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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Edited By Ramraj Saini | Updated on Sep 21, 2023 09:10 PM IST

Trigonometric Functions Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are provided here. Trigonometry class 11 has various real-time applications. It is used to solve height and distance problems. This chapter gives an introduction to basic properties and identities of trigonometric functions and questions based on the same are answered in class 11 trigonometric functions NCERT solutions. Students can practice NCERT Solutions that are prepared by Careers360 expert team according to CBSE latest syllabus 2023. The chapter 3 maths class 11 trigonometric identities are used in multiple other chapters of mathematics and throughout the NCERT physics for both class 11 and 12 also. So it is very important to memories and understand the basic identities and properties of trigonometric functions. Trigonometric functions class 11 NCERT solutions will help you for the same.

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This Story also Contains
  1. Trigonometric Functions Class 11 Questions And Answers
  2. Trigonometric Functions Class 11 Questions And Answers PDF Free Download
  3. Trigonometric Functions Class 11 Solutions - Important Formulae
  4. Trigonometric Functions Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. Trigonometric Functions Class 11 Solutions - Topics
  6. NCERT Solutions for Class 11 Mathematics - Chapter Wise
  7. Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
  8. NCERT Solutions for Class 11 - Subject Wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Command on this NCERT syllabus is required to understand the concepts of inverse trigonometric functions which will be taught in class 12. You should practice more in order to get clarity of the concepts. First, try to solve NCERT textbook exercise problems. If you getting difficulties in doing so, you can take help from solutions of NCERT for class 11 maths chapter 3 trigonometric functions. Check all NCERT solutions at a single place which will be helpful when you are not able to solve the NCERT questions. Here you will get NCERT solutions for class 11 also.

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Trigonometric Functions Class 11 Questions And Answers PDF Free Download

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Trigonometric Functions Class 11 Solutions - Important Formulae

Angle Conversion:

Radian Measure = π/180 × Degree Measure

Degree Measure = 180/π × Radian Measure

Trigonometric Ratios:

  • sin θ = P / H

  • cos θ = B / H

  • tan θ = P / B

  • cosec θ = H / P

  • sec θ = H / B

  • cot θ = B / P

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Reciprocal Trigonometric Ratios:

  • sin θ = 1 / cosec θ

  • cosec θ = 1 / sin θ

  • cos θ = 1 / sec θ

  • sec θ = 1 / cos θ

  • tan θ = 1 / cot θ

  • cot θ = 1 / tan θ

Trigonometric Ratios of Complementary Angles:

  • sin (90° – θ) = cos θ

  • cos (90° – θ) = sin θ

  • tan (90° – θ) = cot θ

  • cot (90° – θ) = tan θ

  • sec (90° – θ) = cosec θ

  • cosec (90° – θ) = sec θ

Periodic Trigonometric Ratios:

  • sin(π/2-θ) = cos θ

  • cos(π/2-θ) = sin θ

  • sin(π-θ) = sin θ

  • cos(π-θ) = -cos θ

  • sin(π+θ) = -sin θ

  • cos(π+θ) = -cos θ

  • sin(2π-θ) = -sin θ

  • cos(2π-θ) = cos θ

Trigonometric Identities:

  • sin² θ + cos² θ = 1

  • cosec² θ – cot² θ = 1

  • sec² θ – tan² θ = 1

Product to Sum Formulas:

  • sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

  • cos x cos y = 1/2[cos(x–y) + cos(x+y)]

  • sin x cos y = 1/2[sin(x+y) + sin(x−y)]

  • cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas:

  • sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

  • sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

  • cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

  • cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

General Trigonometric Formulas:

  • sin (x+y) = sin x × cos y + cos x × sin y

  • cos(x+y) = cosx × cosy − sinx × siny

  • cos(x–y) = cosx × cosy + sinx × siny

  • sin(x–y) = sinx × cosy − cosx × siny

Sum and Difference Formulas for tan:

  • tan (x+y) = (tan x + tan y) / (1 − tan x tan y)

  • tan (x−y) = (tan x − tan y) / (1 + tan x tan y)

Double Angle Formulas for tan:

  • tan 2θ = (2 tan θ) / (1 - tan²θ)

Triple Angle Formulas for sin, cos, and tan:

  • sin 3θ = 3sin θ – 4sin³θ

  • cos 3θ = 4cos³θ – 3cos θ

  • tan 3θ = [3tan θ – tan³θ] / [1 − 3tan²θ]

Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions for CBSE Exam.

Trigonometric Functions Class 11 NCERT Solutions (Intext Questions and Exercise)

NCERT trigonometric functions class 11 questions and answers - Exercise: 3.1

Question:1 Find the radian measures corresponding to the following degree measures:

(i) 25
(ii) 47 30
(iii) 240
(iv) 520

Answer:

It is solved using relation between degree and radian

(i) 25
We know that 180 = π radian

So, 1=π180 radian


25=π180×25 radian =5π36 radian
(ii) 4730
We know that
4730=4712degree=952

Now, we know that 180=π1=π180 radian

So, 952=π180×(952) radian 19π72 radian
(iii) 240
We know that

180=π1=π180 radian

So, 240=π180×2404π3 radian
(iv) 520
We know that

180=π1=π180 radian

So, 520π180×520 radian 26π9 radian

Question:2 Find the degree measures corresponding to the following radian measures. (Use π=227 )

(i)1116
(ii)4
(iii)5π3
(iv)7π6

Answer:

