In this Solution, we assume that, as functions #tan^-1=arc tan.#
Let us first find, #arc cos(-1/2).#
Recall the following Definition of #arc cos# function :
# arc cos x=theta, |x| <=1 iff costheta=x, theta in [0,pi].#
We have, #cos(2pi/3)=cos (pi-pi/3)=-cos(pi/3)=-1/2, &, (2pi/3) in [0,pi].#
# rArr arc cos(-1/2)=2pi/3..........."[Defn.]"#
Therefore, the Reqd. Value=#tan(A-arc cos(-1/2)),#
#=tan(A-2pi/3),#
#:." the Reqd. Value="(tanA-tan(2pi/3))/(1+tanAtan(2pi/3))........(ast).#
Here, #tan(2pi/3)=tan(pi-pi/3)=-tan(pi/3)=-sqrt3.....(ast_1).#
Given that, #2A=tan^-1(4/3) rArr tan2A=4/3, &, 2A in (-pi/2,pi/2).#
But, #tan2A >0, 2A in (0,pi/2).#
#because, tan2A=(2tanA)/(1-tan^2A)=(2t)/(1-t^2), t=tanA,#
#:. (2t)/(1-t^2)=4/3 :. 2-2t^2=3t :. 2t^2+3t-2=0.#
#:. (t+2)(2t-1)=0 :. t=-2, or, t=1/2.#
Since, # 2A in (0,pi/2), t=tanA > 0 rArr t=1/2....(ast_2).#
Utilising #(ast_1) and (ast_2)" in "(ast),# we have,
#" The Reqd. Value="(1/2-(-sqrt3))/(1+(1/2)(-sqrt3))=(1+2sqrt3)/(2-sqrt3)#
#=(1+2sqrt3)(2+sqrt3)=8+5sqrt3.#
Enjoy Maths.!
