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Question

Prove that:(sec8A-1)(sec4A-1)=tan8Atan2A.


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Solution

Given: we have(sec8A-1)(sec4A-1)=tan8Atan2A.

Consider the LHS

(sec8A-1)(sec4A-1)(1cos8A)cos8A(1cos4A)cos4A[secA=1cosA]2sin²4A2sin²2A×cos4Acos8A[cos2A=1-2sin2a](2sin4A×cos4A)×sin4Acos8A2sin²2Asin8Acos8A×sin4A2sin²2Atan8A×sin4A2sin²2Atan8A×2sin2A×cos2A2sin²2Atan8A×cos2Asin2Atan8A×cot2Atan8Atan2A

Hence Proved.


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