NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.2

Chapter 3, Trigonometric Functions of Class 11 Maths, is categorised under the CBSE Syllabus for 2023-24. Understand the method of solving the problems given in the second exercise of Chapter 3, Class 11 Maths, from the NCERT Solutions given here. Exercise 3.2 of NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions is based on the following topics:

  1. Trigonometric Functions
    1. Sign of trigonometric functions
    2. Domain and range of trigonometric functions

The NCERT solutions are prepared with the utmost care by the subject matter experts at BYJU’S. Students can view as well as download the NCERT Solutions for Class 11 and boost their exam preparation.

NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions Exercise 3.2

Download PDF Download PDF

Access other exercise solutions of Class 11 Maths Chapter 3 – Trigonometric Functions

For more NCERT Class 11 Maths Solutions of Chapter 3, refer to the links below.

Exercise 3.1 Solutions 7 Questions

Exercise 3.3 Solutions 25 Questions

Exercise 3.4 Solutions 9 Questions

Miscellaneous Exercise on Chapter 3 Solutions 10 Questions

Access Solutions for Class 11 Maths Chapter 3.2 Exercise

Find the values of the other five trigonometric functions in Exercises 1 to 5.

1. cos x = -1/2, x lies in third quadrant.

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 1

2. sin x = 3/5, x lies in second quadrant.

Solution:

It is given that

sin x = 3/5

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 2

We know that

sin2 x + cos2 x = 1

We can write it as

cos2 x = 1 – sin2 x

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 3

3. cot x = 3/4, x lies in third quadrant.

Solution:

It is given that

cot x = 3/4

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 4

We know that

1 + tan2 x = sec2 x

We can write it as

1 + (4/3)2 = sec2 x

Substituting the values

1 + 16/9 = sec2 x

cos2 x = 25/9

sec x = ± 5/3

Here x lies in the third quadrant, so the value of sec x will be negative

sec x = – 5/3

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 5

4. sec x = 13/5, x lies in fourth quadrant.

Solution:

It is given that

sec x = 13/5

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 6

We know that

sin2 x + cos2 x = 1

We can write it as

sin2 x = 1 – cos2 x

Substituting the values

sin2 x = 1 – (5/13)2

sin2 x = 1 – 25/169 = 144/169

sin2 x = ± 12/13

Here x lies in the fourth quadrant, so the value of sin x will be negative

sin x = – 12/13

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 7

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 8

5. tan x = -5/12, x lies in second quadrant.

Solution:

It is given that

tan x = – 5/12

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 9

We know that

1 + tan2 x = sec2 x

We can write it as

1 + (-5/12)2 = sec2 x

Substituting the values

1 + 25/144 = sec2 x

sec2 x = 169/144

sec x = ± 13/12

Here x lies in the second quadrant, so the value of sec x will be negative

sec x = – 13/12

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 10

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°

Solution:

We know that values of sin x repeat after an interval of 2π or 360°

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 11

By further calculation

= sin 45o

= 1/ √ 2

7. cosec (–1410°)

Solution:

We know that values of cosec x repeat after an interval of 2π or 360°

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 12

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 13

= cosec 30o = 2

8. NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 14

Solution:

We know that values of tan x repeat after an interval of π or 180°

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 15

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 16

We get

= tan 60o

= √3

9. NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 17

Solution:

We know that values of sin x repeat after an interval of 2π or 360°

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 18

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 19

10. NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 20

Solution:

We know that values of tan x repeat after an interval of π or 180°

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 21

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 22

Comments

Leave a Comment

Your Mobile number and Email id will not be published.

*

*