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NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:41 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 3 Trigonometric Functions is considered a very important chapter for practical use, and application in various different fields and also for the exams. NCERT Exemplar Class 11 Maths chapter 3 solutions give a brief procedure and explanation about angles along with degree and radian measure. It also establishes a relationship between radian and real numbers, and radian and degree measure. Class 11 Maths NCERT Exemplar solutions chapter 3 also covers a variety of questions relating to a conversion of degrees to radian and vice versa through established relation between them through the use of notational convention.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Suggested: JEE Main: high scoring chaptersPast 10 year's papers

This Story also Contains
  1. NCERT Exemplar Class 11 Maths Solutions Chapter 3: Exercise - 1.3
  2. Important Notes of NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions
  3. Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 3
  4. What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 3?
  5. NCERT Solutions for Class 11 Mathematics Chapters
  6. Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 3: Exercise - 1.3

Question:1

Prove that
tanA+secA1tanAsecA+1=1+sinAcosA

Answer:

L.H.S=tanA+secA1tanAsecA+1=tanA+secA(sec2Atan2A)tanAsecA+1=tanA+secA[(secA+tanA)(secAtanA)]tanAsecA+1=(secA+tanA)[1(secAtanA)]tanAsecA+1


=(secA+tanA)[1secA+tanA]tanAsecA+1=secA+tanA=1cosA+sinAcosA=1+sinAcosA=R.H.S


Question:2

If [2sin \alpha  / (1+cos \alpha  +sin \alpha  )] = y, then prove that [(1- cos \alpha  +sin \alpha  ) / (1+sin \alpha  )] is also equal to y.

\left[\mathrm{Hint}: \text { Express } \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}\right]

Answer:

y=2sinα1+cosα+sinα=2sinα1+cosα+sinα1+sinαcosα1+sinαcosα=2sinα(1cosα+sinα)(1+sinα)2cos2α=2sinα(1cosα+sinα)1+sin2α+2sinαcos2α
=2sinα(1cosα+sinα)(1cos2α)+sin2α+2sinα=2sinα(1cosα+sinα)2sin2α+2sinα=2sinα(1cosα+sinα)2sinα(1+sinα)=(1cosα+sinα)(1+sinα)=y


Question:3

If msinθ=nsin(θ+2α), then prove that
tan(θ+α)cotα=(m+n)/(mn)
[Hints: Express sin(θ+2α)/sinθ=m/n and apply componendo and dividend]

Answer:

Given that  msinθ=nsin(θ+2α)   sin(θ+2α)sinθ=mn    sin(θ+2α)+sinθsin(θ+2α)sinθ=m+nmn   


2sin(θ+2α+θ2)cos(θ+2αθ2)2cos(θ+2α+θ2)sin(θ+2αθ2)=m+nmn


tan(θ+α).cotα=m+nmn


Question:4

If cos(α+β)=45 and sin(αβ)=513 where α lie between 0 and π/4, find value of tan 2α
[Hint: Express tan2αastan(α+β+αβ]

Answer:

cos(α+β)=45~ so,tan(α+β)=34     ~~~~ Andsin(αβ)=513  tan(αβ)=512 tan2α=tan[α+β+αβ]  
=tan[(α+β)+(αβ)] =tan(α+β)+tan(αβ)1tan(α+β)tan(αβ)


= 34+512134512=5633 


Question:5

If tanx=b/a then find the value of a+bab+aba+b

Answer:

tanx=ba     a+bab+aba+b=a+b+ab(ab)(a+b)=2aa2b2=2aa1b2a2


=21tan2x=21(sin2x)/(cos2x)=2cosx/(cos2x)


Question:6

Prove that cosθcosθ2cos3θcos9θ2=sin7θsin4θ
[Hint:Express L.H.S. =12[2cosθcosθ/22cos3θcos9θ/2]

Answer:

L.H.S=cosθcos(θ2)cos3θcos(9θ2)  =12[2cosθcos(θ2)]12[2cos3θcos(9θ2)]=12[cos(θ+θ2)+cos(θθ2)]12[cos(3θ+9θ2)+cos(3θ9θ2)]

=12[cos3θ2+cosθ2cos15θ2cos(3θ2)]

=12[cos3θ2+cosθ2cos15θ2cos3θ2] 

=12[cosθ2cos15θ2] =12[2sin(θ2+15θ22)sin(θ215θ22)]

=12[2sin4θsin(7θ2)]
=sin(4θ)sin(7θ2)=sin(4θ)sin(7θ2)


Question:7

If acosθ+bsinθ=mandasinθbcosθ=n, then show that m2+n2=a2+b2

Answer:

a cosθ+b sinθ=m  asinθbcosθ=n R.H.S=m2+n2=(acosθ+bsinθ)2+(asinθbcosθ)2 =a2cos2θ+b2sin2θ+2absinθcosθ+a2sin2θ+b2cos2θ2absinθcosθ=a2(cos2θ+sin2θ)+b2(cos2θ+sin2θ)=a2+b2 


Question:8

Find the value of tan2230.


