# Find the principal value of sec^{-1} (2/√3)

**Solution:**

Inverse trigonometric functions as a topic of learning are closely related to the basic trigonometric ratios.

The domain (θ value) and the range(answer) of the trigonometric ratio are changed to the range and domain of the inverse trigonometric function.

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios.

Here the basic trigonometric function of Sin θ = y can be changed to θ = sin^{-1} y

Let

sec^{-1} (2/√3) = y

Hence,

sec y = (2/√3)

= sec (π / 6)

Range of the principal value of sec^{- 1} (x)

= [0, π] - {π/2}

Thus,

principal value of sec^{-1} (2/√3)

= (π / 6)

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.1 Question 7

## Find the principal value of sec^{-1} (2/√3)

**Summary:**

The principal value of sec^{-1} (2/√3) = (π / 6). Range of the principal value of sec^{- 1} (x) = [0, π] - {π/2}