If sin (sin^{- 1} 1/5 + cos^{- 1} x) = 1, find the value of x

Solution:

Inverse trigonometric functions as a topic of learning are closely related to the basic trigonometric ratios.

It is given that

sin (sin^{- 1} 1/5 + cos^{- 1} x) = 1

Since we know that sin (x + y)

= sin x cos y + cos x sin y [ using trigonometric identities]

Therefore,

sin (sin^{- 1} 1/5 cos(cos^{- 1} x) + cos (sin^{- 1} 1/5) sin (cos^{- 1} x) = 1

(1/5) x (x) + cos (sin^{- 1} 1/5) sin (cos^{- 1} x) = 1

x/5 + cos (sin^{- 1} 1/5) sin (cos^{- 1} x) = 1 ....(1)

Now, let sin^{- 1} 1/5 = y

⇒ sin y = 1/5

Then,

cos y = √1 - (1/5)²

= (2√6/5)

= cos^{- 1} (2√6/5)

Therefore,

sin^{- 1} 1/5 = cos^{- 1} (2√6/5) ....(2)

Now, let cos^{- 1} x = z

⇒ cos z = x

Then,

sin z = √1 - x²

z = sin^{- 1}√1 - x²

Therefore,

cos^{- 1} x = sin^{- 1}√1 - x² ....(3)

From (1), (2) and (3) , we have

⇒ x/5 + cos (cos^{- 1} (2√6/5)) sin (sin^{- 1}√1 - x²) = 1

⇒ x/5 + (2√6/5)√1 - x² = 1

⇒ x + (2√6)√1 - x² = 5

⇒ 5 - x = (2√6)√1 - x²

On squaring both the sides

25 + x² -10x = 24 - 24x²

25x² - 10x + 1 = 0

(5x - 1)² = 0

(5x - 1) = 0

x = 1 / 5

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.2 Question 14