Prove that secA+tanA=(1+tanA/2)/(1-tanA/2)

 Steps to prove secA+tanA=(1+tanA/2)/(1-tanA/2)

Take L.H.S

=secA+tanA

=1/cosA+sinA/cosA

=1+sinA/cosA  [Taking L.C.M]

We know that sin^2A/2+cos^2A/2=1

and 2.sinA/2.cosA/2=sinA

and cos^2A/2-sin^2A/2=cosA

=(sin^2A/2+cos^2A/2+2.sinA/2.cosA/2)/(cos^2A/2-sin^2A/2)

=(cosA/2+sinA/2)^2/(cosA/2+sinA/2)(cosA/2-sinA/2)

=(cosA/2+sinA/2)/(cosA/2-sinA/2)

Dividing cosA/2 in both Numenitor and Denomenitor 

={(cosA/2+sinA/2)/cosA/2}/{(cosA/2-sinA/2)/cosA/2}

=(1+tanA/2)/(1-tanA/2)

=R.H.S

Hence secA+tanA=(1+tanA/2)/(1-tanA/2) is proved.

Detail Information:-

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