Question:
prove that sec8A - 1/sec4A - 1 = tan 8A / tan 2 A .
Answer:
Given, (sec 8A - 1)/(sec 4A - 1)
= (1/cos 8A - 1)/(1/cos 4A - 1)
= {(1 - cos 8A)/cos 8A}/{(1 - cos 4A)/cos 4A}
= {(1 - cos 8A) * cos 4A}/{(1 - cos 4A) * cos 8A}
= (2sin2 4A * cos 4A}/{2sin2 2A * cos 8A} {since cos 2A = 1 - 2sin2 A}
= (2sin 4A * sin 4A * cos 4A}/{2sin 2A * sin 2A * cos 8A}
= (sin 8A * sin 4A}/{2sin 2A * sin 2A * cos 8A} {since sin 2A = 2*sin A * cos A}
= (sin 8A * 2sin 2A * cos 2A}/{2sin 2A * sin 2A * cos 8A}
= (sin 8A * cos 2A}/{sin 2A * cos 8A}
= (sin 8A/cos 8A)/(sin 2A/cos 2A)
= tan 8A/tan 2A
So, (sec 8A - 1)/(sec 4A - 1) = tan 8A/tan 2A
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