(1) 1116

We know that
π radian =1801radian=180πdegree

So, 1116radian=180π×1116degree (we need to take π=227 )

1116radian=180×722×1116degree3158degree

(we use 1=60 and 1' = 60'')

Here 1' represents 1 minute and 60" represents 60 seconds
Now,

3158degree=3938degree=39+3×608minutes39+22+12minutes39+22+303158degree=392230

(ii) -4
We know that

π radian =1801radian=180πdegree (we need to take π=227 )


So, -4 radian = 4×180π4×180×722252011degree


(we use 1=60 and 1' = 60'')

252011degree=229111degree=229+1×6011minutes229+5+511minutes=229+5+27252011=229527

(iii) 5π3

We know that
π radian =1801radian=180πdegree (we need to take π=227 )


So, 5π3radian=180π×5π3degree=300
(iv) 7π6

We know that
π radian =1801radian=180πdegree (we need to take π=227 )


So, 7π6radian=180π×7π6=210

Question:3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Answer:
Number of revolutions made by the wheel in 1 minute = 360
Number of revolutions made by the wheel in 1 second = 36060=6
( 1 minute = 60 seconds)
In one revolutions wheel will cover 2π radian
So, in 6 revolutions it will cover = 6×2π=12π radian

In 1 the second wheel will turn 12π radian

Question:4 Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (use π=227 )

Answer:

We know that
l=rΘ ( where l is the length of the arc, r is the radius of the circle and Θ is the angle subtended)

here r = 100 cm
and l = 22 cm
Now,
Θ=lr=22100radian

We know that
πradian=180So,1radian=180πdegree22100radian=180π×22100degree180×722×22100=635degreeSo,635degree=1235degree=12+3×605minute=12+36635degree=1236
So,
Angle subtended at the centre of a circle Θ=1236

Question:5 In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Answer:

Given :- radius (r)of circle = Diameter2=40cm2=20cm
length of chord = 20 cm

We know that
θ=lr (r = 20cm , l = ? , θ = ?)

Now,
1654678632225 AB is the chord of length 20cm and OA and OB are radii of circle 20 cm each
The angle subtended by OA and OB at centre = θ
OA = OB = AB
Δ OAB is equilateral triangle
So, each angle equilateral is 60
θ=60 =π3radian
Now, we have θ and r
So,
l=rθ=20×π3=20π3
the length of the minor arc of the chord (l) = 20π3 cm

Question:6 If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Answer:

Given:-
θ1=60θ2=75 and l1=l2

We need to find the ratio of their radii r1r2=?

We know that arc length l=rθ
So,
l1=r1θ1l2=r2θ2
Now,
l1l2=r1θ1r2θ2 ( l1=l2 )
So,
r1r2=θ2θ1=7560=54 is the ratio of their radii

Question:7 Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm

Answer:

(i) We know that

l=rθ
Now,
r = 75cm
l = 10cm

So,
θ=lr=1075=215radian

(ii) We know that

l=rθ
Now,
r = 75cm
l = 15cm

So,
θ=lr=1575=15radian

(iii) We know that

l=rθ
Now,
r = 75cm
l = 21cm

So,
θ=lr=2175=725radian


NCERT class 11 maths chapter 3 question answer - Exercise: 3.2

Question:1 Find the values of other five trigonometric functions cosx=12 , x lies in third quadrant.

Answer:

Solution
cosx=12
secx=1cosx=112=2
x lies in III quadrants. Therefore sec x is negative

sin2x+cos2x=1sin2x=1cos2xsin2x=1(12)2sin2x=114=34sinx=34=±32
x lies in III quadrants. Therefore sin x is negative
sinx=32

cosec x=1sinx=132=23

x lies in III quadrants. Therefore cosec x is negative

tanx=sinxcosx=3212=3
x lies in III quadrants. Therefore tan x is positive

cotx=1tanx=13
x lies in III quadrants. Therefore cot x is positive

Question:2 Find the values of other five trigonometric functions sinx=35 x lies in second quadrant.

Answer:

Solution

sinx=35

cosec x=1sinx=135=53
x lies in the second quadrant. Therefore cosec x is positive

sin2x+cos2x=1cos2x=1sin2xcos2x=1(35)2cos2x=1925=1625cosx=1625=±45
x lies in the second quadrant. Therefore cos x is negative
cosx=45

secx=1cosx=145=54
x lies in the second quadrant. Therefore sec x is negative

tanx=sinxcosx=3545=34
x lies in the second quadrant. Therefore tan x is negative

cotx=1tanx=134=43
x lies in the second quadrant. Therefore cot x is negative

Question:3 Find the values of other five trigonometric functions cotx=34 , x lies in third quadrant.

Answer:

Solution

cotx=34

tanx=1cotx=134=43
1+tan2x=sec2x1+4232=sec2x1+169=sec2x259=sec2xsecx=259=±53
x lies in x lies in third quadrant. therefore sec x is negative
secx=53

cosx=1secx=153=35
sin2x+cos2x=1sin2x=1cos2xsin2x=1(35)2sin2x=1925sin2x=1625sinx=1625=±45
x lies in x lies in third quadrant. Therefore sin x is negative
sinx=45
cosecx=1csc=145=54

Question:4 Find the values of other five trigonometric functions secx=135 , x lies in fourth quadrant.

