8 A-1 4 A-1 = - Vidyadip

MCQ

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

A
$\frac{\tan 2 A}{\tan 8 A}$
B
$\frac{\tan 8 A}{\tan 2 A}$
C
$\frac{\cot 8 A}{\cot 2 A}$
✓
$\frac{\tan 6 A}{\tan 2 A}$

✓

Answer

Correct option: D.

$\frac{\tan 6 A}{\tan 2 A}$

(D)
$\frac{\sec 8 A-1}{\sec 4 A-1}=\frac{1-\cos 8 A}{\cos 8 A} \cdot \frac{\cos 4 A}{1-\cos 4 A}$
$=\frac{2 \sin ^2 4 A}{\cos 8 A} \cdot \frac{\cos 4 A}{2 \sin ^2 2 A}$
$=\frac{2 \sin 4 A \cos 4 A}{\cos 8 A} \cdot \frac{\sin 4 A}{2 \sin ^2 2 A}$
$=\tan 8 A \frac{2 \sin 2 A \cos 2 A}{2 \sin ^2 2 A}$
$=\frac{\tan 8 A}{\tan 2 A}$

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