NCERT Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

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NCERT Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

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Inverse Trigonometric Functions

Chapter – 2

Exercise 2.1

Find the principal values of the following:

Ans: 

Ans: 

3. cosec⁻¹ (2).

Ans: Let, cosec⁻¹ (2) = y 

Hence, 

cosecy = 2

4. tan⁻¹ (-√3)

Ans: Let, tan⁻¹ (- √3) = y 

Hence, 

tan y = – √3

Ans: 

6. tan⁻¹ (-1).

Ans: Let, tan⁻¹ (- 1) = y 

Hence,

tan y = – 1

Ans: 

8. cot⁻¹ (√3).

Ans: Let, cot⁻¹ (√3) = y 

Hence, 

cot y = √3

Range of the principal value of cot⁻¹ (x) = (0, π)

Thus, principal value of

Ans: Let, 

Hence,

10. cosec⁻¹ (- √2)

Ans: Let, cosec⁻¹ (-√2) = y

Hence,

cosec y = -√2

Find the values of the following:

Ans: Let, tan⁻¹ (1) = x

Hence,

tan x = 1

Ans: 

13. If sin⁻¹ x = y, then

(A) 0 ≤ y ≤ π

(C) 0 < y < π

Ans:

14. tan⁻¹ √3 – sec⁻² (- 2) is equal to

Ans: Let tan⁻¹ (√3) = x 

Hence, tan x = √3 

= tan(π/3)

EXERCISE 2.2

Prove the following:

Ans: Let x = sinθ

⇒ sin⁻¹ x = θ 

Thus we have, 

R. H. S.= sin⁻¹ (3x – 4x³) 

= sin⁻¹ (3 sin θ – 4sin³ θ) 

= sin⁻¹ (sin 3 θ)[∵ sin 3A = 3 sin A – 4sin³ A] 

= 3 θ[∵ sin⁻¹ (sin A) = A] 

= 3sin⁻¹x 

= L.H.S. 

Therefore, 3sin⁻¹ x = sin⁻¹ (3x – 4x³)

Ans: We have to prove that 3 cos⁻¹ x = cos⁻¹ (4x³ – 3x) 

First consider RHS, cos⁻¹ (4x³ – 3x)

Let us take x = cosθ, we know that cos(3θ) = 4cos³ θ – 3 cos θ 

⇒ cos⁻¹ (4x³ – 3x) = cos⁻¹ (4cos³ θ – 3cos θ) = cos⁻¹ (cos(3θ) = 3 θ 

∵ θ = cos⁻¹ x, ⇒ cos⁻¹ (4x³ – 3x) = 3 cos⁻¹ x

∴ LHS = RHS 

0 ≤ 3 cos⁻¹ x ≤ π 

0 ≤ cos¹ x ≤ π / 3 

∴ χ∈[½,1]

Hence proved

Ans:  Let x = tan θ ⇒ θ = tan⁻¹ x

Hence,

Ans: 

Ans:  

Ans: 

Ans: 

Find the values of each of the following:

Ans: 

Ans: 

Find the values of each of the expressions in Exercises 16 to 18.

Ans:  

Ans: 

Ans: 

Ans: 

Ans: 

Ans: Let tan⁻¹ √3 = x

Hence,

Miscellaneous Exercise on Chapter 2

Find the value of the following:

Ans:

Ans: 

Prove that

Ans:

Ans: 

Ans: 

Ans: 

Ans: 

Prove that

Ans: Let x = tan² θ

Then, 

√x = tanθ 

θ = tan⁻¹√x

Therefore,

Ans: 

Ans: Let x = cos 2θ ⇒ θ = 1/2 cos⁻¹ x

Thus,

Solve the following equations:

11. 2tan⁻¹ (cos x) = tan⁻¹ (2 cosec x).

Ans: It is given that 2 tan⁻¹ (cos x) = tan⁻¹ (2 cosec x).

since,

Hence,

Ans:  

Ans: 

Ans: It is given that sin⁻¹ (1 – x) – 2sin⁻¹ x = π/2 

⇒ sin⁻¹ (1 – x) – 2sin⁻¹ x = π/2 

⇒ – 2sin⁻¹ x = π/2 – sin⁻¹ (1 – x) 

⇒ – 2 sin⁻¹ x = cos⁻¹ (1 – x)  …(1)

Let sin⁻¹ x = y ⇒ sin y = x 

Hence,

cos y = √1 – x² 

y = cos⁻¹ (√1 – x²) 

sin⁻¹ x = cos⁻¹ √1 – x² 

From equation (1), we have 

– 2 cos⁻¹ √1 – x² = cos⁻¹ (1 – x)

Put x = sin y

⇒ – 2 cos⁻¹ √1 – sin²y = cos⁻¹ (1 – sin y) 

⇒ – 2 cos⁻¹ (cos y) = cos⁻¹ (1 – sin y) 

⇒ – 2y = cos⁻¹ (1 – sin y) 

⇒ 1 – sin y = cos(- 2y) 

⇒ 1 – sin y = cos 2y 

⇒ 1 – sin y = 1 – 2sin² y 

⇒ 2sin² y – sin y = 0 

⇒ sin y (2sin y – 1) = 0

⇒ sin y = 0 , 1/ 2

Therefore,

x = 0, ½

When x = ½, it does not satisfy the equation.

Hence, x = 0 is the only solution

Thus, the correct option is C.