

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2: Exercise 3.2 of Chapter 3 in Class 11 Maths focuses on understanding the signs of trigonometric functions in different quadrants, along with their domain and range.
This exercise helps students analyze the behavior of trigonometric functions such as sine, cosine, tangent, and others based on the angle's position on the coordinate plane. The solutions provide step-by-step explanations, ensuring students grasp these fundamental concepts effectively. By practicing these problems, students can develop a strong foundation for advanced trigonometry topics.Class 11 Maths Chapter 3 Exercise 3.2 Questions and Answers PDF
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:
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
sin
2
x + cos
2
x = 1
We can write it as
cos
2
x = 1 – sin
2
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 + tan
2
x = sec
2
x
We can write it as
1 + (4/3)
2
= sec
2
x
Substituting the values
1 + 16/9 = sec
2
x
cos
2
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
sin
2
x + cos
2
x = 1
We can write it as
sin
2
x = 1 – cos
2
x
Substituting the values
sin
2
x = 1 – (5/13)
2
sin
2
x = 1 – 25/169 = 144/169
sin
2
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 + tan
2
x = sec
2
x
We can write it as
1 + (-5/12)
2
= sec
2
x
Substituting the values
1 + 25/144 = sec
2
x
sec
2
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 45
o
= 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 30
o
= 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 60
o
=
√
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
