Can Trigonometric Identities Validate This Proof?

[Sudharaka]

srirahulan's question on Math Help Forum,

Prove that, \(\dfrac{\sec(8A)-1}{\sec(4A)-1}=\dfrac{\tan(8A)}{\tan(4A)}\)

Hi srirahulan,

Consider the left hand side of the above equation,

\begin{eqnarray}
\frac{\sec 8A-1}{\sec 4A-1}&=&\frac{\cos 4A}{\cos 8A}\left(\frac{1-\cos 8A}{1-\cos 4A}\right)\\

&=&\frac{\cos 4A}{\cos 8A}\left(\frac{2\sin^{2} 4A}{2 \sin^{2} 2A}\right)\\

&=&\frac{\sin 8A\,\sin 4A}{2\sin^{2}2A\,\cos 8A}\\

&=&\frac{\tan 8A}{\left(\dfrac{2\sin^{2}2A}{\sin 4A}\right)}\\

&=&\frac{\tan 8A}{\left(\dfrac{2 \sin^{2}2A}{2\sin 2A\cos 2A}\right)}\\

&=&\frac{\tan 8A}{\left(\dfrac{\sin 2A}{\cos 2A}\right)}\\

&=&\frac{\tan 8A}{\tan 2A}\\

\end{eqnarray}

\[\therefore\dfrac{\sec(8A)-1}{\sec(4A)-1}=\dfrac{\tan(8A)}{\tan(2A)}\]

So your question has a mistake in it. The denominator of the right hand side should be \(\tan(2A)\).