Answer:

Solution
secx=135
cosx=1secx=1135=513
sin2x+cos2x=1sin2x=1cos2xsin2x=1513sin2x=125169=144169sinx=144169=±1213
lies in fourth quadrant. Therefore sin x is negative
sinx=1213
cscx=1sinx=11213=1312
tanx=sinxcosx=1213513=125
cotx=1tanx=1125=512

Question:5 Find the values of the other five trigonometric functions tanx=512 , x lies in second quadrant.

Answer:
tanx=512
cotx=1tanx=1512=125
1+tan2x=sec2x1+(512)2=sec2x1+25144=sec2x169144=sec2xsecx=169144=±1312
x lies in second quadrant. Therefore the value of sec x is negative
secx=1312
cosx=1secx=11312=1213
sin2x+cos2x=1sin2x=1cos2xsin2x=1(1213)2sin2x=1144169sin2x=25169sinx=25169=±513
x lies in the second quadrant. Therefore the value of sin x is positive
sinx=513
csc=1sinx=1513=135

Question:6 Find the values of the trigonometric functions sin765

Answer:
We know that values of sin x repeat after an interval of 2π or 360 degree

sin765=sin(2×360+45)=sin45sin45=12

Question:7 Find the values of the trigonometric functions cosec (1410)

Answer:

We know that value of cosec x repeats after an interval of 2π or 360
cosec(1410)=cosec(1410+360×4)cosec 30=2

or

cosec(1410)=cosec(1410)=cosec(4×36030)=cosec(30)=2

Question:8 Find the values of the trigonometric functions tan19π3

Answer:

We know that tan x repeats after an interval of π or 180 degree
tan(19π3)=tan(6π+π3)=tanπ3=tan60=3

Question:9 Find the values of the trigonometric functions sin(11π3)

Answer:

We know that sin x repeats after an interval of 2πor360
sin(11π3)=sin(4π+π3)=sinπ3=32

Question:10 Find the values of the trigonometric functions cot(15π4)

Answer:

We know that cot x repeats after an interval of πor180
cot(15π4)=cot(4π+π4)=cot(π4)=1

NCERT class 11 maths chapter 3 question answer - Exercise: 3.3

Question:1 Prove that sin2(π6)+cos2(π3)tan2(π4)=12

Answer:

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:


sin(π6)=(12)cos(π3)=(12)tan(π4)=1
sin2π6+cos2π3tan2π4= (12)2+(12)212

=14+141=12
= R.H.S.

Question:2 Prove that 2sin2(π6)+cosec2(7π6)cos2π3=32

Answer:

The solutions for the given problem is done as follows.

sinπ6=12cosec7π6=cosec(π+π6)=cosecπ6=2cosπ3=12
2sin2π6+cosec27π6cos2π3=2(12)2+(2)2(12)22×14+4×14=12+1=32
R.H.S.

Question:3 Prove that cot2(π6)+csc(5π6)+3tan2(π6)=6

Answer:

We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)

cotπ6=3cosec5π6=cosec(ππ6)=cosecπ6=2tanπ6=13

cot2π6+cosec5π6+3tan2π6=((3))2+2+3×(13)23+2+1=6
R.H.S.

Question:4 Prove that 2sin2(3π4)+2cos2(π4)+2sec2(π3)=10

Answer:

sin3π4=sin(ππ4)=sinπ4=12cosπ4=12secπ3=2
Using the above values

2sin23π4+2cos2π4+2sec2π3=2×(12)2+2×(12)2+2(2)21+1+8=10
R.H.S.

Question:5(i) Find the value of (i)sin75

Answer:

sin75=sin(45+30)
We know that
(sin(x+y)=sinxcosy + cosxsiny)
Using this idendity

sin75=sin(45+30)=sin45cos30+cos45sin3012×32+12×12322+122=3+122

Question:5(ii) Find the value of
(ii)tan15

Answer:

tan15=tan(4530)
We know that,

[tan(xy)=tanxtany1+tanxtany]
By using this we can write

tan(4530)=tan45tan301+tan45tan301131+1(13)=3133+13=313+1=(31)2(3+1)(31)=3+123(3)2(1)242331=2(23)2=23

Question:6 Prove the following: cos(π4x)cos(π4y)sin(π4x)sin(π4y)=sin(x+y)

Answer:

cos(π4x)cos(π4y)sin(π4x)sin(π4y)

Multiply and divide by 2 both cos and sin functions
We get,

12[2cos(π4x)cos(π4y)]+12[2sin(π4x)sin(π4y)]

Now, we know that

2cosAcosB = cos(A+B) + cos(A-B) -(i)
-2sinAsinB = cos(A+B) - cos(A-B) -(ii)
We use these two identities

In our question A = (π4x)

B = (π4y)
So,

12[cos{(π4x)+(π4y)}+cos{(π4x)(π4y)}]+12[cos{(π4x)+(π4y)}cos{(π4x)+(π4y)}]

2×12[cos{(π4x)+(π4y)}]

=cos[π2(x+y)]

As we know that

(cos(π2A)=sinA)
By using this

=cos[π2(x+y)] =sin(x+y)

R.H.S

Question:7 Prove the following tan(π4+x)tan(π4x)=(1+tanx1tanx)2

Answer:

As we know that

(tan(A+B)=tanA+tanB1tanAtanB) and tan(AB)=tanAtanB1+tanAtanB

So, by using these identities

tan(π4+x)tan(π4x)=tanπ4+tanx1tanπ4tanxtanπ4tanx1+tanπ4tanx=1+tanx1tanx1tanx1+tanx=(1+tanx1tanx)2
R.H.S

Question:8 Prove the following cos(π+x)cos(x)sin(πx)cos(π2+x)=cot2x

Answer:

As we know that,
cos(π+x)=cosx , sin(πx)=sinx , cos(π2+x)=sinx
and
cos(x)=cosx

By using these our equation simplify to

cosx×cosxsinx×sinx=cos2xsin2x=cot2x (cotx=cosxsinx)
R.H.S.

