NCERT Solutions Class 11 Maths Chapter 3

NCERT Solutions Class 11 Maths Chapter 3 – Trigonometric Functions

Class 11 and Class 12 Mathematics forms the fundamentals for many future subjects related to engineering, statistics, and artificial intelligence models. So it’s very important for students to get an in-depth understanding of all topics covered in Class 11 and Class 12 Mathematics.

Class 11 Chapter 3 Trigonometric functions is an essential topic as it prepares students for higher grades. It deals with the domain and range of trigonometric functions. In addition, it covers the sum and difference of two angles within the trigonometric functions. 

Extramarks  Mathematics faculty experts have prepared detailed NCERT Mathematics solutions to help students with their Mathematics preparation. Students can access  NCERT Solutions Class 11 Mathematics Chapter 3  on the Extramarks’ website. It comprises a stepwise and detailed description of the various concepts covered in the chapter and also includes basic and advanced level questions. Students are advised to refer to NCERT Solutions Class 11 Mathematics Chapter 3   to perform better in the examination.

These NCERT Solutions notes are prepared by the Extramarks subject experts team with extensive teaching experience. Students of Class 11 can learn and revise essential points, definitions, and Q&A from the study material offered by NCERT Solutions for Class 11 Mathematics Chapter 3. 

Key Topics Covered In NCERT Solution for Class 11 Maths Chapter 3

The knowledge of Trigonometric Functions is an essential part of Class 11 Mathematics as it involves studying various relations with sides and angles. Trigonometric functions are used for basic geometric calculations and to explain numeric solutions.  Students must have a thorough knowledge of trigonometry as it  is a regular feature in many competitive exams. It is important to work out complex angles and dimensions in very little time,  therefore, students must apply test taking strategies  and they can refer to NCERT Solutions Class 11 Mathematics Chapter 3 to supplement their learning outcome and be satisfied with their preparation. 

The key topics covered in NCERT Solutions Class 11 Mathematics Chapter 3 are as follows: 

Exercise  Topics 
3.1 Introduction
3.2 Trigonometric Equation and Angles
3.3 Trigonometric Functions and Ranges
3.4 Trigonometric Functions of Sum of Two Angles and Difference of Two Angles
Others Miscellaneous

3.1 Introduction

In this section, the students are presented with basic trigonometric calculations and their use. It is expected that you calculate distances using these ratios. The degrees and the radians are two of the most commonly used measurement units for angles. Students studying mathematics and other subjects must understand the fundamentals of measuring angles with trigonometry. The questions provided in NCERT Solutions Class 11 Mathematics Chapter 3   are structured to help students understand core concepts of Trigonometry better.

3.2 Trigonometric Equation and Angles

Students will study measurements of angles in various units, degrees, and radians and their relationships—a solution to the numerical equation that requires converting measures of angles from one format to another. The exercise in Chapter 3 Mathematics Class 11 includes questions on graphs of trigonometric function graphs, domains, signs, and the range. 

The exercise includes numerous examples and problems that will help you discover how a specific trigonometric relationship behaves in one or the other of four quadrants. Students can also examine graphs of various trigonometric ratios and how they perform under particular circumstances. Thus, Class 11 Mathematics NCERT Solutions Chapter 3 will benefit the students. 

3.3 Trigonometric Functions and Ranges

Trigonometric functions connect the angle of a straight-angled triangle with the proportion of side lengths. Sin, Cos, and Tan are the three main functions. This practice consists of problems based on the trigonometric functions of sum and differences between two angles. 

3.4 Sum and Difference of Two Angles

While solving problems with trigonometry, students will come across many problems and situations where they are required to calculate the trigonometric solutions for the sum of two angles or differences of two angles.

3.5 Miscellaneous

The questions in this exercise can be beneficial in revising the essential concepts that are related to Applications of Trigonometric functions.

The identities for the sum and difference are represented by the formulas below:

  • sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
  • cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
  • tan(x+y) = (tan x + tan y)/ (1−tan x • tan y)
  • sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
  • cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
  • tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

NCERT Solutions Class 11 Maths Chapter 3: Exercise & Solutions

NCERT Solutions Class 11 Mathematics Chapter 3 explains the key concepts based on the trigonometric functions. Students can get access to the NCERT exercise and solutions prepared by Extramarks subject matter experts. It will help students to grasp the topic and essential concepts and also improve their problem-solving speed with accuracy. There are 61 questions in the NCERT Solutions Class 11 Mathematics Chapter 3 : Exercise and Solutions. Furthermore, it has 21 identity-based sums, 12 are intermediate level and 28 of high difficulty level. 

 Students can get  the exercise  NCERT Solutions Class 11 Mathematics Chapter 3 by clicking on the links  below.

  • Class 11 Maths Chapter No. 3 Exercise 3.1 – 7 Questions
  • Class 11 Maths Chapter No. 3 Exercise 3.2 – 10 Questions
  • Class 11 Maths Chapter No. 3 Exercise 3.3 – 25 Questions
  • Class 11 Maths Chapter No. 3 Exercise 3.4 – 9 Questions
  • Class 11 Maths Chapter No. 3 Miscellaneous Exercise – 10 Questions

Students may access NCERT Solutions Class 11 Mathematics other chapters by clicking here. In addition, students can also explore NCERT solutions for other Classes below.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT Solutions Class 10
  • NCERT Solutions Class 11
  • NCERT Solutions Class 12

NCERT Exemplar Class 11 Maths 

The NCERT Exemplar is designed for the students of Class 11 Mathematics under the guidance of CBSE. The Exemplar consists of multiple-choice questions, long-form type questions and short-form type questions. It helps students test their knowledge and preparation. So,  they are confident that they will get good marks in their annual exams. Furthermore, Exemplar solutions help to clear the toughest competitive exams. NCERT Exemplar covers all essential topics and concepts prescribed in the Class 11 Mathematics Chapter 3 syllabus. 

