Prove that $tan(pi/4+A/2)=secA+tanA$ ?

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Answer 1 · 2016-05-03T08:43:19

Please see below.

Let us start from Right Hand Side

$tan(pi/4+A/2)$

= $(tan(pi/4)+tan(A/2))/(1-tan((pi)/4)tan(A/2)$

= $(1+tan(A/2))/(1-1*tan(A/2)$

= $(1+sin(A/2)/cos(A/2))/(1-sin(A/2)/cos(A/2)$

= $(cos(A/2)+sin(A/2))/(cos(A/2)-sin(A/2))$

= $(cos(A/2)+sin(A/2))/(cos(A/2)-sin(A/2))xx(cos(A/2)+sin(A/2))/(cos(A/2)+sin(A/2))$

= $(cos(A/2)+sin(A/2))^2/(cos^2(A/2)-sin^2(A/2))$

= $(cos^2(A/2)+sin^2(A/2)+2sin(A/2)cos(A/2))/cosA$

= $(1+sinA)/cosA$

= $1/cosA+sinA/cosA$

= $secA+tanA$

Prove that tan(pi/4+A/2)=secA+tanAtan(pi/4+A/2)=secA+tanA?