How do you verify the identity # tan(x/2 + pi/4) = secx + tanx#?

1 Answer
Mar 9, 2018

See Below

Explanation:

We will use the following Properties:

#color(blue)(sin(x+y)=sinxcosy+cosxsiny#

#color(blue)(cos(x+y)=cosxcosy-sinxsiny#

#color(blue)(sin2x=2sinxcosx#

#color(blue)(cos2x=cos^2x-sin^2x#

#color(blue)(sin^2x+cos^2x=1#

#LHS: tan(x/2 + pi/4)#

#=sin(x/2+pi/4)/(cos(x/2+pi/4)#

#=(sin (x/2)cos(pi/4)+cos(x/2)sin(pi/4))/(cos(x/2)cos(pi/4)-sin(x/2)sin(pi/4))#

#=(1/sqrt2 sin(x/2)+1/sqrt2 cos (x/2))/(1/sqrt2cos(x/2)-1/sqrt2sin(x/2)#

#=(1/sqrt2( sin(x/2)+ cos (x/2)))/(1/sqrt2(cos(x/2)-sin(x/2))#->factor out common factor

#=(cancel(1/sqrt2)( sin(x/2)+ cos (x/2)))/(cancel(1/sqrt2)(cos(x/2)-sin(x/2))#-> cancel common factor

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

#=( sin(x/2)+ cos (x/2))/(cos(x/2)-sin(x/2)) * (cos(x/2)+sin(x/2))/(cos(x/2)+sin(x/2))#->multiply by conjugate

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

#=([sin^2(x/2)+cos^2(x/2)]+2sin(x/2)cos(x/2))/(cos^2(x/2)-sin^2(x/2))#

#=(1+2sin(x/2)cos(x/2))/(cos^2(x/2)-sin^2(x/2))#

#=(1+sin 2(x/2))/(cos 2(x/2))#

#=(1+sin cancel2(x/cancel2))/(cos cancel2(x/cancel2))#

#=(1+sin x)/cos x#

#=1/cosx+sin x/cos x#

#=secx + tanx#

#=RHS#