To score better in the exam, the students can also refer to the NCERT Solutions Class 11 Mathematics Chapter 3 prepared by the Extramarks experienced faculty. Students who have clear theoretical concepts or a strong foundational base can start practising with Exemplar. Both solutions booklet and  Exemplar’s notes are available on the Extramarks’ website. 

Key Features of NCERT Solutions Class 11 Maths Chapter 3         

NCERT Solutions Class 11 Mathematics Chapter 3 promotes core fundamentals to have a firm grip on the topics. The solution set covers topics such as trigonometric functions, equations and formulas.  

The key features of Extramarks NCERT Solutions for Class 11 Mathematics Chapter 3 include: 

  • The solutions will help the students to get a quick review of the chapter ahead of the exam. 
  • It covers essential questions for the board exams and the competitive examinations. 
  • With the help of NCERT Solutions Class 11 Mathematics Chapter 3, the students can develop a strong foundation in Trigonometric Functions. 
  • It discusses crucial topics such as trigonometric ratios of acute angles, general solutions, sine, cosine, and other trigonometry formulas. Students can learn the trigonometric ratios and definitions of right angles and hypotenuse sides. 

Q.1 Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) – 47°30′
(iii) 240°
(iv) 520°


Since, 1°=π180°i    25°=25°×π180°             =5π36°   radiansii47°30=47.5° 30=0.5°                  =47.5°×π180°  radians                   =19π72  radiansiii       240°=240°×π180°  radians                   =4π3  radiansiv       520°=520°×π180°  radians                   =26π9  radians


Find the degree measures corresponding to the followingradian measures  Use π=227.i1116ii4iii5π3iv7π6


Since, 1 radian=180°π(i)1116  radians=1116×180°π =114×45°π =114×45°×722 =14×45°×72 =(3938)° =39°+38×60 =39°+22+12 =39°+22+12×60 =39°2230 (ii)4  radians=4×180°π =4×180°×722 =(229111)° =(229°+111×60) =(229°+5+511×60) =(229°+5+27) =229°527 (Approx.)(iii) 5π3  radians=5π3×180°π =53×180° =300°(iv)7π6  radians=7π6×180°π =76×180° =210°

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


Number of revolutions made by a wheel in 1 minute=360Number of revolutions made by a wheel in 1 second=36060                                                                         =6Angle covered in 1 revolution by wheel=2π RadiansAngle covered in 6 revolutions by wheel=6×2π                                                         =12π RadiansThus, wheel turns 12π radians in 1 second.


Find the degree measure of the angle subtended at thecentre of a circle of radius 100 cm by an arc of length 22 cm(Use π=227).


Radius of circle=100 cm   Length​ of arc =22cmAngle(θ) subtended at the centre by an arc    =ArcRadiusradiansθ=22100radiansθ=22100×180°π[ 1radian=180°π]θ=22100×180°×722    =1.8°×7    =12.6°    =12°+0.6×60     =12°+36    =12°36Thus, angle subtended by an arc of 22 cm at the centre is 12°36′ .

Q.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.


Diameter of a circle=40cm      Radius​ of a circle=402cm =20cm          Length of chord=20cm

Since,ΔOAB is equilateral triangle.So,​            AOB=60° =π3 radiansSince,        angle=Length of arcRadius π3 Radian=Length of arc AB20 cm Length of arc AB=20 cm × π3 =20π3 cm

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


Let radii of circles be r1 and r2 respectively.Angle subscribed by arc in first circle=60°  =60°×π180°  radians  =π3  radiansAngle subscribed by arc in second circle  =75°  =75°×π180°  radians  =5π12  radiansSince, length of both arcs is same.So,             Length of arc of first circle=Length of arc of second circleπ3  radian×r1=5π12  radian×r2      r1r2=(5π12)(π3) =54 r1:r2=5:4Thus, the ratio of radii of circles is 5:4.

Q.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


Length of pendulum=75 cmiLength of arc described by pendulum=10cmAngle swept by pendulum=1075Rad. θ=arcr                                     =215radiansiiLength of arc described by pendulum=15cmAngle swept by pendulum=1575Rad. θ=arcr                                     =15radiansiiiLength of arc described by pendulum                                     =21cmAngle swept by pendulum=2175Rad. θ=arcr                                     =725radians

Q.8 Find the values of other five trigonometric functions in Exercises 1 to 5.

1.cos x=12, x lies in third quadrant. 2.sin x=35, x lies in second quadrant. 3.cot x=34, x lies in third quadrant. 4.sec x=135, x lies in fourth quadrant. 5.tan x=512, x lies in second quadrant.


