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

Chapter 3 Trigonometric Functions of Class 11 Maths is categorized under the term – II CBSE Syllabus for 2021-22. 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 present at BYJU’S. Students can view as well as download the NCERT solutions for class 11 and boost their second term exam preparation.

## Download PDF of NCERT Solutions for Class 11 Maths Chapter 3- Trigonometric Functions Exercise 3.2      ### Access other exercise solutions of Class 11 Maths Chapter 3- Trigonometric Functions

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 other five trigonometric functions in Exercises 1 to 5.

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

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

Solution:

It is given that

sin x = 3/5

We can write it as We know that

sin2 x + cos2 x = 1

We can write it as

cos2 x = 1 – sin2 x 3. cot x = 3/4, x lies in third quadrant.

Solution:

It is given that

cot x = 3/4

We can write it as 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 4. sec x = 13/5, x lies in fourth quadrant.

Solution:

It is given that

sec x = 13/5

We can write it as 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  5. tan x = -5/12, x lies in second quadrant.

Solution:

It is given that

tan x = – 5/12

We can write it as 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 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 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 By further calculation = cosec 30o = 2

8. Solution:

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

So we get By further calculation We get

= tan 60o

= 3

9. Solution:

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

So we get By further calculation 10. Solution:

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

So we get By further calculation 