Question

Prove that:$\frac{(sec8A-1)}{(sec4A-1)}=\frac{\tan 8A}{\tan 2A}.$

Answer 1

Given: we have$\frac{(sec8A-1)}{(sec4A-1)}=\frac{\tan 8A}{\tan 2A}.$

Consider the LHS

$$\begin{gathered} \frac{(sec8A-1)}{(sec4A-1)} \\ \Rightarrow \frac{\left[{\displaystyle \frac{(1–\cos 8A)}{\cos 8A}}\right]}{\left[{\displaystyle \frac{(1–\cos 4A)}{\cos 4A}}\right]}\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{A}}\mathbf{]} \\ \Rightarrow \left[\frac{2\sin ²4A}{2\sin ²2A}\right]\times \left[\frac{\cos 4A}{\cos 8A}\right]\mathbf{[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{2}\mathbf{A}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{a}\mathbf{]} \\ \Rightarrow \frac{(2\sin 4A\times \cos 4A)\times {\displaystyle \frac{\sin 4A}{\cos 8A}}}{2\sin ²2A} \\ \Rightarrow \frac{\left({\displaystyle \frac{\sin 8A}{\cos 8A}}\right)\times \sin 4A}{2\sin ²2A} \\ \Rightarrow \tan 8A\times \frac{\sin 4A}{2\sin ²2A} \\ \Rightarrow \tan 8A\times \frac{2\sin 2A\times \cos 2A}{2\sin ²2A} \\ \Rightarrow \tan 8A\times \left(\frac{\cos 2A}{\sin 2A}\right) \\ \Rightarrow \tan 8A\times \cot 2A \\ \Rightarrow \frac{\tan 8A}{\tan 2A} \end{gathered}$$

Hence Proved.