1.secx=1cosx       =112       =2xliesinthirdquadrant.sin2x+cos2x=1             sinx=1cos2x                   =1122            sinx=34sinx=32  x lies in third quadrant.cosecx=1sinx          =132          =23   tanx=sinxcosx          =3212          =3x lies in third quadrant.   cotx=1tanx          =13x lies in third quadrant.Thus, sinx=32, cosecx=23,secx=2,tanx=3andcotx=13. 2. sinx=3/5cosecx=1/sinx           =135cosecx=53sin2x+cos2x=1cosx=±1sin2x        =±1352cosx=±1625        =45           x lies in II quadrant.secx=1cosx        =145        =54tanx=sinxcosxtanx=3545        =34  x lies in II quadrant.cotx=1tanx        =134        =43        x lies in II quadrant.Thus, cos x=45,secx=54 , cosecx=53,tan x=34andcotx=43. 3.tanx=1cotx       =134       =43cosec2x=1+cot2x            =1+342            =1+916cosecx=±2516           =54x lies in third quadrant.    sinx=1cosecx           =154           =45   secx=1+tan2x           =1+432           =1+169   secx=±259           =53x lies in third quadrant.   cos x=35x lies in third quadrant.Thus,  sinx=45,cosecx=54,cosx=35,secx=53andtanx=43. 4.cos x=1sec x       =1135       =513 x lies in fourth quadrant.sin x=1cos2x       =±15132       =1213 x lies in fourth quadrant.cosec x=1sin x           =11213           =1312    tan x=sin xcos x =(1213)(513)=125 [x lies in fourth quadrant.]cot x=1tanx =1(125) =512Thus, sin x=1213,cos x=513,tan x=125,cot x=512and cosec x=1312. 5.cot  x=1512 cot x=1tan  x        =125sec2x=1+tan2x         =1+5122         =1+25144secx=±169144        =1312 x lies in second quadrant.cosx=1secx        =11312        =1213 x lies in second quadrant.cosec2x=1+cot2x           =1+1252           =1+14425cosecx=±16925            =135 x lies in second quadrant.     sinx=1cosecx             =1135sinx=513 x lies in second quadrant.Thus, sinx=513,cosx=1213,cotx=125,secx=1312andcosecx=135.

Q.9 Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°
7. cosec (– 1410°)

8. tan19π3

9. sin(11π3)

10. cot(15π4)


6.sin 765° = sin 2 x 360° + 45°               = sin 45° sinn×360°+θ=sinθ               =12 7.cosec 1410°=cosec1410°        cosecθ=cosecθ   =cosec4×360°30°   =cosec 30°cosec n×360°θ=cosecθ   =2 8.tan19π3=tan6π+π3 tan19π3=tanπ3[ tan (2+θ)=tanθ]       =3 9.sin11π3=sin11π3 sinx=sin x    =sin4ππ3    =sinπ3         sin 2x=sin xsin11π3=32 10.cot 15π4=cot 15π4 cot x=cot x    =cot 4ππ4    =cotπ4 cot 2x=cot x    =1


Prove that:sin2π6+cos2π3tan2π4=12


L.H.S.:sin2π6+cos2π3tan2π4=(12)2+(12)2(1)2=14+141=1+144=24=12=R.H.S.Hence proved.


Prove that:2sin2(π6)+cosec2(7π6)cos2(π3)=32


L.H.S.:2sin2π6+cosec27π6cos2π3=2(12)2+cosec2(π+π6)×(12)2      =2×14+14×cosec2(π6)       =12+14×(2)2      =12+1      =32=R.H.S.Hence proved.


Prove that:cot2π6+cosec5π3+3tan2π6=6


L.H.S:cot2π6+cosec5π6+3tan2π6=(3)2+cosec(ππ6)+3(13)2        =3+cosec(π6)+3×13[cosec(πx)=cosec x]        =3+2+1        =6=R.H.S.Hence proved.


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


L.H.S.:2sin2(3π4)+2cos2(π4)+2sec2(π3)=2sin2(ππ4)+2(12)2+2(2)2 =2sin2(π4)+2(12)+8=2(12)2+1+8=2(12)+1+8=1+1+8=10=R.H.S.

Q.14 Find the value of: (i) sin75° (ii) tan15°


(i) sin 75°=sin(45°+30°)=sin 45° cos 30° + sin 30° cos 45°[ sin(x+y)=sin x cos y+sin y cos x]=12×32+12×12=322+122=3+122(ii) tan 15°=tan (45°30°)=tan 45° tan 30°1 + tan 45° tan 30°=1131+1×13 =313+1×3131=(31)231=323+12   tan 15°=4232=23


cos (π4x)cos(π4y)sin(π4x)sin(π4y)=sin (x+y)


Since,cosA cosB sinA sinB = cos (A + B)L.H.S. =cos (π4 x) cos (π4 y) sin (π4 x) sin (π4 y)= cos {(π4 x) + (π4 y)}= cos (π2 x y)= cos {π2 (x + y)}= sin (x + y) [ cos (π2 θ) = sin θ]= R.H.S. Hence proved.


Prove the following:tan (π4 + x)tan (π4 x) = (1 + tan x1 tan x)2


L.H.S:tan (π4 + x)tan (π4 x) = (tan π4 + tan x1 tan π4 tan x)(tan π4 tan x1 + tan π4 tan x)      [ tan (A + B) = tan A + tan B1 tan A tan Btan (A B) = tan A tan B1 + tan A tan B]= (1 + tan x1 1.tan x)(1 tan x1 + 1.tan x)= (1 + tan x1 tan x) × (1 + tan x1 tan x)= (1 + tan x1 tan x)2= R.H.S. Hence proved.


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


L.H.S.:cos(π+x)cos(x)sin(πx)cos(π2+x)=cos x cos xsin x×sin x[cos(π+x)=cosx         cos(x)=cosx      sin(πx)=sinx    cos(π2+x)=sinx]    =cot2x    =R.H.S.Hence proved.


cos(3π2 +x)cos (2π+x)[cot(3π2x)+cot(2π + x)] = 1


L.H.S. = cos (3π2+ x)cos(2π+x)[cot (3π2 x) +cot (2π+x)]= sin x.cosx [tanx +cot x]= sin x.cos x [sinxcosx+cosxsinx]= sin x.cos x[sin2x +cos2xsin x.cosx] = sin2x+ cos2x=1 =R.H.S.

