prove that sin4a+sin2a/1+cos4a+cos2a=2tana/1-tan^2a
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RHS
=sin4a+sin2a/1+cos4a+cos2a
= (2sin2acos2a+sin2a)/(1+2cos^22a-1+cos2a)
= sin2a(2cos2a+1)/ cos2a(2cos2a+1)
= sin2a/cos2a= tan2a
= 2tana /1-tan^a= RHs
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