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

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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