 [Hint: Let θ=45 , use tanθ2=sinθ2cosθ2=2sinθ2cosθ22cos2θ2=sinθ1+cosθ]

Answer:

Let 22030=θ2   tan22030=tanθ2 =sinθ2cosθ2=2sinθ2cosθ22cos2θ2=sinθ1+cosθ     Puttingθ=45sinθ1+cosθ=121+12=12+1=(2 1)[Onrationalising]


Question:9

Prove that sin4A=4sinAcos\textsuperscript3A4cosAsin\textsuperscript3A .

Answer:

L.H.Ssin4A=sin(A+3A) =sinAcos3A+cosAsin3A =sinA(4cos3A3cosA)+cosA(3sinA4sin3A)=4sinAcos3A3sinAcosA+3sinAcosA4cosAsin3A=4sinAcos3A4cosAsin3A=R.H.S


Question:10

If tanθ+sinθ=m and tanθsinθ=n, then prove that m\textsuperscript2n\textsuperscript2=4sinθtanθ
[Hint: m+n=2tanθ,mn=2sinθ, then use [m\textsuperscript2n2]=(m+n)(mn)]

Answer:

tanθ+sinθ=m andtanθsinθ=nL.H.S=m2n2=(m+n)(mn)=[(tanθ+sinθ)+(tanθsinθ)]. [(tanθ+sinθ)(tanθsinθ)]=(tanθ+sinθ+tanθsinθ).(tanθ+sinθtanθ+sinθ)=2tanθ.2sinθ=4sinθtanθ=R.H.S


Question:11

If tan(A+B)=p,tan(AB)=q, then show that tan2A=(p+q)/(1pq).
[Hint: Use 2A=(A+B)+(AB)]

Answer:

tan(A+B)=p,tan(AB)=q    tan2A=tan(A+B+AB)=tan[(A+B)+(AB)]=tan(A+B)+tan(AB)1tan(A+B).tan(AB)=p+q1pq


Question:12

If cosα+cosβ=0=sinα+sinβ, then prove that cos2α+cos2β=2cos(α+β). [Hint: (cosα+cosβ)\textsuperscript2(sinα+sinβ)\textsuperscript2=0]

Answer:

Given that: cosα+cosβ=0  andsinα+sinβ=0


 So,(cosα+cosβ)2(sinα+sinβ)2=0  (cos2α+cos2β+2cosαcosβ)(sin2α+sin2β+2sinαsinβ)=0 (cos2αsin2α)+(cos2βsin2β)+2(cosαcosβsinαsinβ)=0cos2α+cos2β+2cos(α+β)=0cos2α+cos2β=2cos(α+β)


Question:13

If sin(x+y)sin(xy)=a+bab , then show that tanxtany=ab [Hint: Use componendo and Dividendo]

Answer:

sin(x+y)sin(xy)=a+bab      sin(x+y)+sin(xy)sin(xy)sin(xy)=a+b+aba+ba+b

2sinx+y+xy2cosx+yx+y22cosx+y+xy2sinx+yx+y2=2a2b

sinxcosycosxsiny=ab

tanxtany=ab


Question:14

If tanθ=sinαcosαsinα+cosα then show that sinα+cosα=2cosθ
[Hint: Express tanθ=tan(απ/4), θ=απ/4]

Answer:

tanθ=sinαcosαsinα+cosαtanθ=sinαcosαcosαsinα+cosαcosα=tanα1tanα+1=(tanαtanπ4)/(1+tanπ4tanα)    tanθ=tan(απ4)   


θ=απ4

cosθ=cos(απ4)

cosθ=cosαcosπ4+sinαsinπ4  cosθ=cosα.12+sinα.12  

 2cosθ=cosα+sinα

Hence, proved


Question:15

If sinθ+cosθ=1 , then find the general value of θ

Answer:

Given that sinθ+cosθ=1Dividing both sides by12+12=2


  12sinθ+12 cosθ=12  cos(θπ4)=cosπ4    θπ4=2nπ±π4 , nϵZ

θ=2nπ±π4+π4   θ=2nπ+π4+π4   orθ=2nππ4+π4    θ=2nπ+π2 orθ=2nπ,nϵZ


Question:16

Find the most general value of θ satisfying the equation tanθ=1andcosθ=1/2

Answer:

Given that

tanθ=1andcosθ=12  cosθ is positive and tanθis negative in fourth quadranttanθ=1  tanθ=tan(π4)  tanθ=tan(2ππ4)

 tanθ=tan7π4 θ=7π4  cosθ=12  cosθ=cosπ4     

   cosθ=cos(2ππ4) cosθ=cos7π4      θ=7π4    So, general solution is θ=2nπ+7π4


Question:17

If cotθ+tanθ=2cosecθ, then find the general value of θ.

Answer:

Given thatcotθ+tanθ=2cosecθ   

  cosθsinθ+sinθcosθ=2sinθ      sin2θ+cos2θsinθcosθ=2sinθ           1sinθcosθ=2sinθ        2sinθcosθ=sinθ
 sinθ(2cosθ1)=0 sinθ=0 or2cosθ1=0orcosθ=12        Now, sinθ=0θ=nπ,nϵZ cosθ=12cosθ=cosπ3   θ=2nπ±π3 
Hence, general values of θ is 2nπ±π3 andnπ,nϵZ


Question:18

If 2sin\textsuperscript2θ=3cosθ,where0θ2π , then find the value of θ.

Answer:

Given that 2sin2θ=3cosθ  2(1cos2θ)=3 cosθ22cos2θ3 cosθ=0 2cos2θ+3cosθ2=0 

(cosθ+2)(2cosθ1)=0   [On factorising]   

cosθ+2=0or2cosθ1=0

cosθ2 

So,2cosθ1=0,cosθ=12 
orθ=π3or 2ππ3θ=π3or  5π3


Question:19

If secxcos5x+1=0,where0<xπ/2, then find the value of x.

Answer:

Given that  secxcos5x+1=0 1cosx .cos5x+1=0cos5x+cosx=0~ 2cos(5x+x2)cos(5xx2)=0
 cos3x.cos2x=0  cos3x=0 orcos2x=0 3x=π2 or2x=π2    x=π6orx=π4   


Question:20

If sin(θ+α)=a andsin(θ+β)=b, then prove that cos2(αβ)4abcos(αβ)=12a\textsuperscript22b\textsuperscript2

Answer:

Given that sin(θ+α)=a andsin(θ+β)=b (i)cos(αβ)=cos[θ+αθβ]=cos[(θ+α)(θ+β)]


=cos(θ+α)cos(θ+β)+sin(θ+α)sin(θ+β)=1sin2(θ+α).1sin2(θ+β)+sin(θ+α)sin(θ+β) =(1a2)(1b2)+ab  cos(αβ)=ab+1a2b2+a2b2 

 Now,cos2(αβ)4abcos(αβ) \)=2cos2(αβ)14abcos(αβ)=2[ab+1a2b2+a2b2]214ab[ab+1a2b2+a2b2] 

=2[a2b2+1a2b2+a2b2+2ab1a2b2+a2b2]14a2b24ab1a2b2+a2b2  =12a22b2


Question:21

If cos(θ+ϕ)=mcos(θϕ), then prove that tanθ=((1m)/(1+m))cotϕ
[Hint: Express cos(θ+ϕ)/cos(θϕ)=m/1 and apply Componendo and Dividendo]

Answer:

Given thatcos(θ+ϕ)=mcos(θϕ)     cos(ϕ+θ)cos(θϕ)=m1

~~Using~componendo~and dividend rule, cos(ϕ+θ)+cos(θϕ)cos(ϕ+θ)cos(θϕ)=m+1m1 
  cosθcosϕsinθsinϕ=m+1m1  cotθcotϕ=m+1m1  cotϕtanθ=m+1m1  tanθ(1+m)=(1m)cotϕ 

 tanθ=1m1+mcotϕ


Question:22

Find the value of the expression
=3[sin4(3π2α)+sin4(3π+α)]2[sin6(π2+α)+sin6(5πα)]

Answer:

=3[sin4(3π2α)+sin4(3π+α)]2[sin6(π2+α)+sin6(5πα)]=3[cos4α+sin4(π+α)]2[cos6α+sin6(πα)] =3[cos4α+sin4α]2[cos6α+sin6α]=3[cos4α+sin4α+2sin2αcos2α2sin2αcos2α]2[(cos2α+sin2α)33cos2αsin2α(cos2α+sin2α)]

=3[(cos2α+sin2α)22sin2αcos2α]2[13cos2αsin2α]=3[12sin2αcos2α]2[13cos2αsin2α]=36sin2αcos2α2+6cos2αsin2α=32=1


Question:23

If acos2θ+bsin2θ=c has α and β as its roots, then prove that tanα+tanβ=2b/(a+c)[Hint: Use the identities cos2θ=((1tan\textsuperscript2θ)/(1+tan2θ)andsin2θ=2tanθ/(1+tan2θ)]

Answer:

Given that acos2θ+bsin2θ=c(i)    aatan2θ+2btanθ=c(1+tan2θ)  aatan2θ+2btanθ=c+ctan2θ(a+c)tan2θ2btanθ+(ca)=0..(ii) 
Since αandβ are the roots of the equation i)we have tanα and tanβ are the roots of ii)
tanα+tanβ=(2b)a+c    [Sum of roots of a quadratic equation]  tanα+tanβ=2b/(a+c)


Question:24

If x=secϕtanϕandy=cosecϕ+cotϕ , then show that xy + x – y + 1 = 0. [Hint: Find xy+1 and then show tan xy=(xy+1)]

Answer:

x=secϕtanϕ   y=cosecϕ+cotϕL.H.S=xy+xy+1=(secϕtanϕ)(cosecϕ+cotϕ)+(secϕtanϕ)(cosecϕ+cotϕ)+1=(1cosecϕsinϕcosϕ)(1sinϕ+cosϕsinϕ)+(1cosϕsinϕcosϕ)(1sinϕ+cosϕsinϕ)+1

=(1sinϕcosϕ)(1+cosϕsinϕ)+1sinϕcosϕ1+cosϕsinϕ+1  

=1sinϕ+cosϕsinϕcosϕcosϕsinϕ+sinϕsin2ϕcosϕcos2ϕcosϕsinϕ+1


=1sinϕ+cosϕsinϕcosϕ+sinϕsin2ϕcosϕcos2ϕ+cosϕsinϕcosϕsinϕ=11cosϕsinϕ=0L.H.S=R.H.S


Question:25

If θ lies in the first quadrant and cos θ = 8/17, then find the value of cos(30+θ)+cos(45θ)+cos(120θ)

Answer:

Given that cosθ=817   sinθ=1(817)2=164289=28964289=±1517 Butθ lies in I quadrant and sosinθ is positive.sinθ=1517  Nowcos(300+θ)+cos(450θ)+cos(1200θ)=cos300cosθsin300sinθ+cos450cosθ+sin450sinθ+cos1200cosθ+sin1200sinθ


=32cosθ12sinθ+12cosθ+12sinθ12cosθ+32sinθ=32(cosθ+sinθ)12(sinθ+cosθ)+12(cosθ+sinθ)=(3212+12)(cosθ+sinθ)=(312+12)(817+1517)

=31+2 22317=2334(31+2)


Question:26

Find the value of the expression cos\textsuperscript4(π/8)+cos\textsuperscript4(3π/8)+cos\textsuperscript4(5π/8)+cos\textsuperscript4(7π/8).
[Hint: Simplify the expression to

2(cos4π8+cos43π8)=2[(cos2π8+cos23π8)22cos2π8cos23π8]

Answer:

cos4π8+cos43π8+cos45π8+cos47π8 =cos4π8+cos43π8+cos4(π3π8)+cos4(ππ8)


=cos4π8+cos43π8+cos43π8+cos4π8
=2cos4π8+2cos43π8=2[cos4π8+cos43π8]=2[cos4π8+cos4(π2π4)]=2[cos4π8+sin4π8]

=2[(cos2π8+sin2π8)22sin2π8cos2π8]=2[12sin2π8cos2π8]=2(sinπ4)2  =2(12)2=212=32


Question:27

Find the general solution of the equation
5cos\textsuperscript2θ+7sin2θ6=0

Answer:

