Write the value of tan⁻¹[2sin(2cos⁻¹√3/2)]

First solve the inside bracket value
2cos⁻¹√3/2
The range of the principal value of cos⁻¹x is [0, π]
Let cos⁻¹√3/2 = x
cosx = √3/2
x = π/6 where x ∈ [0, π]
2cos⁻¹√3/2 = 2(π/6) = π/3
Then the expression reduces to tan⁻¹(2sin(π/3))
Now we are going to reduce the term 2sin(π/3)
We know that sin(π/3) = √3/2
2sin(π/3) = 2(√3/2) = √3
The expression becomes, tan⁻¹(√3)
We know that tan⁻¹(√3) = π/3
Therefore, tan⁻¹[2sin(2cos⁻¹√3/2)] = π/3.