Question:9 Prove the following cos(3π2+x)cos(2π+x)[cot(3π2x)+cot(2π+x)]=1

Answer:

We know that

cos(3π2+x)=sinxcos(2π+x)=cosxcot(3π2x)=tanxcot(2π+x)=cotx

So, by using these our equation simplifies to

cos(3π2+x)cos(2π+x)[cot(3π2x)+cot(2π+x)]=sinxcosx[tanx+cotx]=sinxcosx[sinxcosx+cosxsinx]sinxcosx[sin2x+cos2xsinxcosx]=sin2x+cos2x=1 R.H.S.

Question:10 Prove the following sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx

Answer:

Multiply and divide by 2

=2sin(n+1)xsin(n+2)x+2cos(n+1)xcos(n+2)x2

Now by using identities


-2sinAsinB = cos(A+B) - cos(A-B)
2cosAcosB = cos(A+B) + cos(A-B)

{(cos(2n+3)xcos(x))+(cos(2n+3)+cos(x))}2(cos(x)=cosx)=2cosx2=cosx

R.H.S.

Question:11 Prove the following cos(3π4+x)cos(3π4x)=2sinx

Answer:

We know that

[ cos(A+B) - cos (A-B) = -2sinAsinB ]

By using this identity

cos(3π4+x)cos(3π4x)=2sin3π4sinx=2×12sinx=2sinx R.H.S.

Question:12 Prove the following sin26xsin24x=sin2xsin10x

Answer:

We know that
a2b2=(a+b)(ab)

So,
sin26xsin24x=(sin6x+sin4x)(sin6xsin4x)

Now, we know that


sinA+sinB=2sin(A+B2)cos(AB2)sinAsinB=2cos(A+B2)sin(AB2)
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx

sin26xsin24x=(2cos5xsin5x)(2sinxcosx)

Now,

2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)

by using these identities

2cos5x sin5x = sin10x - 0
2sinx cosx = sin2x + 0

hence
sin26xsin24x=sin2xsin10x

Question:13 Prove the following cos22xcos26x=sin4xsin8x

Answer:

As we know that

a2b2=(ab)(a+b)

cos22xcos26x=(cos2xcos6x)(cos2x+cos6x)
Now
cosAcosB=2sin(A+B2)sin(AB2)cosA+cosB=2cos(A+B2)cos(AB2)
By using these identities

cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( sin(-x) = -sin x
cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes


R.H.S.

Question:14 Prove the following sin2x+2sin4x+sin6x=4cos2xsin4x

Answer:

We know that

sinA+sinB=2sin(A+B2)cos(AB2)
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x

sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( cos(-x) = cos x)

So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( cos2x=2cos2x1 )
=2sin4x( 2cos2x1 +1 )
=2sin4x( 2cos2x )
= 4sin4xcos2x
R.H.S.

Question:15 Prove the following cot4x(sin5x+sin3x)=cotx(sin5xsin3x)

Answer:

We know that
sinx+siny=2sin(x+y2)cos(xy2)
By using this , we get

sin5x + sin3x = 2sin4xcosx

cos4xsin4x(2sin4xcosx)=2cos4xcosx

now nultiply and divide by sin x

 2cos4xcosxsinxsinx          =cotx(2cos4xsinx)                (cosx sinx=cotx)

Now we know that

2cosxsiny=sin(x+y)sin(xy)

By using this our equation becomes

=cotx(sin5xsin3x)
R.H.S.

Question:16 Prove the following cos9xcos5xsin17xsin3x=sin2xcos10x

Answer:

As we know that

cosxcosy=2sinx+y2sinxy2cos9xcos5x=2sin7xsin2xsinxsiny=2cosx+y2sinxy2sin17xsin3x=2cos10xsin7xcos9xcos5xsin17xsin3x=2sin7xsin2x2cos10xsin7x=sin2xcos10x
R.H.S.

Question:17 Prove the following sin5x+sin3xcos5x+cos3x=tan4x

Answer:

We know that

sinA+sinB=2sinA+B2cosAB2andcosA+cosB=2cosA+B2cosAB2

We use these identities

sin5x+sin3x=2sin4xcosxcos5x+cos3x=2cos4xcosxsin5x+sin3xcos5x+cos3x=2sin4xcosx2cos4xcosx=sin4xcos4x=tan4x
R.H.S.

Question:18 Prove the following sinxsinycosx+cosy=tan(xy)2

Answer:

We know that
sinxsiny=2cosx+y2sinxy2andcosx+cosy=2cosx+y2cosxy2

We use these identities

We use these identitiessinxsinycosx+cosy=2cosx+y2sinxy22cosx+y2cosxy2=sinxy2cosxy2=tanxy2

R.H.S.

Question:19 Prove the following sinx+sin3xcosx+cos3x=tan2x

Answer:

We know that

sinx+siny=2sinx+y2cosxy2andcosx+cosy=2cosx+y2cosxy2We use these equationssinx+sin3x=2sin2xcos(x)=2sin2xcosx     (cos(x)=cosx)cosx+cos3x=2cos2xcos(x)=2cos2xcosx     (cos(x)=cosx)sinx+sin3xcosx+cos3x=2sin2xcosx2cos2xcosx=sin2xcos2x=tan2x R.H.S.

