8A-1 4A-1 is equal to: - Vidyadip

MCQ

$\frac{\sec8\text{A}-1}{\sec4\text{A}-1}$ is equal to:

A
$\frac{\tan2\text{A}}{\tan8\text{A}}$
✓
$\frac{\tan8\text{A}}{\tan2\text{A}}$
C
$\frac{\cot8\text{A}}{\cot2\text{A}}$
D
None of these

✓

Answer

Correct option: B.

$\frac{\tan8\text{A}}{\tan2\text{A}}$

We have,
$\frac{\sec8\text{A}-1}{\sec4\text{A}-1}=\frac{\frac{1}{\cos2\text{A}}-1}{\frac{1}{\cos}4\text{A}-1}$
$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{1-\cos8\text{A}}{1-\cos4\text{A}}$
$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{2\sin^24\text{A}}{2\sin^22\text{A}}(2\sin^2\theta-1-\cos2\theta)$
$=\frac{(2\cos4\text{A}\sin4\text{A})\sin4\text{A}}{2\times\cos8\text{A}\sin^22\text{A}}$
$=\frac{\sin8\text{A}\sin4\text{A}}{\cos8\text{A}\times\text{2}\sin2\text{A}\times\sin2\text{A}}$
$=\tan8\text{A}\times\frac{2\sin2\text{A}\times\cos2\text{A}}{2\sin2\text{A}\times\sin2\text{A}}$
$=\tan8\text{A}\times\cot2\text{A}$
$=\frac{\tan8\text{A}}{\tan2\text{A}}$

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