{2}2\theta ]$
$=\frac{\left(2\sin 4\theta \cos 4\theta \right)}{\cos 8\theta }\times \frac{\sin 4\theta }{2{\sin }^{2}2\theta }$
$=\left(\frac{2\sin 4\theta \cos 4\theta }{\cos 8\theta }\right)\times \left(\frac{2\sin 2\theta \cos 2\theta }{2{\sin }^{2}2\theta }\right)$
$=\left(\frac{\sin 2\left(4\theta \right)}{\cos 8\theta }\right)\times \left(\frac{\cos 2\theta }{\sin 2\theta }\right)=\left(\frac{\sin 8\theta }{\cos 8\theta }\right)\times \left(\frac{\cos 2\theta }{\sin 2\theta }\right)$
$\tan 8\theta \cot 2\theta =\frac{\tan 8\theta }{\tan 2\theta }=$ R.H.S.
Ans: 8