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

where π / 3 ∈ (- π/2, π/2)

Therefore,

tan^{- 1} √3 = π/3

Now,

let us assume cot^-1 (- √3) = y

Hence,

cot y = (- √3)

= - cot (π / 6)

= cot (π - π / 6)

= cot (5π / 6)

Since, Range of principal value of cot^{- 1} x = (0, π)

Therefore,

cot^{- 1}(- √3) = (5π / 6)

Then,

tan^{- 1} √3 - cot^{- 1} (- √3)

= π / 3 - 5π / 6

= - π / 2

Thus, the correct option is B

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.2 Question 21

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

Summary:

The value of tan^{- 1} √3 - cot^{- 1} (- √3) is equal to - π / 2. Thus, the correct option is B