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

Question

17464 views around the world

You can reuse this answer
Creative Commons License

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#