5cos2θ+7sin2θ6=0 5cos2θ+7(1cos2θ)6=05cos2θ+77cos2θ6=0    2cos2θ+1=0


 2cos2θ=1cos2θ=12  cos2θ=cos2π4   1+cos2θ2=1+cosπ22 


cos2θ=cosπ2    2θ=2nπ±π2  θ=nπ±π4 


Question:28

Find the general solution of the equation sinx3sin2x+sin3x=cosx3cos2x+cos3x

Answer:

sinx3sin2x+sin3x=cosx3cos2x+cos3x  

(sin3x+sinx)3sin2x=(cos3x+cosx)3cos2x2sin(3x+x2).cos(3xx2)3sin2x=2cos(3x+x2).cos(3xx2)3cos2x2sin2xcosx3sin2x=2cos2x.cosx3cos2x  


2sin2xcosx2cos2xcosx=3sin2x3cos2x  2cosx(sin2xcos2x)=3(sin2xcos2x)  (sin2xcos2x)(2cosx3)=0 sin2xcos2x=0
 2cosx30sin2xcos2x1=0tan2x=1tan2x=tan(π/4) 2x=nπ+π4    x=nπ/2+π/8


Question:29

Find the general solution of the equation (31)cosθ+(3+1)sinθ=2
[Hint: Put Put31=rsinα,3+1=rcosα which gives tanα=tan((π/4)(π/6))α=π/12]
.

Answer:

(31)cosθ+(3+1)sinθ=2~ put 31=rsinα, 3+1=rcosα~squaring~and adding we get r2=3+123+3+1+23    r2=8,  r=22  


  rsinαcosθ+rcosαsinθ=2~ r(sinαcosθ+cosαsinθ)=2 22sin(α+θ)=2 sin(α+θ)=222 sin(θ+α)=sinπ4  


α+θ=nπ+(1)nπ4.(i) ~ Now, rsinθrcosα=313+1  tanα=tanπ3tanπ41+tanπ4tanπ3  tanα=tan(π3π4)  tanα=tanπ12


α=π12  Putting the value of αinequation(i)~we get π12+θ=nπ+(1)n.π4         θ=nπ+(1)n.π4π12 


Question:30

If sinθ+cosecθ=2, then sin\textsuperscript2θ+cosec\textsuperscript2θ is equal to
A. 1
B. 4
C. 2
D. None of these

Answer:

The answer is the option (c).

sinθ+cosecθ=2 (sinθ+cosecθ)2=22   sin2θ+cosec2θ+2sinθcosecθ=4  sin2θ+cosec2θ+2sinθcosecθ=4  


   sin2θ+cosec2 θ+2=4   sin2θ+cosec2θ=2


Question:31

If f(x)=cos\textsuperscript2x+sec\textsuperscript2x, then

A. f(x)<1

B. f(x)=1

C. 2<f(x)<1

D. f(x)2

[Hint: A.MG.M.]

Answer:

The answer is the option (d)
f(x)=cos2x+sec2x  We know thatAMGM(cos2x+sec2x)2cos2xsec2 x      (cos2x+sec2x)21   cos2x+sec2x2 ~ f(x)2


Question:32

If tanθ=1/2and tanϕ=1/3, then the value of θ+ϕ is

A. π/6

B. π

C.0

D. π/4

Answer:

The answer is the option (d)
tan(θ+ϕ)=tanθ+tanϕ1tanθtanϕ=12+1311213 =5656=1tan(ϕ+θ)=tanπ4   (θ+ϕ)=π4 


Question:33

Which of the following is not correct?
A. sinθ=1/5
B.cosθ=1
C. secθ=1/2
D. tanθ=20

Answer:

The answer is the option (c)
sinθ=15 is correct since 1sinθ1
cosθ=1is true forθ=1 secθ=12    cosθ=2is not correct as 1cosθ1


Question:34

The value of tan1tan2tan3tan89is
A. 0
B. 1
C. 1/2
D. Not defined

Answer:

The answer is the option (b).
Given that tan10 tan20.tan890  =tan10 tan20tan450tan(90440)tan(90430).tan(9010)=tan10cot10tan20cot20.tan890cot890=1.111.1=1


Question:35

The value of (1tan\textsuperscript215\textsuperscripto)/(1+tan\textsuperscript215\textsuperscripto) is
A. 1

B. 3

C. 3/2
D. 2

Answer:

The answer is the option (c).
Given that  1tan21501+tan2150    Letθ=150 2θ=300   cos2θ=1tan2θ1+tan2θ   cos300=1tan21501+tan2150=32