Q.19 sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x


L.H.S.:sin (n + 1) x sin (n +2)x + cos (n +1)x cos (n + 2) x           = cos (n +1)x cos (n +2)x +sin (n +1) x sin (n + 2) x           = cos {(n +1) x (n+2)x}[ cosA cosB sinA sinB=cos(A +B)]           =cos (nx + x nx2x)           =cos (x)                  [ cos (A)=cosA]           =cos x=R.H.S.


cos(3π4 + x) cos(3π4 x) =2 sin x


L.H.S.:cos(3π4+x)cos(3π4 x)=2sin((3π4 +x)+(3π4x)2) sin((3π4 +x)(3π4 x)2)=2 sin (2(3π4)2)sin (2x2) =2sin(π π4)sin x=2 sin (π π4)sinx=2 sinπ4×sin x= 2× 12× sin x= 2sinx = R.H.S.

Q.21 Prove that:
sin2 6x – sin2 4x = sin 2x sin 10x


L.H.S.=sin2 6x sin 2 4x  =(sin 6x+sin4x)(sin6x sin 4x) [a2b2=(ab)(a+b)]  = {2sin(6x+ 4x2) cos(6x4x2)} {2 cos(6x + 4x2) sin(6x 4x2)}  ={2sin5x cos x}{2cos5xsin x}  =(2 sinx cos x) (2sin5xcos5x)  = sin2x.sin 10x [ sin 2x = 2 sin x cos x]=R.H.S.Hence proved.

Q.22 Prove that:
cos2 2x – cos2 6x = sin 4x sin 8x


L.H.S.= cos22x cos26x  =(cos2x+cos 6x) (cos 2xcos 6x)   ={2 cos(2x+ 6x2) cos (2x 6x2)}{2sin (2x 6x2)sin(2x+ 6x2)}  ={2 cos4xcos (2x)}{2sin(2x)sin 4x}  =(2sin2xcos2x)(2sin4xcos 4x)  =sin4x.sin8x[sin 2x=2 sinx cos x]=R.H.S.Hence proved.

Q.23 Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x


sin 2x + 2 sin 4x + sin 6x =2 sin 4x +sin 6x +sin 2x  = 2 sin 4x +2sin (6x+ 2x2) cos(6x 2x2)  =2 sin 4x + 2 sin4xcos2x  = 2 sin 4x (1 + cos2x)  = 2 sin 4x (2cos2x)  =4 cos2x sin 4x  = R.H.S.

Q.24 Prove that:
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)


We have, cot 4x (sin 5x +sin 3x)=cot x (sin 5x  sin 3x)The given equation can be written as:sin 5x+sin 3xsin 5xsin 3x=cot xcot 4x L.H.S:sin 5x +sin 3xsin 5x sin 3x =2 sin(5x+ 3x2)cos(5x 3x2)2 cos 5x+3x2 sin 5x 3x2  = sin 4xcos xcos4xsin x  = (cosxsin x)(cos 4xsin 4x)  =cot xcot4x  = R.H.S.Hence proved.


Prove that:cos9xcos5xsin17xsin3x=sin2xcos10x


L.H.S.:cos9xcos5xsin17xsin3x=2sin(9x+5x2)sin(9x5x2)2cos(17x+3x2)sin(17x3x2)[cosAcosB=2sin(A+B2)sin(AB2)      sinAsinB=2cos(AB2)sin(AB2)]   =sin(7x)sin(2x)cos(10x)sin(7x) =sin2xcos10x=R.H.S.Hence​ proved.




L.H.S.:sin5x+sin3xcos5x+cos3x=2sin(5x+3x2)cos(5x3x2)2cos(5x+3x2)cos(5x3x2)    =sin4xcosxcos4xcosx    =tan4x    =R.H.S.Hence proved.




L.H.S.:sinxsinycosx+cosy=2cosx+y2sinxy22cosx+y2cosxy2                 =sinxy2cosxy2                =tanxy2                =R.H.S.Hence proved.




L.H.S.:sinx+sin3xcosx+cos3x=sin3x+sinxcos3x+cosx=2sin(3x+x2)cos(3xx2)2cos(3x+x2)cos(3xx2)=sin2xcos2x=tan2x=R.H.S.Hence proved.


Provethat:sinx sin3xsin2x cos2x = 2sinx


L.H.S.:sinx sin 3xsin2xcos2x = (sin3x sin x)(cos2x sin2x)= 2cos (3x+ x2)sin (3x x2)cos2x=2cos2xsin xcos 2x=2sin x=R.H.S.


Provethat:cos 4x+cos3x+cos 2xsin4x +sin3x+ sin2x=cot 3x


L.H.S.:cos4x+ cos 3x + cos2xsin 4x + sin3x + sin 2x= cos 4x + cos 2x+ cos 3xsin 4x+ sin 2x+ sin 3x= 2cos(4x + 2x2) cos(4x 2x2) + cos 3x2sin (4x + 2x2) cos(4x 2x2)+ sin 3x= 2cos 3x cos x +cos3x2 sin 3x cos x+ sin 3x = cos 3x (2 cosx + 1)sin3x (2 cos x+ 1)cos 4x +cos3x + cos 2xsin 4x+ sin 3x+sin2x = cot3x =R.H.S. Hence​ proved.

Q.31 Prove that:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1


cot 3x = cot (2x + x) =cot 2x cot x 1cot 2x + cot xcot 3x (cot 2x + cot x)= cot 2x cot x 1cot 3x cot 2x + cot 3x cot x = cot 2x cot x 11 = cot x cot 2x cot 2x cot 3x cot 3x cot xor cot x cot 2x cot 2x cot 3x cot 3x cot x = 1.Hence​ proved.


