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

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!

#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#