(i) Given as tan^-1 (1/√3)

As we know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

Therefore, tan^-1 (1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

we know that the value is equal to π/6

So, tan^-1 (1/√3) = π/6

Hence the principal value of tan^-1 (1/√3) = π/6

(ii) Given as tan^-1 (-1/√3)

As we know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

Therefore, tan^-1 (1-/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

But we know that the value is equal to -π/6

So, tan^-1 (-1/√3) = -π/6

Hence the principal value of tan^-1 (-1/√3) = – π/6

(iii) Given tan^-1 (cos(π/2))

we know that cos(π/2) = 0

As we know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

So, tan^-1 (0) = 0

Hence the principal value of tan^-1 (cos(π/2) is 0.

(iv) Given tan^-1 (2 cos(2π/3))

we know that cos π/3 = -1

So, tan^-1 (2 cos(2π/3)) = tan^-1 (2 × – ½)

= tan^-1(-1)

= – π/4

Thus, the principal value of tan^-1 (2 cos(2π/3)) is – π/4