Question:20 Prove the following sinxsin3xsin2xcos2x=2sinx

Answer:
We know that

sin3x=3sinx4sin3x   ,  cos2sin2x=cos2xandcos2x=12sin2x

We use these identities

sinxsin3x=sinx(3sinx4sin3x)=4sin3x2sinx.                                                         =2sinx(2sin2x1)cos2xsin2=cos2xcos2x=12sin2x

sinxsin3x=sinx(3sinx4sin3x)=4sin3x2sinx.                                                         =2sinx(2sin2x1)sin2cos2x=cos2x          (cos2x=12sin2x)sin2cos2x=(12sin2x)=2sin(2)x1sinxsin3xsin2cos2x=2sinx(2sin2x1)2sin(2)x1=2sinx

sinxsin3x=sinx(3sinx4sin3x)=4sin3x2sinx.                                                         =2sinx(2sin2x1)sin2cos2x=cos2x                         (cos2x=12sin2x)sin2cos2x=(12sin2x)=2sin(2)x1sinxsin3xsin2cos2x=2sinx(2sin2x1)2sin2x1=2sinx
R.H.S.

Question:21 Prove the following cos4x+cos3x+cos2xsin4x+sin3x+sin2x=cot3x

Answer:

We know that

cosx+cosy=2cosx+y2cosxy2andsinx+siny=2sinx+y2cosxy2
We use these identities

(cos4x+cos2x)+cos3x(sin4x+sin2x)+sin3x=2cos3xcosx+cos3x2sin3xcosx+sin3x=2cos3x(1+cosx)2sin3x(1+cosx)         =cot3x

=RHS

Question:22 prove the following cotxcot2xcot2xcot3xcot3xcotx=1

Answer:

cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)

and we know that

cot(a+b)=cotacotb1cota+cotb
So,
cotx cot2xcot2xcotx1cot2x+cotx(cot2x+cotx)
= cotx cot2x - (cot2xcotx -1)
= cotx cot2x - cot2xcotx +1
= 1 = R.H.S.

Question:23 Prove that tan4x=4tanx(1tan2x)16tan2x+tan4x

Answer:

We know that

tan2A=2tanA1tan2A

and we can write tan 4x = tan 2(2x)
So, tan4x=2tan2x1tan22x = 2(2tanx1tan2x)1(2tanx1tan2x)2


= 2(2tanx)(1tan2x)(1tanx)2(4tan2x)

= (4tanx)(1tan2x)(1)2+(tan2x)22tan2x(4tan2x)

= (4tanx)(1tan2x)12+tan4x6tan2x = R.H.S.

Question:24 Prove the following cos4x=18sin2xcos2x

Answer:

We know that
cos2x=12sin2x
We use this in our problem
cos 4x = cos 2(2x)
= 12sin22x
= 12(2sinxcosx)2 (sin2x=2sinxcosx)
= 18sin2xcos2x = R.H.S.

Question:25 Prove the following cos6x=32cos6x48cos4x+18cos2x1

Answer:

We know that
cos 3x = 4 cos3x - 3cos x
we use this in our problem
we can write cos 6x as cos 3(2x)
cos 3(2x) = 4 cos32x - 3 cos 2x
= 4(2cos2x1)3 - 3(2cos2x1) (cos2x=2cos2x1)
= 4[(2cos2x)3(1)33(2cos2x)2(1)+3(2cos2x)(1)2] 6cos2x+3 ((ab)3=a3b33a2b+3ab2)
= 32 cos6x - 4 - 48 cos4x + 24 cos2x - 6cos2x+3
= 32 cos6x - 48 cos4x + 18 cos2x - 1 = R.H.S.

NCERT class 11 maths chapter 3 question answer - Exercise: 3.4

Question:1 Find the principal and general solutions of the following equations: tanx=3

Answer:

It is given that given
tanx=3
Now, we know that tanπ3=3 and tan4π3=tan(π+π3)=3

Therefore,
the principal solutions of the equation are x=π3,4π3
Now,
The general solution is tanx=tanπ3

x=nπ+π3 where n ϵ Z and Z denotes sets of integer

Therefore, the general solution of the equation is x=nπ+π3 where n ϵ Z and Z denotes sets of integer

Question:2 Find the principal and general solutions of the following equations: secx=2

Answer:

We know that value of secπ3=2 and sec5π3=sec(2ππ3)=secπ3=2

Therefore the principal solutions are x = π3and5π3
secx=secπ3
We know that value of sec x repeats after an interval of 2π
So, by this we can say that

the general solution is x = 2nπ±π3 where n ϵ Z

Question:3 Find the principal and general solutions of the following equations: cotx=3

Answer:

we know that  cotπ6=3 and we know that  cot5π6=cot(ππ6)=cotπ6=3

Similarly , the value for  cot11π6=cot(2ππ6)=cotπ6=3
Therefore, principal solution is x = 5π6 and 11π6


We also know that the value of cot x repeats after an interval of π
There the general solution is x = nπ±5π6 where n ϵ Z

Question:4 Find the principal and general solutions of the following equations: cosecx=2

Answer:

