Question

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

• A. $0$
• B. $\frac{tan8A}{tan2A}$
• C. $\frac{cos8A}{cos2A}$
• D. $\frac{sin8A}{sin2A}$

Answer 1

The correct option is B $\frac{\tan 8A}{\tan 2A}$

Given,

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

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

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

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

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

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

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

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

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

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

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

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