Sin3Acos4A-sinAcos2A/sin4AsinA+cos6AcosA=tan2A
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2sin3Acos4A - 2sinAcos2A =(sin7A - sinA) - ( sin3A - sinA) = sin7A - sin3A = 2sin2Acos5A2sinAsin4A + 2cosAcos6A = ( Cos3A - cos5A) + (cos7A + cos5A) = cos3A + cos7A =2cos5Acos2A
LHS = (2sin2Acos5A) / (2cos2Acos5A) = tan 2A
LHS = (2sin2Acos5A) / (2cos2Acos5A) = tan 2A
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Here is the answer :-)
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