We know that
cosecπ6=2

cosec(π+π6)=cosecπ6=2 and also cosec(2ππ6)=cosec11π6=2
So,
cosecx=cosec7π6 and cosecx=cosec11π6

So, the principal solutions are x=7π6 and 11π6


Now,
cosecx=cosec7π6

sinx=sin7π6 (sinx=1cosecx)

x=nπ+(1)n7π6
Therefore, the general solution is

x=nπ+(1)n7π6

where n ϵ Z

Question:5 Find the general solution for each of the following equation cos4x=cos2x

Answer:

cos4x = cos2x
cos4x - cos2x = 0
We know that
cosAcosB=2sinA+B2sinAB2
We use this identity
cos 4x - cos 2x = -2sin3xsinx
-2sin3xsinx = 0 sin3xsinx=0
So, by this we can that either
sin3x = 0 or sinx = 0
3x = nπ x = nπ
x = nπ3 x = nπ

Therefore, the general solution is

x=nπ3 or nπ where nZ

Question:6 Find the general solution of the following equation cos3x+cosxcos2x=0

Answer:

We know that
cosA+cosB=2cosA+B2cosAB2andcosAcosB=2sinA+B2sinAB2
We use these identities
(cos3x + cosx) - cos2x = 2cos2xcosx -cos2x = 0
= cos2x(2cosx-1) = 0
So, either
cos2x = 0 or cosx=12
2x=(2n+1)π2 cosx=cosπ3
x=(2n+1)π4 x=2nπ±π3

the general solution is

x=(2n+1)π4  or 2nπ±π3

Question:7 Find the general solution of the following equation sin2x+cosx=0

Answer:

sin2x + cosx = 0
We know that
sin2x = 2sinxcosx
So,
2sinxcosx + cosx = 0
cosx(2sinx + 1) = 0
So, we can say that either

cosx = 0 or 2sinx + 1 = 0
x=(2n+1)π2 sinx=sin7π6
x=nπ+(1)n7π6
Therefore, the general solution is

x=(2n+1)π2 or nπ+(1)n7π6 where nZ

Question:8 Find the general solution of the following equation sec22x=1tan2x

Answer:

We know that
sec2x=1+tan2x
So,
1+tan22x=1tan2x
tan22x+tan2x=0tan2x(tan2x+1)=0
either
tan2x = 0 or tan2x = -1 ( tanx=tan(ππ4)=tan3π4 )
2x = nπ 2x=nπ+3π4
x=nπ2 x=nπ2+3π8
Where n ϵ Z

Question:9 Find the general solution of the following equation sinx+sin3x+sin5x=0

Answer:

We know that
sinA+sinB=2sinA+B2cosAB2
We use this identity in our problem
sin5x+sinx=2sin5x+x2cos5xx2=2sin3xcos2x
Now our problem simplifeis to
2sin3xcos2x+sin3x = 0
take sin3x common
sin3x(2cos2x+1)=0
So, either
sin3x = 0 or cos2x=12 (cos2x=cosπ3=cos(ππ3)=cos2π3)
3x=nπ 2x=2nπ±2π3
x=nπ3 x=nπ±π3
Where n ϵ Z

Class 11 maths chapter 3 NCERT solutions - Miscellaneous Exercise

Question:1 Prove that 2cosπ13cos9π13+cos3π13+cos5π13=0

Answer:

We know that

cos A+ cos B = 2cos(A+B2)cos(AB2)

we use this in our problem

2cosπ13cos9π13+2cos(3π13+5π13)2cos(3π135π13)2

2cosπ13cos9π13+2cos4π13cosπ13 ( we know that cos(-x) = cos x )

2cosπ13cos9π13+2cos4π13cosπ13
2cosπ13(cos9π13+cos4π13)
again use the above identity

2cosπ13(2cos(9π13+4π132)cos(9π134π132)
2cosπ132cosπ2cos5π26
we know that

cosπ2 = 0
So,
2cosπ132cosπ2cos5π26 = 0 = R.H.S.

Question:2 Prove that (sin3x+sinx)sinx+(cos3xcosx)cosx=0

Answer:

We know that
sin3x=3sinx4sin3x
and
cos3x=4cos3x3cosx
We use this in our problem
(sin3x+sinx)sinx+(cos3xcosx)cosx
= (3sinx4sin3x+sinx)sinx + (4cos3x3cosxcosx)cosx
= (4sinx - 4 sin3x )sinx + (4 cos3x - 4cos x)cosx
now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4 sin2x (1 - sin2x ) - 4 cos2x (1 - cos2x )
= 4 sin2x cos2x - 4 cos2x sin2x    cos2x=1sin2xandsin2x=1cos2x
= 0 = R.H.S.

Question:3 Prove that (cosx+cosy)2+(sinxsiny)2=4cos2(x+y2)

Answer:

We know that (a+b)2=a2+2ab+b2
and
(ab)2=a22ab+b2
We use these two in our problem

(sinxsiny)2=sin2x2sinxsiny+sin2y
and
(cosx+cosy)2=cos2x+2cosxcosy+cos2y

(cosx+cosy)2+(sinxsiny)2 = cos2x+2cosxcosy+cos2y + sin2x2sinxsiny+sin2y
= 1 + 2cosxcosy + 1 - 2sinxsiny (sin2x+cos2x=1 and sin2y+cos2y=1)
= 2 + 2(cosxcosy - sinxsiny)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
cos(x+y)=2cos2(x+y)21 (cos2x=2cos2x1 cosx=2cos2x21)

= 2(1+2cos2(x+y)21)
=4cos2(x+y)2

= R.H.S.

