How to prove that #tan(pi/4+alpha)-tan(pi/4-alpha)=2tan2alpha#?

We need

#tan(a+b)=(tana+tanb)/(1-tanatanb)#

#tan(pi/4)=1#

#tan(2a)=(2tana)/(1-tan^2a)#

Therefore,

#LHS=tan(pi/4+alpha)-tan(pi/4-alpha)#

#=(tan(pi/4)+tanalpha)/(1-tan(pi/4)tanalpha)-(tan(pi/4)-tanalpha)/(1+tan(pi/4)tanalpha)#

#=(1+tanalpha)/(1-tanalpha)-(1-tanalpha)/(1+tanalpha)#

#=((1+tanalpha)^2-(1-tanalpha)^2)/((1+tanalpha)(1-tanalpha))#

#=(1+2tanalpha+tan^2alpha-1+2tanalpha-tan^2alpha)/(1-tan^2alpha)#

#=(4tanalpha)/(1-tan^2alpha)#

#=4tan(2alpha)#

#=RHS#

#QED#

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)tan(x+-y)=(tanx+-tany)/(1∓tanxtany)#

#•color(white)(x)tan2x=(2tanx)/(1-tan^2x)#

#"consider the left side"#

#(tan(pi/4)+tanalpha)/(1-tan(pi/4)tanalpha)-(tan(pi/4)-tanalpha)/(1+tan(pi/4)tanalpha)#

#=(1+tanalpha)/(1-tanalpha)-(1-tanalpha)/(1+tanalpha)#

#=((1+tanalpha)^2-(1-tanalpha)^2)/((1-tanalpha)(1+tanalpha))#

#=(1+2tanalpha+tan^2alpha-1+2tanalpha-tan^2alpha)/(1-tan^2alpha)#

#=(4tanalpha)/(1-tan^2alpha)#

#=(2tanalpha+2tanalpha)/(1-tan^2alpha)#

#=(2tanalpha)/(1-tan^2alpha)+(2tanalpha)/(1-tan^2alpha)#

#=tan2alpha+tan2alpha#

#=2tan2alpha="right side "rArr"verified"#