 # Find the value of each of the following: (i) tan-1 (1/?3) (ii) tan-1 (-1/?3) (iii) tan-1 (cos (?/2) ) (iv) tan-1 (2 cos (2?/3) )

(i) Given tan-1 (1/√3)

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

So, 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

Therefore tan-1 (1/√3) = π/6

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

(ii) Given tan-1 (-1/√3)

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

So, 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

Therefore tan-1 (-1/√3) = -π/6

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

(iii) Given that tan-1 (cos (π/2))

But we know that cos (π/2) = 0

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

Therefore tan-1 (0) = 0

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

(iv) Given that tan-1 (2 cos (2π/3))

But we know that cos π/3 = 1/2

So, cos (2π/3) = -1/2

Therefore tan-1 (2 cos (2π/3)) = tan-1 (2 × – ½)

= tan-1(-1)

= – π/4

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