Question:4 Prove that (cosxcosy)2+(sinxsiny)2=4sin2(xy2)

Answer:

We know that (a+b)2=a2+2ab+b2
and
(ab)2=a22ab+b2
We use these two in our problem

(sinxsiny)2=sin2x2sinxsiny+sin2y
and
(cosxcosy)2=cos2x2cosxcosy+cos2y

(cosxcosy)2+(sinxsiny)2 = cos2x2cosxcosy+cos2y + sin2x2sinxsiny+sin2y
= 1 - 2cosxcosy + 1 - 2sinxsiny (sin2x+cos2x=1 and sin2y+cos2y=1)
= 2 - 2(cosxcosy + sinxsiny)
= 2 - 2cos(x - y) (cos(xy)=cosxcosy+sinxsiny)
= 2(1 - cos(x - y) )
Now we can write
cos(x+y)=12sin2(x+y)2 (cos2x=12sin2x cosx=12sin2x2)

so

2(1cos(xy))=2(1(12sin2(x+y)2))


=4sin2(xy)2 = R.H.S.

Question:5 Prove that sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x

Answer:
we know that
sinA+sinB=2sinA+B2cosAB2
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and sin5x tp get sin4x

(sin7x+sinx)+(sin5x+sin3x)=2sin7x+x2cos7xx2 +2sin5x+3x2cos5x3x2
= 2sin4xcos3x+2sin4xcosx
take 2sin4x common
= 2sin4x(cos3x + cosx)
Now,
We know that
cosA+cosB=2cosA+B2cosAB2
We use this
cos3x+cosx=2cos3x+x2cos3xx2
= 2cos2xcosx
= 2sin4x( 2cos2xcosx )
= 4cosxcos2xsin4x = R.H.S.

Question:6 Prove that (sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x)=tan6x

Answer:

We know that

sinA+sinB=2sinA+B2cosAB2
and
cosA+cosB=2cosA+B2cosAB2

We use these two identities in our problem

sin7x + sin5x = 2sin7x+5x2cos7x5x2 = 2sin6xcosx

sin 9x + sin 3x = 2sin9x+3x2cos9x3x2 = 2sin6xcos3x

cos 7x + cos5x = 2cos7x+5x2cos7x5x2 = 2cos6xcosx

cos 9x + cos3x = 2cos9x+3x2cos9x3x2 = 2cos6xcos3x


(sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x) = (2sin6xcosx)+(2sin6xcos3x)(2cos6xcosx)+(2cos6xcos3x)

= 2sin6x(cosx+cos3x)2cos6x(cosx+cos3x)=tan6x = R.H.S. (sinxcosx=tanx)

Question:7 Prove that sin3x+sin2xsinx=4sinxcosx2cos3x2

Answer:

We know that
cosA+cosB=2cosA+B2cosAB2
sinAsinB=2cosA+B2sinAB2

we use these identities
sin3xsinx=2cos3x+x2sin3xx2

=2cos2xsinx


sin2x + 2cos2xsinx = 2sinx cosx + 2cos2xsinx
take 2 sinx common
2sinx(cosx+cos2x)=2sinx(2cos2x+x2cos2xx2)

=2sinx(2cos3x2cosx2)
=4sinxcos3x2cosx2

= R.H.S.

Question:8 Find sinx2,cosx2,andtanx2 in tanx=43 , x in quadrant II

Answer:

tan x = 43
We know that ,
sec2x=1+tan2x
=1+(43)2
=1+169 = 259
secx=259 = ±53
x lies in II quadrant thats why sec x is -ve
So,

secx=53
Now, cosx=1secx = 35
We know that,
cosx=2cos2x21 ( cos2x=2cos2x1cosx=2cos2x21 )
35+1=2 cos2x2

= 3+55 = 2cos2x2

25 = 2cos2x2
cos2x2 = 15
cosx2 = 15 = ±15
x lies in II quadrant so value of cosx2 is +ve

cosx2 = 15=55
we know that
cosx=12sin2x2

2sin2x2 = 1 - (35) = 85

sin2x2=45=sinx2=45=±25
x lies in II quadrant So value of sin x is +ve

sinx2=25=255

tanx2=sinx2cosx2=255(55)=2

Question:9 Find sinx2,cosx2,andtanx2 in cosx=13 , x in quadrant III

Answer:

π<x<3π2π2<x2<3π4

We know that
cos x = 2cos2x21
2cos2x2= cos x + 1
= (13) + 1 = (1+33) = 23

cosx2=13=±13

cosx2=13=33
Now,
we know that
cos x = 12sin2x2
2sin2x2=1cosx
= 1 - (13) = 3+13 = 43

2sin2x2=43sin2x2=23sinx2=23=±23=63
Because sinx2 is +ve in given quadrant

tanx2=sinx2cosx2=6333=2

Question:10 Find sinx2,cosx2,andtanx2 in sinx=14 ,x in quadrant II

Answer:

π2<x<ππ4<x2<π2 all functions are positive in this range
We know that
cos2x=1sin2x
= 1 - (14)2 = 1116 = 1516

cos x = 1516=±154=154 (cos x is -ve in II quadrant)

