Question -
Answer -
(i) Given tan-1 (1/√3)
We know that for any x∈ R, tan-1 representsan 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 thevalue is equal to π/6
Therefore tan-1 (1/√3)= π/6
Hence the principalvalue of tan-1 (1/√3) = π/6
(ii) Given tan-1 (-1/√3)
We know that for any x∈ R, tan-1 representsan 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 thevalue is equal to -π/6
Therefore tan-1 (-1/√3)= -π/6
Hence the principalvalue 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 representsan angle in (-π/2, π/2) whose tangent is x.
Therefore tan-1 (0)= 0
Hence the principalvalue of tan-1 (cos (π/2) is 0.
(iv) Given that tan-1 (2cos (2π/3))
But we know that cosπ/3 = 1/2
So, cos (2π/3) = -1/2
Therefore tan-1 (2cos (2π/3)) = tan-1 (2 × – ½)
= tan-1(-1)
= – π/4
Hence, the principalvalue of tan-1 (2 cos (2π/3)) is – π/4