NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions

Class 12: Maths Chapter 2 solutions. Complete Class 12 Maths Chapter 2 Notes.

NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions

Class 12: Maths Chapter 2 solutions. Complete Class 12 Maths Chapter 2 Notes.

Page No: 41

Exercise 2.1

Find the principal values of the following:
1. sin^-1(-1/2)

Answer

1. Let sin^-1(−1/2) = y, then

sin y = −1/2 = −sin(π/6) = sin(−π/6)

Range of the principal value of sn^-1 is [-π/2, π/2] and sin -π/6) = -1/2

Therefore, the principal value of sin^-1(-1/2) is -π/6.

2. cos^-1(√3/2)

Answer

Let cos^-1(√3/2) = y,

cos y = √3/2 = cos (π/6)

We know that the range of the principal value branch of cos^-1 is [0, π] and cos (π/6) = √3/2

Therefore, the principal value of cos^-1(√3/2) is π/6.

3. cosec^-1(2)

Answer

Let cosec^-1(2) = y.

Then, cosec y = 2 = cosec (π/6)

We know that the range of the principal value branch of cosec^-1 is [-π/2, π/2] - {0} and cosec (π/6) = 2.

Therefore, the principal value of cosec^-1(2) is π/6.

4. tan^-1(√3)

Answer

Let tan^-1(-√3) = y,

then tan y = -√3 = -tan π/3 = tan (-π/3)

We know that the range of the principal value branch of tan^-1 is (-π/2, π/2) and tan (-π/3)

= -√3

Therefore, the principal value of tan^-1 (-√3) is -π/3

5. cos^-1(-1/2)

Answer

Let cos^-1(-1/2) = y,

then cos y = -1/2 = -cos π/3 = cos (π-π/3) = cos (2π/3)

We know that the range of the principal value branch of cos^-1 is [0, π] and cos (2π/3) = -1/2

Therefore, the principal value of cos^-1(-1/2) is 2π

6. tan^-1(-1)

Answer

Let tan^-1(-1) = y. Then, tan y = -1 = -tan (π/4) = tan (-π/4)

We know that the range of the principal value branch of tan^-1 is (-π/2, π/2) and tan (-π/4) = -1.

Therefore, the principal value of tan^-1(−1) is -π/4.

7. sec^-1(2/√3)

Answer

Let sec^-1(2/√3) = y, then sec y = 2/√3 = sec (π/6)

We know that the range of the principal value branch of sec^-1 is [0, π] − {π/2} and sec (π/6) = 2/√3.

Therefore, the principal value of sec^-1(2/√3) is π/6.

8. cot^-1(√3)

Answer

Let cot^-1√3 = y, then cot y = √3 = cot (π/6).

We know that the range of the principal value branch of cot^-1 is (0, π) and cot (π/6) = √3.

Therefore, the principal value of cot^-1√3 is π.

9. cos^-1(-1/√2)

Answer

Let cos^-1(-1/√2) = y,

then cos y = -1/√2 = -cos (π/4) = cos (π - π/4) = cos (3π/4).

We know that the range of the principal value branch of cos^-1 is [0, π] and cos (3π4) = -1/√2.

Therefore, the principal value of cos^-1(-1/√2) is 3π/4.

10. cosec^-1(-√2)  

Answer

Let cosec^-1(−√2) = y, then cosec y = −√2 = −cosec (π/4) = cosec (−π/4)

We know that the range of the principal value branch of cosec^-1 is [-π/2, π/2]-{0} and cosec(-π/4) = -√2.

Therefore, the principal value of cosecc^-1(-√2) is -π/4.

Page No. 42

Find the values of the following:

11. tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2)

Answer

Let tan^-1(1) = x,

then tan x = 1 = tan(π/4)

We know that the range of the principal value branch of tan^-1 is (−π/2, π/2).

∴ tan^-1(1) = π/4

Let cos^-1(−1/2) = y,

then cos y = −1/2 = −cosπ/3 = cos (π − π/3)

= cos (2π/3)

We know that the range of the principal value branch of cos^-1 is [0, π].

∴ cos^-1(−1/2) = 2π/3

Let sin^-1(−1/2) = z,

then sin z = −1/2 = −sin π/6 = sin (−π/6)

We know that the range of the principal value branch of sin^-1 is [-/π2, π/2].