We know that
cosx = 2cos2x21
2cos2x2=cosx+1=154+1=15+44

cos2x2=15+48
cosx2=±15+48=15+422=82154 (because all functions are posititve in given range)

similarly,
cos x = 12sin2x2
2sin2x2=1cosx2sin2x2=1(154)=4+154
sinx2=±15+48=15+422=8+2154 (because all functions are posititve in given range)
tanx2=sinx2cosx2=8+215482154=8+2156415×4=8+2154=4+15

Trigonometric Functions Class 11 Solutions - Topics

Interested students can study Class 11 NCERT Solutions at a single place. Here are some highlights of Maths NCERT Solutions for Class 11 Chapter 3 encompasses the following topics and sub-topics:

  • Introduction (3.1): This section of class 11 chapter 3 covers the basic trigonometric ratios and identities, along with their applications in solving word problems related to heights and distances.
  • Angles (3.2): This section of class 11 chapter 3 maths discusses different terminologies used in trigonometry, such as terminal side, initial sides, measuring an angle in degrees and radians, etc. It also covers degree and radian measures and their relation to real numbers.
  • Trigonometric Functions (3.3): Students can understand the generalised trigonometric functions with signs and learn about the domain and range of trigonometric functions with examples in this section.
  • Trigonometric Functions of Sum and Difference of Two Angles (3.4): This section provides formulas related to the sum and difference of two angles in trigonometric functions.

Interested students can study Trigonometric Functions Exercise using following links-

NCERT Solutions for Class 11 Mathematics - Chapter Wise

Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Comprehensive Coverage: The solutions provide a comprehensive explanation of all the topics covered in Chapter 3, ensuring a strong understanding of trigonometric functions.

Step-by-Step Solutions: Each problem and example in the NCERT textbook is solved step by step, making it easier for students to follow the logical progression of trigonometric concepts.

Clear and Concise Language: The solutions are presented in clear and concise language, making complex trigonometric concepts more accessible to students.

NCERT Solutions for Class 11 - Subject Wise

Trigonometric Functions Basic Identities

1)cos2x+sin2x=12) 1+tan2x    =sec2x3)1+cot2x      =cosec2x4)cos(2nπ+x) =cosx5)sin(2nπ+x) =sinx6)sin(x)       =sinx7)cos(x)       =cosx

The above identities you may have studied in your high school classes also. Here are a few more identities that you have to remember and understand from the NCERT class 11 maths chapter 3 trigonometric functions

8)cos(x+y)=cosxcosysinxsiny9)cos(xy)=cosxcosy+sinxsiny10)sin(x+y)=sinxcosy+cosxsiny11)sin(xy)=sinxcosycosxsiny

Some conditional identities from the NCERT solutions for class 11 maths chapter 3 Trigonometric Functions

If angles x, y and (x ± y) is not an odd multiple of π 2, then

a)tan(x+y)=tanx+tany1tanxtanyb)tan(xy)=tanxtany1+tanxtany

If angles x, y and (x ± y) is not a multiple of π, then

a)cot(x+y)=cotxcoty1cotx+cotyb)cot(xy)=1+cotxcotycotycotx

There are a few more identities used in the NCERT solutions for class 11 maths chapter 3 trigonometric functions which can be derived using the above identities. Try to derive it by your self.

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. What are important topics of the chapter Trigonometric Functions ?

In the chapter 3 class 11 maths Trigonometric functions, identities of trigonometric functions,  trigonometric functions of some standard angles, trigonometric functions of sum and difference of two angles and trigonometric equations are the important topics of this chapter. 

2. How does the NCERT solutions are helpful ?

NCERT solutions will help the students, if the stuck while solving the NCERT problems. Trigonometry class 11 solutions provides practice problems for students that help them to command and get indepth understanding. finally class 11 maths trigonometry help to build strong foundation for higher classes. For ease Interested students can study trigonometric functions class 11 pdf both online and offline. therefore students can practice class 11 maths chapter 3 solutions. 

3. Which are the most difficult chapters in the NCERT class 11 maths ?
  1. Complex Numbers and Quadratic Equations : This chapter introduces students to complex numbers, which can be a new and abstract concept for many. Understanding operations with complex numbers and solving quadratic equations involving complex roots can be challenging.

  2. Limits and Derivatives : Calculus, in general, tends to be difficult for many students, especially when they encounter the concept of limits for the first time. Understanding the concept of limits and their applications, as well as learning differentiation rules and techniques, can be challenging.

  3. Permutations and Combinations : This chapter involves various counting principles and problem-solving techniques, which can be confusing for some students. Understanding when to use permutations versus combinations and solving problems involving both can be tricky.

  4. Binomial Theorem : The binomial theorem involves expanding expressions raised to positive integer powers, which can require a good understanding of algebraic manipulation and pattern recognition.

  5. Statistics : This chapter involves understanding various measures of central tendency and dispersion, as well as probability distributions. It can be challenging for some students to grasp the concepts and apply them to solve problems.

  6. Trigonometric Functions : Trigonometry, particularly when it involves proving identities and solving trigonometric equations, can be challenging due to the abstract nature of the concepts and the numerous identities to remember.

Ultimately, the toughest chapter can vary based on individual preferences and strengths. Some students may find certain chapters easier than others based on their prior knowledge or interests in specific topics. Regular practice, seeking clarification when needed, and utilizing resources such as textbooks, online tutorials, and practice problems can help students overcome challenges in any chapter.

4. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

5. Does CBSE provides the solutions of NCERT class 11 ?

No, CBSE doesn’t provided NCERT solutions for any class or subject.

6. Where can I find the complete solutions of NCERT class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths  by clicking on the link.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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