# NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 12 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 12 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 12 Mathematics are an important part of exams for Class 12 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 12 Mathematics and also download more latest study material for all subjects. Chapter 2 Inverse Trigonometric Functions is an important topic in Class 12, please refer to answers provided below to help you score better in exams

## Chapter 2 Inverse Trigonometric Functions Class 12 Mathematics NCERT Solutions

Class 12 Mathematics students should refer to the following NCERT questions with answers for Chapter 2 Inverse Trigonometric Functions in Class 12. These NCERT Solutions with answers for Class 12 Mathematics will come in exams and help you to score good marks

### Chapter 2 Inverse Trigonometric Functions NCERT Solutions Class 12 Mathematics

Exercise 2.1

Find the principal values of the following:
Question 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.

Question 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.

Question 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.

Question 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

Question 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

Question 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.

Question 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.

Question 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.

Question 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.

Question 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.

Find the values of the following:

Question 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

Question 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

Question 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.

Question 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.

Exercise 2.1

Prove the following

Question 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

Question 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

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

Answer :

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

Answer :

Write the following functions in the simplest form:

Question 5

Answer :

Question 6.

Answer :

Question 7.

Answer :

Question 8.

Answer :

Question 9.

Answer :

Question 10.

Answer :

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 :

Question 14. If

Answer :

Question 15. If

Answer :

Question 16.

Answer :

Question 17

Answer :

Question 18.

Answer :

Question 19.

Answer :

Question 20.

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1

Answer :

Question 21. tan-1√3 - cot-1 (-√3) is equal to:
(A) π
(B) -π/2
(C) 0
(D) 2√3

Answer :

Miscellaneous Solutions

Question 1.

Answer :

Question 2.

Answer :

Question 3.

Answer :

Question 4.

Answer :

Question 5.

Answer :

Question 6.

Answer :

Question 7.

Answer :

Question 8.

Answer :

Question 9.

Answer :

Question 10.

Answer :

Question 11.

Answer :

Question 12.

Answer :

Question 13. 2 tan-1 (cos x) = tan-1 (2 cosec z) is equal to

Answer :

Question 14.

Answer :

Question 15. sin (tan-1 x) x <1 is equal to

Answer :

Question 16. sin–1(1 – x) – 2 sin–1x = π/2, then x is equal to
(A) 0, 1/2
(B) 1, 1/2
(C) 0
(D) 1/2

Answer :

Given that sin–1(1 − x) − 2sin–1x = π/2

Let x = sin y

∴ sin–1(1 − sin y) − 2y = π/2

⇒ sin–1(1 − sin y) = π/2 + 2y

⇒ 1 − sin y = sin (π/2 + 2y)

⇒ 1 − sin y = cos 2y

⇒ 1 − sin y = 1 − 2sin2y [as cos2y = 1−2sin2y]

⇒ 2sin2y − sin y = 0

⇒ 2x2 − x = 0 [as x = sin y]

⇒ x(2x − 1) = 0

⇒ x = 0 or,  x = 1/2

But x = 1/2 does not satisfy the given equation.

∴ x = 0 is the solution of the given equation.

The correct option is C.

Question 17. tan–1(x/y) − tan–1(x-y/x+y) is equal to
(A) π/2
(B) π/3
(C) π/4
(D) -3π/4

Answer :

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

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### NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

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#### Class 12 NCERT Solution Mathematics Chapter 2 Inverse Trigonometric Functions

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