In this Solution, we assume that, as functions tan^-1=arc tan.tan−1=arctan.
Let us first find, arc cos(-1/2).arccos(−12).
Recall the following Definition of arc cosarccos function :
arc cos x=theta, |x| <=1 iff costheta=x, theta in [0,pi].arccosx=θ,|x|≤1⇔cosθ=x,θ∈[0,π].
We have, cos(2pi/3)=cos (pi-pi/3)=-cos(pi/3)=-1/2, &, (2pi/3) in [0,pi].cos(2π3)=cos(π−π3)=−cos(π3)=−12,&,(2π3)∈[0,π].
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.!
