Question

Prove that $\frac{\sec 8A-1}{\sec 4A-1}=\frac{\tan 8A}{\tan 2A}$.

Answer 1

L.H.S $=\frac{\sec 8A-1}{\sec 4A-1}$

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

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

$=\frac{\cos 4A(1-\cos 8A)}{\cos 8A(1-\cos 4A)}$

$=\frac{\cos 4A[1-(1-2{\sin }^{2}4A\left)\right]}{\cos 8A[1-(1-2{\sin }^{2}2A\left)\right]}$

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

$=\frac{\left(2\sin 4A\cos 4A\right)\sin 4A}{\cos 8A{\sin }^{2}2A}$ [ $\sin 2\theta =2\sin \theta \cos \theta$ ]

$=\tan 8A\frac{2\sin 2A\cos 2A}{{\sin }^{2}2A}$ [ $\sin 2\theta =2\sin \theta \cos \theta$ ]

$=\tan 8A.\cot 2A$

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

$=R.H.S$

Hence proved.