NCERT Exemplar Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Referring to the NCERT Exemplar textbooks for various classes subject-wise, students will get a thorough knowledge of the concepts covered in their class. Further, students can accessNCERT Exemplar Solutions to learn the steps involved in solving the exercise problems of every chapter. All these solutions are created by subject experts following the latest CBSE patterns.

Chapter 2 of NCERT Exemplar Solutions for Class 12 Maths is Inverse Trigonometric Functions. In this chapter, students will understand concepts related to inverse trigonometry functions and their properties. Also, they will learn the restrictions on ranges and domains of trigonometric functions. Further, they will also record their behaviour via graphical representations. Students who wish to learn the right methods of solving these exercise problems can make use of the solutions PDF ofNCERT Exemplar Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions from the link given below.

Download the PDF of NCERT Exemplar Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions

NCERT Exemplar Solutions Class 12 Mathematics Chapter 2
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 1
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 2
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 3
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 4
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 5
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 6
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 7
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 8
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 9
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 10
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 11
NCERT Exemplar Solutions Class 12 Mathematics Chapter 2 12

Access answers to NCERT Exemplar Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Exercise 2.3 Page No: 35

Short Answer (S.A)

1. Find the value of tan-1 [tan (5π/6)] + cos-1 [cos (13π/6)]

Solution:

We know that,

tan-1 tan x = x, x ∈ (-π/2, π/2)

And, here

tan-1 tan (5π/6) ≠ 5π/6 as 5π/6 ∉ (-π/2, π/2)

Also,

cos-1 cos x = x; x ∈ [0, π]

So,

cos-1 cos (13π/6) ≠ 13π/6 as 13π/6 ∉ [0, π]

Now,

tan-1 [tan (5π/6)] + cos-1 [cos (13π/6)]

= tan-1 [tan (π – π /6)] + cos-1 [cos (2π + π/6)]

= tan-1 [ -tan π /6] + cos-1 [ -cos (7π/6)]

= – tan-1 [tan π /6] + cos-1 [cos (π/6)]

= – π /6 + π /6

= 0

2. Evaluate cos[cos-1(-√3/2) + π/6]

Solution:

NCERT Exemplar Solutions Chapter 2 - 1

3. Prove that cot (π/4 – 2 cot-1 3) = 7

Solution:

Re-writing the given,

NCERT Exemplar Solutions Chapter 2 - 2

NCERT Exemplar Solutions Chapter 2 - 3

L.H.S = R.H.S

– Hence Proved

4. Find the value of tan-1 (-1/√3) + cot-1(1/√3) + tan-1(sin (-π/2))

Solution:

Given,

tan-1 (-1/√3) + cot-1(1/√3) + tan-1(sin (-π/2))

NCERT Exemplar Solutions Chapter 2 - 4

5. Find the value of tan-1 (tan 2π/3).

Solution:

We know that,

tan-1 tan x = x, x ∈ (-π/2, π/2)

NCERT Exemplar Solutions Chapter 2 - 5

6. Show that 2 tan-1(-3) = – π/2 + tan-1(-4/3)

Solution:

Taking L.H.S = 2 tan-1(-3) = -2 tan-1 3 (∵ tan-1 (-x) = – tan-1 x, x ∈ R)

NCERT Exemplar Solutions Chapter 2 - 6

= R.H.S

– Hence Proved.

NCERT Exemplar Solutions Chapter 2 - 7

7. Find the real solution of the equation

Solution:

Given equation,

NCERT Exemplar Solutions Chapter 2 - 8

Hence, the real solutions of the given trigonometric equation are 0 and -1.

8. Find the value of the expression sin (2 tan-1 1/3) + cos (tan-1 2√2).

Solution:

Given expression, sin (2 tan-1 1/3) + cos (tan-1 2√2)

NCERT Exemplar Solutions Chapter 2 - 9

9. If 2 tan-1(cos θ) = tan-1 (2 cosec θ), then show that θ = π/4.

Solution:

Given, 2 tan-1(cos θ) = tan-1 (2 cosec θ)

So,

NCERT Exemplar Solutions Chapter 2 - 10

10. Show that cos (2 tan-1 1/7) = sin (4 tan-1 1/3).

Solution:

Taking L.H.S, we have

NCERT Exemplar Solutions Chapter 2 - 11

NCERT Exemplar Solutions Chapter 2 - 12

Thus, L.H.S = R.H.S

– Hence proved

11. Solve the equation cos (tan-1 x) = sin (cot-1 3/4).

Solution:

Given equation, cos (tan-1 x) = sin (cot-1 3/4)

Taking L.H.S,

NCERT Exemplar Solutions Chapter 2 - 13

On squaring on both sides,

16(x2 + 1) = 25

16x2 + 16 = 25

16x2 = 9

x2 = 9/16

Therefore, x = ± 3/4

NCERT Exemplar Solutions Chapter 2 - 14

12. Prove that

Solution:

NCERT Exemplar Solutions Chapter 2 - 15

Taking L.H.S,

NCERT Exemplar Solutions Chapter 2 - 16

= R.H.S

– Hence Proved

13. Find the simplified form of cos-1 [3/5 cos x + 4/5 sin x], x ∈ [-3π/4, π/4].

Solution:

We have,

cos-1 [3/5 cos x + 4/5 sin x], x ∈ [-3π/4, π/4]

Now, let cos α = 3/5

So, sin α = 4/5 and tan α = 4/3

cos-1 [3/5 cos x + 4/5 sin x]

⇒ cos-1 [3/5 cos x + 4/5 sin x] = cos-1 [cos α cos x + sin α sin x]

= cos-1 [cos (α – x)]

