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