Solve sin (tan^{- 1} x), |x| < 1 is equal to
(A) x/(√1 - x²)   (B) 1/(√1 - x²)   (C) 1/(√1 + x²)   (D) x/(√1 + x²)

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 y = x

Therefore,

sin y = x/(√1 + x²)

Now, let tan^{- 1} x = y

Therefore, sin^{- 1} (1 - x) - 2x sin^{- 1} x

= π / 2

y = sin^{- 1} x / (√1 + x²)

Hence,

tan^{- 1}x = sin^{- 1}x / (√1 + x²)

Thus,

sin (tan^{- 1}x)

= sin (sin^{- 1}x/(√1 + x²))

= x/(√1 + x²)

Thus, the correct option is D

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 15

Solve sin (tan^{- 1} x), |x| < 1 is equal to (A) x/(√1 - x²) (B) 1/(√1 - x²) (C) 1/(√1 + x²) (D) x/(√1 + x²)

Summary:

For the given function: sin (tan^{- 1} x), |x| < 1,the correct option is D. Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios