Using #t=tan(1/2theta)#, how do you prove that #(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha#?

Using the formula above we get
#(2tan(x)/(1-tan^2(x))+1/tan(x))/(2tan(x)/(1-tan^2(x))--tan(x))#
Multiplying numerator and denominator by
#1-tan^2(x)#

we get

#(2tan(x)+((1-tan^2(x))/tan(x)))/(2tan(x)-tan(x)(1-tan^2(x))#
multiplying numerator and denominator by

#tan(x)#
#(2tan^2(x)+1-tan^2(x))/(2tan^2(x)-tan^2(x)(1-tan^2(x))#
and this is

#(tan^2(x)+1)/(tan^2(x)(1+tan^2(x)))=cot^2(x)#

We know that,

#color(red)((1)cos(A-B)=cosAcosB+sinAsinB#

#color(blue)((2)sin(A-B)=sinAcosB-cosAsinB#

Here,

#(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha#

We take,

#LHS=(tan2alpha+cotalpha)/(tan2alpha-tanalpha)#

#color(white)(LHS)= ((sin2alpha)/(cos2alpha)+cosalpha/sinalpha)/((sin2alpha)/(cos2alph a)-sinalpha/cosalpha)#

#color(white)(LHS)= ((sin2alphasinalpha+cos2alphacosalpha)/(cancelcos2alphasinalpha ))/((sin2alphacosalpha- cos2alphasinalpha)/(cancelcos2alphacosalpha#

#color(white)(LHS)= (color(red)((cos2alphacosalpha+sin2alphasinalpha))xxcosalpha)/(color(blue)((sin2alphaco salpha-cos2alphasinalpha))xxsinalpha#

Using #(1) and(2)#, we get

#LHS=(color(red)(cos(2alpha-alpha))xxcosalpha)/(color(blue)(sin(2alpha- alpha))xxsinalpha)#

#color(white)(LHS)=(cosalphaxxcosalpha)/(sinalphaxxsinalpha)#

#color(white)(LHS)=cos^2alpha/sin^2alpha#

#color(white)(LHS)=cot^2alpha#

#LHS=RHS#