How do I prove the identity without having to change the right side at all?

tan (t/2) (cos^2t) - tan(t/2) = sint/sect - sint

2 Answers
Apr 15, 2018

Use one of the tan half-angle identity:

tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))

color(white)(tan(theta/2))=(1-costheta)/sintheta

color(white)(tan(theta/2))=(sintheta)/(costheta+1)

I will use the third one, because it works out well in this case. First, factor out the tan(t/2), then use the identity, and lastly rewrite cost as 1/(1/cost):

LHS=tan (t/2) (cos^2t) - tan(t/2)

color(white)(LHS)=tan(t/2)(cos^2t-1)

color(white)(LHS)=tan(t/2)(cost-1)(cost+1)

color(white)(LHS)=sint/(cost+1)*(cost-1)(cost+1)

color(white)(LHS)=sint/color(red)cancelcolor(black)((cost+1))*(cost-1)color(red)cancelcolor(black)((cost+1))

color(white)(LHS)=sint*(cost-1)

color(white)(LHS)=sintcost-sint

color(white)(LHS)=sint*1/(1/cost)-sint

color(white)(LHS)=sint*1/(sect)-sint

color(white)(LHS)=sint/(sect)-sint

color(white)(LHS)=RHS

That's the proof. Hope this helped!

Apr 15, 2018

LHS=tan (t/2) (cos^2t) - tan(t/2)

=tan(t/2)(cos^2t-1)

=tan(t/2)((2cos^2(t/2)-1)^2-1)

=tan(t/2)(4cos^4(t/2)-4cos^2(t/2)+1-1)

=sin(t/2)/cos(t/2)xx4cos^2(t/2)(cos^2(t/2)-1)

=2xx2sin(t/2)cos(t/2)(cos^2(t/2)-1)

=sint(2cos^2(t/2)-2)

=sint(1+cost-2)

=sint(cost-1)

=sint(1/sect-1)

=sint/sect-sint=RHS