## Question

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

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

## Answers

Please see below.

## Explanation:

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#

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