Question

$\frac{\sec 8A-1}{\sec 4A-1}=$

(a) $\frac{\tan 2A}{\tan 8A}$


(b) $\frac{\tan 8A}{\tan 2A}$

(c) $\frac{\cot 8A}{\cot 2A}$

(d) none of these.

Answer 1

(b) $\frac{\tan 8A}{\tan 2A}$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{\sec 8A-1}{\sec 4A-1}=\frac{{\displaystyle \frac{1}{\cos 8A}}-1}{{\displaystyle \frac{1}{\cos 4A}}-1} \\ =\frac{\cos 4A}{\cos 8A}\times \frac{1-\cos 8A}{1-\cos 4A} \end{gathered}$$

$$\begin{gathered} =\frac{\cos 4A}{\cos 8A}\times \frac{2{\sin }^{2}4A}{2{\sin }^{2}2A}\left(2{\sin }^2\mathrm{\theta }=1-\cos 2\mathrm{\theta }\right) \\ =\frac{\left(2\cos 4A\sin 4A\right)\sin 4A}{2\times \cos 8A{\sin }^{2}2A} \\ =\frac{\sin 8A\sin 4A}{\cos 8A\times 2\sin 2A\times \sin 2A} \\ =\tan 8A\times \frac{2\sin 2A\times \cos 2A}{2\sin 2A\times \sin 2A} \\ =\tan 8A\times \cot 2A \\ =\frac{\tan 8A}{\tan 2A} \end{gathered}$$