Prove that sin8A-sin4A/cos8A+cos4A = tan2A
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Answer:
Given, (sec8A - 1) / (sec4A - 1)
= (1/cos8A) - 1) / (1/cos4A) - 1
= (1 - cos8A)/cos8A) / (1 - cos4A) / cos4A)
= cos4A (1 - cos8a) / (cos8A (1 - cos4A))
= cos4A(1 - (1 - 2sin²4A)) / cos8A (1 - (1 - 2sin²2A))
= cos4A sin²4A / (cos8A sin²2A)
= (2 cos4A sin4A) sin4A / (2 cos8A sin²2A)
= sin8A sin4A / (2 cos8A sin²2A)
= tan 8A * (sin 4A / 2 sin^2 2A)
= tan 8A * (cos 2A / sin 2A)
= tan 8A/tan 2A
Hope this helps!
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