Prove that:tan 4x = 4 tan x (1 tan2x)1 6 tan2x + tan4x


L.H.S:tan 4x = tan 2 (2x)    = 2 tan 2x1 tan2 2x     = 2 (2 tan x1 tan2x)1 (2 tan x1 tan2x)2    = (4 tan x1 tan2x){(1 tan2x)2 4 tan2x(1 tan2x)2}    = 4 tan x1 tan2x × (1 tan2x)21 2 tan2x + tan4x 4 tan2x    = 4 tan x(1 tan2x)1 6 tan2x + tan4x = R.H.S.Hence proved.

Q.33 Prove that:
cos 4x = 1 – 8sin2 x cos2 x


L.H.S. : cos 4x = cos 2 (2x)        = 1 2sin2 (2x) [ cos 2x = 1 sin2 x]        = 1 2(2 sin x cos x)2 [ sin 2x = 2 sin x cos x]        = 1 2 (4 sin2x cos2x)        = 1 8 sin2x cos2x        = R.H.S.

Q.34 Prove that:
cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1


L.H.S.=cos 6x=cos3(2x)=4cos32x3cos2x=4(2cos2x1)33(2cos2x1)=4(8cos6x12cos4x+6cos2x1)6cos2x+3=32cos6x48cos4x+24cos2x46cos2x+3=32cos6x24cos4x+18cos2x1=R.H.S.Hence proved.




Since, tanπ3=3 and tan4π3=tanπ+π3=tanπ3=3Therefore, principal solutions are x=π3 and 4π3.General solution of tanx=3tanx=3       =tanπ3tanx=tan+π3 x=+π3,where nZ.




Since, secx=2secπ3=2 and sec5π3=sec(2ππ3)=secπ3=2Therefore, principal solutions are x=π3 and 5π3.General solution:secx=2=secπ3=sec(2±π3)x=2±π3,where​ nZ.




Since, cot x=3     cot x=cotπ6      =cot(ππ6)      =cot(5π6)and     cot x=cot(2ππ6)      =cot(11π6)Therefore, the principal solution are x=5π6 and 11π6.General solution:              cotx=3     =cot(5π6)     =cot(+5π6)x=+5π6,where​ nZ.


Find the principal and general solutions of the followingequation : cosec x = 2


Since, cosec x = cosec (π6) cosec x = 2 = cosec (π6) = cosec (π + π6) = cosec (7π6)andcosec x = cosec (2π π6) = cosec 11π6Therefore, the principal solutions are 7π6, 11π6.General solution:cosec x = cosec (7π6) = cosec ( + (1)n 7π6) x = + (1)n 7π6, n Z.


Find the general solution for each of the followingequation : cos 4x = cos 2x


We​​  have,cos 4x = cos 2x = cos (2 ± 2x) [The general solution of cos θ = cos α θ = 2 ± α] 4x = 2 ± 2xFor(+) ve sign:     4x = 2 + 2x     2x = 2 x = For () ve sign:     4x = 2 2x     6x = 2 x = 13 The general solutions are x = 13 and , where n Z.


Find the general solution for each of the followingequation : cos 3x + cos x cos 2x = 0


We have, cos 3x + cos x cos 2x = 02 cos (3x + x2) cos (3x x2) cos 2x = 02 cos 2x cos x cos 2x = 0cos 2x (2 cos x 1) = 0 cos 2x = 0 2x = (2n + 1) π2      x = (2n + 1) π4, n Z or 2 cos x 1 = 0 cos x = 12 = cos π3      x = 2 ± π3, n Z.


Find the general solution for each of the followingequation : sin 2x + cos x = 0


We have, sin 2x + cos x = 0 2 sin x cos x + cos x = 0 cos x (2 sin x + 1) = 0Either cos x = 0 x = (2n + 1) π2, n Z.or (2 sin x + 1) = 0 sin x = 12 sin x = sin 7π2 x = + (1)n 7π2, n Z.Thus, the general solution of given equation is: + (1)n 7π2 ​  or (2n + 1) π2, n Z.




We have, sec2 2x = 1 tan 2x 1 + tan2 2x = 1 tan 2x tan2 2x + tan 2x = 0 tan 2x (tan 2x + 1) = 0Either tan 2x = 0 2x =        x = 2, where n  Z.or   tan 2x + 1 = 0 tan 2x = 1       tan 2x = tan π4      = tan (π π4)       tan 2x = tan 3π4       2x = + 3π4       x = 2 + 3π8, where n Z.The general solution of the given equation is:x = 2, 2 + 3π8, where n Z.


Find the general solution for each of the followingequation : sin x + sin 3x + sin 5x = 0


We have,   sin x+sin 3x+sin 5x=0   sin 5x+sin x+sin 3x=0   2sin(5x+x2)cos(5xx2)+sin 3x=0     2sin 3x cos 2x+sin 3x=0          sin 3x (2cos 2x+1)=0Either sin3x=03x=x=3,nZ.or 2 cos 2x+1=0cos 2x=12           2x=2±2π3             x=±π3,nZ.Therefore,​​ the general solution of the given equations is:x=3​​ or±π3,nZ.

Q.44 In any triangle ABC, if a = 18, b = 24, c = 30, find

1. cos A, cos B, cos C
2. sin A, sin B, sin C


1.Since,cosA=b2+c2a22bc                  =242+30218222430                  =576+9003241440                  =11521440                  =45cosB=c2+a2b22ca        =302+18224223018        =900+3245761080        =6481080        =35cosC=a2+b2c22ab         =182+24230221824         =324+576900864         =0864         =0

2.Since,cosA=b2+c2a22bc                  =242+30218222430                  =576+9003241440                  =11521440                  =45andsinA=1cos2A             =1452             =11625             =35Since,sinAa=sinBb=sinCc        3518=sinB24=sinC30        3518=sinB24 and 3518=sinC30      3×245×18=sinB and 3×305×18=sinC            45=sinB and  1=sinCThus, sinA=35, sinB=45, sinC=1.


