Prove that: (sec8A - 1) / (sec4A - 1) = tan8A/tan2A.

L.H.S. = (sec 8A -1) / (sec 4A -1)

=> [(1 – cos 8A)/ cos 8A] / [(1 – cos 4A)/ cos 4A]

=> [2 sin² 4A / 2 sin² 2A] * [cos 4A / cos 8A]

=> [(2 sin 4A * cos 4A) * sin 4A / cos 8A] / [2 sin² 2A]

=> [(sin 8A / cos 8A) * sin 4A] / [2 sin² 2A]

=> tan 8A * (sin 4A / 2 sin² 2A)

=> tan 8A * (2sin 2A * cos 2A / 2 sin² 2A)

=> tan 8A * ( cos 2A / sin 2A)

=> tan 8A * cot 2A

=> tan 8A / tan 2A

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