Find the value of tan^{- 1} (1) + cos^{- 1} (- 1/2) + sin^{- 1} (- 1/2)

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} (1) = x

Hence,

tan x = 1

= tan (π / 4)

Therefore,

tan^{- 1} (1) = (π / 4)

Now, let cos^{- 1} (- 1/2) = y

Hence,

cos y = - 1/2

= - cos (π/3)

= cos (π - π/3)

= cos (2π/3)

Therefore,

cos^{- 1} (- 1/2) = (2π / 3)

Again, let sin^{- 1} (- 1/2)

= z

Hence,

sin z = - 1 / 2

= - sin (π / 6)

= sin (- π / 6)

Therefore,

sin^{- 1} (- 1/2) = (- π / 6)

Thus,

tan^{- 1} (1) + cos^{- 1} (- 1/2) + sin^{- 1} (- 1/2) =

π / 4 + 2 π / 3 - π / 6

= (3 π + 8 π - 2 π) / 12

= 9 π / 12

= 3 π / 4

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.1 Question 11

Find the value of tan^{- 1} (1) + cos^{- 1} (- 1/2) + sin^{- 1} (- 1/2).

Summary:

The value of tan^{- 1} (1) + cos^{- 1} (- 1/2) + sin^{- 1} (- 1/2) = π / 4 + 2 π / 3 - π / 6 =  3 π / 4