# Write the function in the simplest form : tan^{- 1} [2 cos (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

sin^{- 1} 1/2 = x

Hence,

sin x = 1/2

Using trigonometric formulaes

= sin (π / 6)

x = π / 6

sin^{- 1} 1/2

= π / 6

Therefore,

tan^{- 1} [2 cos (2 sin^{- 1} 1 / 2]

= tan^{- 1} [2 cos (2 × π / 6)]

= tan^{- 1} [2 cos π / 3]

= tan^{- 1} [2 x 1/2]

= tan^{- 1} [1]

= π / 4

tan^{- 1} [2 cos (2 sin^{- 1} 1/2] = π / 4

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.2 Question 11

## Write the function in the simplest form : tan^{- 1} [2 cos (2 sin^{- 1} 1/2]

**Summary:**

The function in the simplest form: tan^{- 1} [2 cos (2 sin^{- 1} 1/2] can be written as π / 4. Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios