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Question

Prove that sec8A1sec4A1=tan8Atan2A.

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Solution

L.H.S =sec8A1sec4A1

=1cos8A11cos4A1

=1cos8Acos8A1cos4Acos4A

=cos4A(1cos8A)cos8A(1cos4A)

=cos4A[1(12sin24A)]cos8A[1(12sin22A)]

=2cos4Asin24A2cos8Asin22A

=(2sin4Acos4A)sin4Acos8Asin22A [ sin2θ=2sinθcosθ ]

=tan8A2sin2Acos2Asin22A [ sin2θ=2sinθcosθ ]

=tan8A.cot2A

=tan8Atan2A

=R.H.S
Hence proved.

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