We know,
#color(blue)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB)=> tan(A±B)*{1∓tanAtanB}=tanA±tanB)#
So,
#(tanunderbrace((π/4+theta))_color(blue)text(A)-tanunderbrace((π/4-theta)_color(blue)text(B)))#
#= tan(cancel(π/4)+theta-cancel(π/4)+theta)*{1+tan(π/4+theta) tan (π/4-theta)}#
Again applying, #color(blue)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB)=> tan(A±B)*1∓tanAtanB=tanA±tanB)#
#=tan(2theta){1+(tan(π/4)+tantheta)/(1-tan(π/4)tantheta)×(tan(π/4)-tantheta)/(1+tan(π/4)tantheta)}#
We know,
#tan(π/4)=1#
So,
#=tan(2theta){1+cancel((1+tantheta))/(cancel((1-tantheta)))×(cancel((1-tantheta)))/(cancel((1+tantheta)))}#
#=2tan2theta#
#=RHS#
hence, proved! :)