Prove that : tan2 A tan2 B = [(cos 2B - cos2A)/(cos2B.cos2A)] = [(sin2A – sin2 B) / (cos2A.cos 2B)] I will best answer as brainliest
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Answer:n2 A tan2 B = [(cos 2B - cos2A)/(cos2B.cos2A)] = [(sin2A – sin2 B) / (cos2A.cos 2B)
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