Question

Solve: $\frac{\sec 8A-1}{\sec 4A-1}-\frac{\tan 8A}{\tan 4A}$

Answer 1

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

$I$ $II$

Let

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

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

$\Rightarrow \frac{1-\cos 8A}{\cos 8A}\times \frac{\cos 4A}{1-\cos 4A}$

$\because \cos 2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$

$=\frac{1-(1-2{\sin }^{2}4A)}{\cos 8A}\times \frac{\cos 4A}{1-(1-2{\sin }^{2}2A)}$

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

$\because \sin 2\theta =2\sin \theta \cos \theta$

So,

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

$\Rightarrow \tan 8A\times \frac{2\sin 2A\cos 2A}{2{\sin }^{2}2A}$

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

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

$\therefore I=II$

$\Rightarrow I-II=0$