Find the value of each of the following:

(i) tan^-1 (1/√3)

(ii) tan^-1 (-1/√3)

(iii) tan^-1 (cos (π /2) )

(iv) tan^-1 (2 cos (2π /3) )

To find the values of the above inverse trigonometric ratios, we will use trigonometric identities.

Answer: The value of inverse tan ratios are (i) is π / 6, (ii) is - π / 6, (iii) is 0 and (iv) is - π / 4.

Let's find the values of inverse trigonometric ratios.

Explanation:

(i) Let  tan^-1 (1/√3) = θ

⇒ tan θ = 1/√3

⇒ tan θ = tan (π / 6)

⇒ θ = π / 6

(ii) Let tan^-1 (-1/√3) = θ

⇒ tan θ = - 1/√3

⇒ tan θ = tan (- π / 6)

⇒ θ = - π / 6

(iii) Let tan^-1 (cos (π/2)) = θ

⇒ tan θ = cos π/2

As we know that cos π/2 = 0

⇒ tan θ = 0

⇒ θ = 0

(iv) Let tan^-1 (2 cos (2π /3) ) = θ

⇒ tan θ = 2 cos (2π /3)

⇒ tan θ = 2 cos (π - π /3)

As we know that cos π  is 180º = -1 and cos π /3 is cos 60º = 1/ 2

⇒ tan θ = -2 cos (π/3)

⇒ tan θ = -2 × (1/2)

⇒ tan θ = - 1

⇒ tan θ = tan (- π / 4)

 ⇒ θ = - π / 4

Thus, the values of inverse tan ratios are (i) is π / 6, (ii) is -π / 6, (iii) is 0 and (iv) is - π / 4.