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Question

Question: \[\frac{\sec 8A - 1}{\sec 4A - 1} =\]...

sec⁡8A−1sec⁡4A−1=\frac{\sec 8A - 1}{\sec 4A - 1} =

A

tan⁡2Atan⁡8A\frac{\tan 2A}{\tan 8A}

B

tan⁡8Atan⁡2A\frac{\tan 8A}{\tan 2A}

C

cot⁡8Acot⁡2A\frac{\cot 8A}{\cot 2A}

D

None of these

Answer

tan⁡8Atan⁡2A\frac{\tan 8A}{\tan 2A}

Explanation

Solution

sec⁡8A−1sec⁡4A−1=1−cos⁡8Acos⁡8A.cos⁡4A1−cos⁡4A\frac{\sec 8A - 1}{\sec 4A - 1} = \frac{1 - \cos 8A}{\cos 8A}.\frac{\cos 4A}{1 - \cos 4A}

=2sin⁡24Acos⁡8Acos⁡4A2sin⁡22A=2sin⁡4Acos⁡4Acos⁡8Asin⁡4A2sin⁡22A= \frac{2\sin^{2}4A}{\cos 8A}\frac{\cos 4A}{2\sin^{2}2A} = \frac{2\sin 4A\cos 4A}{\cos 8A}\frac{\sin 4A}{2\sin^{2}2A}

=tan⁡8A2sin⁡2Acos⁡2A2sin⁡22A=tan⁡8Atan⁡2A.= \tan 8A\frac{2\sin 2A\cos 2A}{2\sin^{2}2A} = \frac{\tan 8A}{\tan 2A}.