NCERT Solutions Class 11th Maths Chapter – 3 Trigonometric Functions Exercise 3.2
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Mathematics |
| Chapter | Chapter – 3 |
| Chapter Name | Trigonometric Functions |
| grade | Class 11th Maths solution |
| Medium | English |
| Source | last doubt |
NCERT Solutions Class 11th Maths Chapter – 3 Trigonometric Functions Exercise 3.2
?Chapter – 3?
✍ Trigonometric Functions✍
?Exercise 3.2?
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 = – 13/1
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 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
