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# 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.

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