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Question: \(\frac{\sec 8A - 1}{\sec 4A - 1}\) equal to...

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

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

1−cos⁡8Acos⁡8A.cos⁡4A1−cos⁡4A=2sin⁡24Acos⁡8A.cos⁡4A2sin⁡22A=2sin⁡4.Acos⁡4A.sin⁡4Acos⁡8A.2sin⁡2⥂2A\frac{1 - \cos 8A}{\cos 8A}.\frac{\cos 4A}{1 - \cos 4A} = \frac{2\sin^{2}4A}{\cos 8A}.\frac{\cos 4A}{2\sin^{2}2A} = \frac{2\sin 4.A\cos 4A.\sin 4A}{\cos 8A.2\sin^{2} ⥂ 2A}=sin⁡8A.2sin⁡2A.cos⁡2Acos⁡8A.2sin⁡22A=tan⁡8Atan⁡2A\frac{\sin 8A.2\sin 2A.\cos 2A}{\cos 8A.2\sin^{2}2A} = \frac{\tan 8A}{\tan 2A}.