# Find the value of tan^{- 1 }√3 sec^{- 1} (- 2) is equal to

(A) 0 (B) - π/3 (C) π/3 (D) 2π/3

**Solution:**

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, tan^{- 1 }√3 = x

Hence,

tan x = √3

= tan (π / 3)

Range of the principal value of tan^{- 1} x

= (- π/2, π/2)

Therefore,

tan^{- 1 }√3 = π/3

sec^{- 1} (- 2) = y

Hence,

sec y = (- 2)

= - sec (π / 3)

= sec (π - π / 3)

= sec (2π / 3)

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

= [0, π] - {π / 2}

Therefore, sec^{-1} (-2) = 2π / 3

Thus,

tan^{- 1 }√3 sec^{- 1} (- 2)

= π / 3 - 2π / 3

= - π/3

The answer is B

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.1 Question 14

## Find the value of tan^{- 1 }√3 sec^{- 1} (- 2) is equal to (A) 0 (B) - π/3 (C) π/3 (D) 2π/3

**Summary:**

The value of tan^{- 1 }√3 sec^{- 1} (- 2) is equal to - π/3. The correct answer is B