L.H.S. = a bc=k sin A k sin Bk sin C [ asin A = bsin B = csin C = k (let)]=sin A sin Bsin C=2 cos (A + B2) sin (A B2)2 sin C2 cos C2=cos (π2 C2) sin (A B2)sin C2 cos C2=sin (C2) sin (A B2)sin C2 cos C2 =sin (A B2)cos C2 = R.H.S.Thus, it is proved.


For any triangle ABC, prove thata bc = sin (A B2)cos C2


L.H.S. = a bc=k sin A k sin Bk sin C [ asin A = bsin B = csin C = k (let)]=sin A sin Bsin C=2 cos (A + B2) sin (A B2)2 sin C2 cos C2=cos (π2 C2) sin (A B2)sin C2 cos C2=sin (C2) sin (A B2)sin C2 cos C2 =sin (A B2)cos C2 = R.H.S.Thus, it is proved.


L.H.S. = a + bc=k sin A + k sin Bk sin C [ asin A = bsin B = csin C = k (let)]=sin A + sin Bsin C=2 sin (A + B2) cos (A B2)2 sin C2 cos C2=2 sin (π2 C2) cos (A B2)2 sin C2 cos C2=cos (C2) cos (A B2)sin C2 cos C2 =cos (A B2)sin C2 = R.H.S.Thus, it is proved.


For any triangle ABC, prove thatsin BC2 = bca cos A2


We have,         b ca=k sin B k sin Ck sin A      [asin A=bsin B=csin C=k (let)]     =sin B sin Asin A     =2 cos (B + C2) sin (B C2)2 sin A2 cos A2     =cos (π2 A2) sin (B C2)sin A2 cos A2     =sin (A2) sin (B C2)sin A2 cos A2     =sin (B C2)cos A2 sin (B C2) = b ca cos A2Therefore, it is proved.

Q.48 For any triangle ABC, prove that a (b cos C – c cos B) = b2 – c2


We have,bcosCccosB=b×a2+b2c22abc×a2+c2b22ac                        =a2+b2c22aa2+c2b22a                        =a2+b2c2a2c2+b22a                        =2b22c22aabcosCccosB=b2c2Hence, it is proved.


For any triangle ABC, prove thata(cosCcosB)=2(bc)cos2A2


We aregiventhat:   acosCcosB=2bccos2A2  cosCcosB2cos2A2=bcaL.H.S.=cosCcosB2cos2A2         =cosA2sinBC2cos2A2sinπ2θ=sinθ         =cosB+C2cosB+C2×sinBC2cosA2  Multiply and divide by cosB+C2         =2cosB+C22cosπ2A2×sinBC2cosA2          =2cosB+C2sinBC22sinA2cosA2          =sinBsinCsinA2cosx+y2sinxy2=sinxsinysinθ=2sinθ2cosθ2          =kbkckasinAa=sinBb=sinCc=k Let          =bca=R.H.S.Thus, acos Ccos B=2bccos2A2 is proved.


For any triangle ABC, prove thatsin(BC)sin(B+C)=b2c2a2


R.H.S.=b2c2a2  =k2sin2Bk2sin2Ck2sin2A [asinA=bsinB=csinC=k (let)]  =sin2Bsin2Csin2A  =sin(B+C)sin(BC)sin2A [ sin2xsin2y=sin(x+y)sin(xy)]   =sin(πA)sin(BC)sin2A [ B+C=πA]  =sin A sin(BC)sin2A [ sin(πA)=sinA]  =sin(BC)sin{π(B+C)} [ A=π(B+C)]  =sin(BC)sin(B+C) [ sin(πθ)=sinθ]  =L.H.S.Thus, it is proved.


For any triangle ABC, prove that(b+c) cos B + C2 = a cos BC2


We are given : (b + c) cos B + C2 = a cos B C2or               (b + c)a = cos B C2cos B + C2L.H.S.=  (b + c)a  =k sin B + k sin Ck sin A [ asin A = bsin B = csin C = k (let)]  =sin B + sin Csin A  =2 sin(B + C2) cos (B C2)2 sin A2 cos A2[sin x + sin y= 2 sin (x + y2) cos (x y2)]  =sin (π2 A2) cos (B C2)sin A2 cos A2  =cos A2 cos (B C2)sin (π2 B + C2) cos (A2) [ sin (π2θ) = cosθ]   =cos (B C2)cos (B + C2) = R.H.S.Thus,  (b + c) cos B + C2 = a cos B C2 is proved.

Q.52 For any triangle ABC, prove that a cos A + b cos B + c cos C = 2 a sin B sin C


L.H.S.=a cos A+b cos B+c cos C  =k sin A cos A+k sin B cos B+k sin C cos C[ asinA=bsinB=csinC=k (let)]  =k2(2 sin A cos A+2 sin B cos B+2 sin C cos C)  =k2 (sin 2A+sin 2B+sin 2C)  =k2 (2 sin2A+2B2.cos2A2B2+sin 2C)  =k2 {2 sin (A+B).cos(AB)+sin 2C}  =k2 {2 sin (πC).cos(AB)+2 sin C cos C}[ A+B+C=π]  =k2 {2 sin C.cos (AB)+2 sin C cos(πAB)}[ sin(πθ)=sinθ]   = k sin C {cos (AB)cos (A+B)}  =k sin C[2sin(AB+A+B2)sin{A+B(AB)2}]  =2 k sin C.sin A.sin B  =2 (k sin A).sin B.sin C  =2a sin B.sin C[sin Aa=k (let)]  =R.H.S.Thus, it is proved.


