The correct option is A 1√3
Put x=tanθ
⇒tan−1x=θ∈(0,π4)⇒2θ∈(0,π2)
Now,
3sin−1(2tanθ1+tan2θ)−4cos−1(1−tan2θ1+tan2θ)+2tan−1(2tanθ1−tan2θ)=π3⇒3sin−1(sin2θ)−4cos−1(cos2θ)+2tan−1(tan2θ)=π3
Since we know that 2θ∈(0,π2), the above expression can be written as
⇒3⋅2θ−4⋅2θ+2⋅2θ=π3⇒2θ=π3⇒θ=π6∴x=tanπ6=1√3