∴ sin^-1(-1/2) = -π/6

Now,

tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2)

= π/4 + 2π/3 − π/6

= (3π + 8π − 2π)/12

= 9π/12 = 3π/4

12. cos^-1(1/2) + 2 sin^-1(1/2)

Answer

Let cos^-1(1/2) = x, then
cos x = 1/2 = cos π/3
We know that the range of the principal value branch of cos−1 is [0, π].
∴ cos^-1(1/2)
= π/3
Let sin^-1(-1/2) = y, then
sin y = 1/2
= sin π/6

We know that the range of the principal value branch of sin^-1 is [-π/2, π/2].
∴ sin^-1(1/2) = π/6

Now,
cos^-1(1/2) + 2sin^-1(1/2)
= π/3 + 2×π/6
= π/3 + π/3
= 2π/3

13. If sin^-1 x = y, then

(A) 0 ≤ y ≤ π

(B) -π/2 ≤ y ≤ π/2

(C) 0 < y < π 

(D) -π/2 < y < π/2

Answer

It is given that sin^-1x = y.

We know that the range of the principal value branch of sin^-1 is [-π/2, π/2].

Therefore, -π/2 ≤ y ≤ π/2.

Hence, the option (B) is correct.

14. tan^-1√3 - sec^-1(-2) is equal to

(A) π

(B) -π/3

(C) π/3

(D) 2π/3

Answer

Let tan^-1√3 = x,then

tan x = √3 = tan π/3

We know that the range of the principal value branch of tan^-1 is (-π/2, π/2).

∴ tan^-1√3 = π/3

Let sec^-1(-2) = y, then

sec y = -2 = -sec π/3

= sec (π - π/3)

= sec (2π/3)

We know that the range of the principal value branch of sec^-1 is [0, π]- {π/2}

∴ sec^-1(-2) =2π/3

Now,

tan^-1√3 - sec^-1(-2)

= π/3 - 2π/3

= -π/3

Hence, the option (B) is correct.

Page No: 41

Exercise 2.1

Prove the following:
1. 3sin^-1x = sin^-1(3x – 4x3), x ∈ [-/2, 1/2]

Answer

To prove:

3sin^-1x = sin^-1(3x − 4x3), x ∈ [−1/2, 1/2]

Let sin^-1x = θ, then x = sin θ.

We have,

RHS = sin-1(3x - 4x³)

= sin^-1(3 sin θ - 4sin3θ)

= sin^-1(sin 3θ) = 3θ

= 3sin^-1x = LHS

2. 3cos^-1x = cos^-1(4x³ – 3x) x ∈ [1, 1/2]

Answer

To prove:

3cos^-1x = cos^-1(4x³ – 3x) x ∈ [1, 1/2].

Let cos^-1x = θ, then x = cos θ.

We have,

RHS = cos^-1(4x³ - 3x)

= cos^-1(4cos³θ - 3cosθ)

= cos^-1(cos 3θ) = 3θ

= 3cos^-1x

= LHS

3. tan^-12/11 + tan^-17/24 = tan^-11/2

Answer

To prove: tan^-12/11 + tan^-17/24 = tan^-11/2
LHS = tan^-12/11 + tan^-17/24

= tan^-1(48 + 77)/(264 − 14)

= tan^-1125/250 = tan^-11/2 = RHS

4. 2tan^-11/2 + tan^-11/7 = tan^-131/17

Answer

To prove: 2tan^-11/2 + tan^-11/7 = tan^-131/17

LHS = 2tan-11/2 + tan-11/7

Write the following functions in the simplest form:

Question: 5

Answer

Question: 6

Answer

Question: 7

Answer

Question: 8

Answer

Page No. 48

Question: 9

Answer

Question: 10

Answer

Find the values of each of the following:

Question: 11

Answer

Question: 12. cot (tan^-1a + cot^-1a)

Answer

The given function is cot(tan^-1a + cot^-1a).

∴ cot(tan^-1a + cot^-1a)

= cot (π/2) [tan^-1x + cot^-1x = π/2]

= 0

Question: 13

Answer

Formula used:

Question: 14

Answer

Question: 15

Answer

Question: 16

Answer

Question: 17

Answer

Question: 18

Answer

Question: 19

Answer

The correct option is B.

Question: 20

Answer

The correct option is D.

Question: 21

Answer

The correct option is B.

Class 12: Maths Chapter 2 solutions. Complete Class 12 Maths Chapter 2 Notes.

NCERT Solutions for 12th Class Maths: Chapter 2: Download PDF

NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions

Download PDF: NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions PDF

Chapterwise NCERT Solutions for Class 12 Maths :

• Chapter 1 – Relations and Functions
• Chapter 2 – Inverse Trigonometric Functions.
• Chapter 3 – Matrices
• Chapter 4 – Determinants.
• Chapter 5 – Continuity and Differentiability.0.0
• Chapter 6 – Application of Derivatives.
• Chapter 7 – Integrals.
• Chapter 8 – Application of Integrals.
• Chapter 9: Differential Equations
• Chapter 10: Vector Algebra
• Chapter 11: Three Dimensional Geometry
• Chapter 12: Linear Programming
• Chapter 13: Probability

About NCERT

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