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

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

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

Class 12:Maths Chapter 2 solutions. Complete Class 12Maths 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 – 4×3), x ∈ [-/2, 1/2]

Answer

To prove:

3sin-1x = sin-1(3x − 4×3), 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 12Maths Chapter 2 Notes.

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

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

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