For any triangle ABC, prove thatcosAa+cosBb+cosCc=a2 + b2 + c22abc


L.H.S.=cosAa+cosBb+cosCc  =1a(b2 + c2 a22bc) + 1b (c2 + a2 b22ca) +1c (a2 + b2 c22ab)  =b2 + c2 a2 + c2 + a2 b2 + a2 + b2 c22abc  =b2 + c2 + a22abcHence, it is proved.

Q.54 For any triangle ABC, prove that
(b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0


(b2c2)cot A=(b2c2)cos Asin A =(b2c2)cos Aka[ sin Aa=sin Bb=sin Cc=k]=(b2 c2)ka × b2 + c2 a22bc=b4 + b2 c2 b2 a2 b2 c2 c4 + c2 a22kabc(c2 a2) cot B=(c2 a2)cos Bsin B=(c2 a2)cos Bkb=(c2 a2)kb × c2 + a2 b22bc=c4 + c2 a2 c2 b2 a2 c2 a4 + a2 b22kabc(a2 b2) cot C=(a2 b2)cos Csin C=(a2 b2) cos Ckc=(a2 b2)kc × a2 + b2 c22ab=a4 + a2 b2 a2 c2 a2 b2 b4 + b2 c22kabcL.H.S. = (b2 c2) cot A + (c2 a2) cot B + (a2 b2) cot C= b4 + b2 c2 b2 a2 b2 c2 c4 + c2 a22kabc +c4 + c2 a2 c2 b2 a2 c2 a4 + a2 b22kabc+a4 + a2 b2 a2 c2 a2 b2 b4 + b2 c22kabc= (b4 + b2 c2 b2 a2 b2 c2 c4 + c2 a2 + c4 + c2 a2 c2 b2a2 c2 a4 + a2 b2 + a4 + a2 b2 a2 c2 a2 b2 b4 + b2 c2)2kabc=02kabc = 0= R.H.S.Thus, it is proved.


For any triangle ABC, prove thatb2c2a2sin  2A+c2a2b2sin  2B+a2b2c2sin  2C=0


b2 c2a2 sin  2A= (b2 c2a2) 2sinA cos A=(b2 c2a2) 2(ka) cosA [ sin Aa=sin Bb=sin Cc = k]=(b2 c2a2) 2(ka) b2 + c2 a22bc=k (b4 c4 a2 b2 + a2 c22abc)Similarly,c2 a2b2sin  2B = k (c4 a4 c2 b2 + a2 b22abc) and a2 b2c2 sin  2C = k(a4 b4 a2 c2 + b2 c22abc)L.H.S. = b2 c2a2 sin  2A + c2 a2b2 sin  2B + a2 b2c2 sin  2C=k (b4 c4 a2 b2 + a2 c22abc) + k (c4 a4 c2 b2 + a2 b22abc)+ k (a4 b4 a2 c2 + b2 c22abc)=k (b4 c4 a2 b2 + a2 c2 + c4 a4 c2 b2 + a2 b2+ a4 b4 a2 c2 + b2 c22abc)=k (02abc)= 0 = R.H.S.Hence proved.

Q.56 A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35 m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.


Let PQ be a tree of height h m and A be a point on ground as AQ = 35 m.

APR=90°60° =30° APQ=30°       PAQ=60°15°     =45°       PQA=180°PAQAPQ     =180°45°30°     =105°Applying sin rule in ΔPQA, we get    sin45°PQ=sin30°AQ=sin105°APsin45°PQ=sin30°35   (12)PQ=(12)35  352=PQ2  PQ=352Thus, the height of the tree is 352m.

Q.57 Two ships leave a port at the same time. One goes 24 km per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.


Speed of first ship = 24 km/hr
Speed of second ship = 32 km/hr
Distance (OP) covered in 3 hours
= 24 × 3
= 72 km

Distance (OQ) covered in 3 hours
= 32 × 3
= 96 km
Angle POQ = (90° – 45°) + (90° – 75°)
= 45° + 15°
= 60°

By using cosine formula in ΔOPQ, we getPQ2=OP2+OQ22×OP×OQ cos60° =(72)2+(96)22×72×96×12 =5184+92166912  PQ=7488 =86.53  km


Two trees, A and B are on the same side of a river. From apoint C in the river the distance of the trees A and B is 250 mand 300 m, respectively. If the angle C is 45°, find thedistance between the trees (use 2=1.44).


Let A and B be the positions of two trees respectively. The distance of first tree from C is 250m and that of second tree is 300 m.
Angle C is 45°.

Applying cosine formula in ΔABC, we get                 cos C=a2 + b2 c22ab         cos 45°=(300)2 + (250)2 c22(300)(250)        12=90000 + 62500 c2150000 1500002×22=152500AB2    15000022=152500AB2150000 × 1.4142=152500AB2           106050=152500AB2                 AB=152500106050                       =46450                  AB=215.5 mThus, distance between two trees is 215.5m.

Q.59 Prove that:



L.H.S.=2cosπ13cos9π13+cos3π13+cos5π13=cos(π13+9π13)+cos(π139π13)+cos(π10π13)+cos(π8π13)=cos(10π13)+cos(8π13)cos10π13cos8π13 =cos(8π13)cos8π13=0=R.H.S.

Q.60 Prove that:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0


L.H.S.=(sin 3x+sin x) sinx+(cos 3xcos x) cos x=sin 3x sin x+sin2x+cos 3x cos xcos2x=cos 3x cos x+sin 3x sin xcos2x+sin2x=cos (3xx)(cos2xsin2x)[cos (AB)=cos A cos B+sin A cos B]=cos 2xcos 2x [ cos2xsin2x=cos 2x]=0=R.H.S.Hence proved.

