# The value of $\large\frac{\sec 8\theta-1}{\sec 4\theta-1}$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{\tan 2\theta}{\tan 8\theta}&(b)\;\large\frac{\tan 8\theta}{\tan 4\theta}\\(c)\;\large\frac{\tan 8\theta}{\tan 2\theta}&(d)\;\large\frac{\tan 6\theta}{\tan 2\theta}\end{array}$

We have $\large\frac{\sec 8\theta-1}{\sec 4\theta-1}$
$\Rightarrow \large\frac{\Large\frac{1}{\cos 8\theta}\normalsize -1}{\Large\frac{1}{\cos 4\theta}\normalsize-1}$
$\Rightarrow \large\frac{1-\cos 8\theta}{\cos \theta}.\frac{\cos 4\theta}{1-\cos 4\theta}$
$\Rightarrow \large\frac{2\sin^24\theta}{\cos 8\theta}.\frac{\cos 4\theta}{1-\cos 4\theta}$
$\Rightarrow \large\frac{2\sin 4\theta\cos 4\theta}{\cos 8\theta}.\frac{\sin 4\theta}{2\sin^2 2\theta}$
$\sin 4\theta=2\sin 2\theta\cos 2\theta$
$\Rightarrow \large\frac{2\sin 4\theta\cos 4\theta}{\cos 8\theta}.\frac{2\sin 2\theta\cos 2\theta }{2\sin^2 2\theta}$
$\Rightarrow \large\frac{\sin 2(4\theta)}{\cos 8\theta}\frac{\cos 2\theta}{\sin 2\theta}$
$\Rightarrow \large\frac{\sin 8\theta}{\cos 8\theta}\frac{\cos 2\theta}{\sin 2\theta}$
$\Rightarrow \tan 8\theta.\cot 2\theta$
$\cot 2\theta=\large\frac{1}{\tan 2\theta}$
$\Rightarrow \large\frac{\tan 8\theta}{\tan 2\theta}$
Hence (c) is the correct answer.