= α – x

= tan-1 4/3 – x

14. Prove that sin-1 8/17 + sin-1 3/5 = sin-1 77/85

Solution:

Taking the L.H.S,

= sin-1 8/17 + sin-1 3/5

NCERT Exemplar Solutions Chapter 2 - 18= tan-1 8/15 + tan-1 3/4

NCERT Exemplar Solutions Chapter 2 - 17

– Hence proved

15. Show that sin-1 5/13 + cos-1 3/5 = tan-1 63/16

Solution:

Here, sin-1 5/13 = tan-1 5/12

And,

cos-1 3/5 = tan-1 4/3

Taking the L.H.S, we have

NCERT Exemplar Solutions Chapter 2 - 19

Thus, L.H.S = R.H.S

– Hence Proved

16. Prove that tan-1 1/4 + tan-1 2/9 = sin-1 1/√5

Solution:

Taking the LHS,

tan-1 1/4 + tan-1 2/9

NCERT Exemplar Solutions Chapter 2 - 20

= RHS

– Hence Proved

17. Find the value of 4 tan-1 1/5 – tan-1 1/239

Solution:

4 tan-1 1/5 – tan-1 1/239

= 2 (tan-1 1/5) – tan-1 1/239

NCERT Exemplar Solutions Chapter 2 - 21

= 2 tan-1 5/12 – tan-1 1/239

NCERT Exemplar Solutions Chapter 2 - 22

Thus,

4 tan-1 1/5 – tan-1 1/239 = π/4

18. Show that tan-1 (1/2 sin-1 3/4) = (4 – √7)/ 3 and justify why the other value (4 + √7)/ 3 is ignored.

Solution:

We have, tan-1 (1/2 sin-1 3/4)

Let ½ sin-1 ¾ = θ ⇒ sin-1 ¾ = 2θ ⇒ sin 2θ = ¾

2 tan θ/ 1 + tan2 θ = ¾

3 tan2 θ – 8 tan θ + 3 = 0

NCERT Exemplar Solutions Chapter 2 - 23

Hence,

NCERT Exemplar Solutions Chapter 2 - 24

19. If a1, a2, a3, …., an is an arithmetic progression with common difference d, then evaluate the following expression.

NCERT Exemplar Solutions Chapter 2 - 25

Solution:

As a1, a2, a3, …., an is an arithmetic progression with common difference d.

d = a2 – a1 = a3 – a2 = a4 – a3 = …… = an – an-1

So,

NCERT Exemplar Solutions Chapter 2 - 26

NCERT Exemplar Solutions Chapter 2 - 27

20. Which of the following is the principal value branch of cos-1 x?

(a) [-π/2, π/2] (b) (0, π) (c) [0. π] (d) [0, π] – {π/2}

Solution:

(c) [0. π]

As we know that the principal value branch cos-1 x is [0, π].

21. Which of the following is the principal value branch of cosec-1 x?

(a) (-π/2, π/2) (b) [0, π] – {π/2} (c) [-π/2, π/2] (d) [-π/2, π/2] – {0}

Solution:

(d) [-π/2, π/2] – {0}

As the principal branch of cosec-1 x is [-π/2, π/2] – {0}.

22. If 3 tan-1 x + cot-1 x = π, then x equals

(a) 0 (b) 1 (c) -1 (d) ½

Solution:

(b) 1

Given, 3 tan-1 x + cot-1 x = π

2 tan-1 x + tan-1 x + cot-1 x = π

2 tan-1 x + π/2 = π (As tan-1 + cot-1 = π/2)

2 tan-1 x = π/2

tan-1 x = π/4

x = 1

23. The value of sin-1 cos 33π/5 is

(a) 3π/5 (b) -7π/5 (c) π/10 (d) -π/10

Solution:

(d) -π/10

NCERT Exemplar Solutions Chapter 2 - 28

24. The domain of the function cos-1 (2x – 1) is

(a) [0, 1] (b) [-1, 1] (c) [-1, 1] (d) [0, π]

Solution:

(a) [0, 1]

Since, cos-1 x is defined for x ∈ [-1, 1]

So, f(x) = cos-1 (2x – 1) is defined if

-1 ≤ 2x – 1 ≤ 1

0 ≤ 2x ≤ 2

Hence,

0 ≤ x ≤ 1

25. The domain of the function by f(x) = sin-1 √(x – 1) is

(a) [1, 2] (b) [-1, 1] (c) [0, 1] (d) none of these

Solution:

(a) [1, 2]

We know that, sin-1 x is defined for x ∈ [-1, 1]

So, f(x) = sin-1 √(x – 1) is defined if

0 ≤ √(x – 1) ≤ 1

0 ≤ x – 1 ≤ 1

1 ≤ x ≤ 2

Hence,

x ∈ [1, 2]

26. If cos (sin-1 2/5 + cos-1 x) = 0, then x is equal to

(a) 1/5 (b) 2/5 (c) 0 (d) 1

Solution:

(b) 2/5

Given,

cos (sin-1 2/5 + cos-1 x) = 0

So, this can be rewritten as

sin-1 2/5 + cos-1 x = cos-1 0

sin-1 2/5 + cos-1 x = π/2

cos-1 x = π/2 – sin-1 2/5

cos-1 x = cos-1 2/5 [Since, cos-1 x + sin-1 x = π/2]

Hence,

x = 2/5

27. The value of sin (2 tan-1(0.75)) is equal to

(a) 0.75 (b) 1.5 (c) 0.96 (d) sin 1.5

Solution:

(c) 0.96

We have, sin (2 tan-1(0.75))

NCERT Exemplar Solutions Chapter 2 - 29

Leave a Comment

Your email address will not be published. Required fields are marked *