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Question
Options
- \[\frac{\tan 2A}{\tan 8A}\]
- \[\frac{\tan 8A}{\tan 2A}\]
- \[\frac{\cot 8A}{\cot 2A}\]
none of these.
Solution
\[\frac{\tan 8A}{\tan 2A}\]
\[\frac{\sec8A - 1}{\sec4A - 1} = \frac{\frac{1}{\cos8A} - 1}{\frac{1}{\cos4A} - 1}\]
\[ = \frac{\cos4A}{\cos8A} \times \frac{1 - \cos8A}{1 - \cos4A}\]
\[ = \frac{\left( 2\cos4A \sin4A \right) \sin4A}{2 \times \cos8A \sin^2 2A}\]
\[ = \frac{\sin8A \sin4A}{\cos8A \times 2\sin2A \times \sin2A}\]
\[ = \tan8A \times \frac{2\sin2A \times \cos2A}{2\sin2A \times \sin2A}\]
\[ = \tan8A \times \cot2A\]
\[ = \frac{\tan8A}{\tan2A}\]
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