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

Question

1912 views around the world

You can reuse this answer
Creative Commons License

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 $=(secA+sinA)/secA=(1/cosA+sinA)/(1/cosA)$

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

$=LHS$

$QED$

How do you verify the identity 1+1/2sin2A=(secA+sinA)/secA1+1/2sin2A=(secA+sinA)/secA?