Q.61 Prove that:

(cos x+cos y)2 + (sin x sin y)2 = 4 cos2x + y2


L.H.S. : (cos x + cos y)2 + (sin x-sin y)2 = (2cos x + y2cos x – y2)2 + (2cos x + y2sin x y2)2= 4cos2 (x + y2)(cos2 x y2+sin2 x y2)= 4 cos2 (x + y2) × 1 [Q sin2x + cos2x = 1] =4cos2 (x + y2)= R.H.S. Hence proved.

Q.62 Prove that:

(cos xcos y)2+(sin xsin y)2=4 sin2 xy2


L.H.S. : (cos x – cos y)2 + (sin x sin y)2={2sin (x + y2)sin (x y2)}2 + {2cos (x + y2)sin( x y2)}2=4sin2 (x y2){sin2(x + y2)+cos2 (x + y2)}= 4 sin2 (x y2) × 1[Q sin2x + cos2x = 1] =4sin2 (x y2)=R.H.S.Hence proved.

Q.63 Prove that:
sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x


L.H.S.=sin x + sin 3x + sin 5x + sin 7x    =sin 7x + sin x + sin 5x + sin 3x     =2 sin (7x + x2) cos (7x x2)+2sin (5x + 3x2) cos (5x 3x2)    =2 sin 4xcos 3x + 2 sin 4x cos x    =2 sin 4x (cos 3x + cos x)    =2 sin 4x {2 cos(3x + x2) cos (3x x2)}    =4 sin 4x cos 2x cos x    =4 cos x cos 2x sin 4x    =R.H.S.Hence proved.


Prove that:

(sin 7x+sin 5x)+(sin 9x+sin 3x)(cos 7x+cos 5x)+(cos 9x+cos 3x)=tan 6x


L.H.S.=(sin 7x+sin 5x)+(sin 9x+sin 3x)(cos 7x+cos 5x)+(cos 9x+cos 3x) =(2sin7x+5x2cos7x5x2)+(2sin9x+3x2cos9x3x2)(2cos7x+5x2cos7x5x2)+(2cos9x+3x2cos9x3x2)=(2 sin 6x cos x) + (2 sin 6x cos 3x)(2 cos 6x cos x) + (2 cos 6x cos 3x)=2 sin 6x(cos x + cos 3x)2 cos 6x(cos x + cos 3x)=tan 6x=R.H.S.Hence proved.


Prove that:sin3x+sin2xsinx=4sinxcosx2cos3x2


L.H.S. = sin 3x+sin 2x sin x         = sin 3x sin x + sin 2x         = 2cos3x + x2sin3x x2+2sin x cos x         = 2cos 2x sin x + 2sin x cos x         = 2sin xcos 2x + cos x         = 2sin x2cos2x + x2cos2x x2         = 4sin x cos3x2 cosx2         = 4sin x cosx2 cos3x2         = R.H.S. Hence proved.


Findsinx2, cosx2 and tanx2 in each of the following: 8. tan x=43, x in quadrant II 9. cos x=13,x in quadrant III 10.sinx=14,x in quadrant II


8.Since,tanx=43,xinquadrantII       sec2x=1+tan2x                 =1+432                 =1+169                 =259         secx=±259                 =53x is in II quadrant.      cosx=35  2cos2x21=35      2cos2x2=135                 =25       cos2x2=15        cosx2=±15cosx2=15x lies in II quadrant,So, x2 lies in I quadrant.        =15×55        =55sinx2=1cos2x2          =±1152          =±115          =25x2 lies in I quadrant.          =25×55          =255  tanx2=sinx2cosx2          =2515=2So, the values of sinx2, cosx2 and tanx2 are 55,255 and2 respectively. 9.We have,        cosx=132cos2x21=13    2cos2x2=113               =23      cos2x2=13       cosx2=±13cosx2=13x lies in III quadrant,So, x2 lies in II quadrant.        =13×33        =33   sinx2=1cos2x2            =±1132            =±113            =23  x2 lies in II quadrant.            =23×33            =63    tanx2=sinx2cosx2            =2313=2So, the values of sinx2, cosx2 and tanx2 are 63,63 and2 respectively.

10.We have,sinx=14,xinquadrantII    cosx=1sin2x      cosx=±1142              =±1116              =154xinquadrantIISince,cosx=12sin2x2        154=12sin2x2       2sin2x2=1+154         sin2x2=12+158          sinx2=±12+158                  =8+2154x2 lies in I quadrant.     andcosx=2cos2x21154=2cos2x21cos2x2=12158cosx2=±12158        =82154x2 lies in I quadrant.tanx2=sinx2cosx2        =8+215482154        =8+2158215×8+2158+215        =8+2156460        =8+2152tanx2=4+15So, the values of sinx2, cosx2 and tanx2 are 8+2154,82154 and 4+15 respectively.

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FAQs (Frequently Asked Questions)

1. What trigonometric functions are provided in NCERT Solutions Class 11 Mathematics Chapter 3 ?

Trigonometric functions are real functions that are related to the angle of a right-angled triangle to the ratio of the length of its sides. Further, it discusses trigonometric identities and sum and difference identities.

2. What are the important formulas mentioned in NCERT Solutions Class 11 Mathematics Chapter 3?

The following are the important formulas in NCERT Solutions Class 11 Mathematics Chapter 3:

  • sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
  • cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
  • tan(x+y) = (tan x + tan y)/ (1−tan x • tan y)
  • sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
  • cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
  • tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

3. Do I need to attempt all questions provided in NCERT Solutions Class 11 Mathematics Chapter 3?

The questions and answers in NCERT Solutions Class 11 Mathematics Chapter 3 provide an effective way to facilitate in-depth learning of concepts. In addition, every question aims to improve the conceptual clarity of the students. Thus they can attempt long and short-form questions along with multiple-choice questions with ease.