Prove that #cosA/(1+sinA)=tan(π/4-A/2)#?

Question

10440 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-30T10:14:21

#LHS=cosA/(1+sinA)#

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

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

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

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

#=(1-tan(A/2))/(1+tan(A/2))#

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

#=tan(π/4-A/2)=RHS#