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:
- Trigonometric Functions
- Sign of trigonometric functions
- 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
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:
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
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