Prove that secθ+tanθ=tan(π/4+θ/2)

Steps to prove secθ+tanθ=tan(π/4+θ/2)

Take L.H.S

=secθ+tanθ

=1/cosθ+sinθ/cosθ

=1+sinθ/cosθ  [Taking L.C.M]

We know that sin^2θ/2+cos^2θ/2=1

and 2.sinθ/2.cosθ/2=sinθ

and cos^2θ/2-sin^2θ/2=cosθ

=(sin^2θ/2+cos^2θ/2+2.sinθ/2.cosθ/2)/(cos^2θ/2-sin^2θ/2)

=(cosθ/2+sinθ/2)^2/(cosθ/2+sinθ/2)(cosθ/2-sinθ/2)

=(cosθ/2+sinθ/2)/(cosθ/2-sinθ/2)

Dividing cosθ/2 in both Numenitor and Denomenitor 

={(cosθ/2+sinθ/2)/cosθ/2}/{(cosθ/2-sinθ/2)/cosθ/2}

=(1+tanθ/2)/(1-tanθ/2)

=(tanπ/4+tanθ/2)/(1-tanπ/4.tanθ/2)  [tanπ/4=1]

=tan(π/4+θ/2)

=R.H.S

Hence secθ+tanθ=tan(π/4+θ/2) is proved

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