How do you verify the identity #1+1/2sin2A=(secA+sinA)/secA#?

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Answer 1 · 2016-12-26T11:20:27

See proof below

We need

#sin2x=2sinxcosx#

#secx=1/cosx#

So,

LHS #=1+1/2sin2A=1+1/2*2sinAcosA=1+sinAcosA#

RHS #=(secA+sinA)/secA=(1/cosA+sinA)/(1/cosA)#

#=(1+sinAcosA)/(cosA*1/cosA)=1+sinAcosA#

#=LHS#

#QED#