Question:36

The value of cos1cos2cos3cos179is
A. 1/2
B. 0
C. 1
D. -1

Answer:

The answer is the option (b).
cos10 cos20.cos1790=cos10 cos20.cos900.cos1790\]=0(as  cos900=0)


Question:37

If tanθ=3 and θ lies in third quadrant, then the value ofsinθis

A. 1/10B.1/10C.3/10D.3/10

Answer:

The answer is the option (c).
tanθ=3, θ lies in third quadrant, it is positive  tanθ=PB=31  Then,hypotenuse=32+12=9+1=10   sinθ=310 whereθ lies in third quadrant


Question:38

The value of tan75cot75 is equal to
A. 23
B.2+3
${C. 2 - \sqrt 3 \\$
D. 1

Answer:

The answer is the option (a).
tan75cot75=tan75cot(9015)=tan75tan15=sin75cos75sin15cos15=(sin75cos15sin15cos75)cos75cos15=sin(7515)12×2cos75cos15=2sin60cos(75+15)+cos(7515)=2sin60cos90+cos60=2×320+12=23


Question:39

Which of the following is correct?

A. sin1>sin1

B. sin1<sin1

C. sin1=sin1

D.sin1=π18sin1

[Hint: 1 radian = 180 ^{\circ} \pi = 57 ^{\circ} 30'approx.]

Answer:

The answer is the option (b).
If  θ increases then the value of sinθ also increases. \\

So, sin1<sin1
Hence, b is correct.


Question:40

If \tan \alpha=\frac{\mathrm{m}}{\mathrm{m}+1}, \tan \beta=\frac{1}{2 \mathrm{m}+1} then \alpha + \beta is equal to
A.π2B.π3C.πcD.π4

Answer:

The answer is the option (d).
tanα=mm+1tanβ=m2m+1tan(α+β)=tanα+tanβ1tanαtanβ=mm+1+12m+11mm+1×12m+1=2m2+m+m+1(m+1)(2m+1)(m+1)(2m+1)m(m+1)(2m+1)=2m2+2m+12m2+2m+m+1m=2m2+2m+12m2+2m+1=1tan(α+β)=tanπ4α+β=π4


Question:41

The minimum value of 3cosx+4sinx+8 is
A. 5
B. 9
C. 7
D. 3

Answer:

The answer is the option (d).

Let y=3cosx+4sinx+8y8=3cosx+4sinxMinimum value of y8=(3)2+(4)2=5y=85=3

Hence, (d) is the correct option.


Question:42

The value of tan3Atan2AtanA is equal to
A. tan3Atan2AtanA
B. tan3Atan2AtanA
C. tanAtan2Atan2Atan3Atan3AtanA
D. None of these

Answer:

The answer is the option (a).

tan3A=tan(2A+A)=tan2A+tanA1tan2AtanAtan3A(1tan2AtanA)=tan2A+tanAtan3Atan3Atan2AtanA=tan2A+tanAtan3Atan2AtanA=tan3Atan2AtanA
Hence, a is correct.


Question:43

The value of sin(45+θ)cos(45θ) is
A. 2cosθ
B. 2sinθ
C. 1
D. 0

Answer:

The answer is the option (d).
sin(45+θ)cos(45θ)sin(45+θ)=sin45cosθ+cos45sinθ=12cosθ+12sinθcos(45θ)=cos45cosθ+sin45sinθ=12cosθ+12sinθsin(45+θ)cos(45θ)=12cosθ+12sinθ12cosθ12sinθ=0


Question:44

The value of cot(π4+θ)cot(π4θ) is
A. –1
B. 0
C. 1
D. Not defined

Answer:

The answer is the option (c).

cot(π4+θ)cot(π4θ)=cotπ4cotθ1cotθ+cotπ4×cotπ4cotθ+1cotθcotπ4=cotθ1cotθ+1×cotθ+1cotθ1=1
(c) is correct.


Question:45

cos2θcos2ϕ+sin2(θϕ)sin2(θ+ϕ) is equal to
A. sin2(θ+ϕ)
B. cos2(θ+ϕ)
C. sin2(θϕ)
D.cos2(θϕ)
[Hint: Use sin2Asin2B=sin(A+B)sin(AB)]

Answer:

The answer is the option (b).
cos2θcos2+sin2(θ)+sin2(θ+)since, sin2Asin2B=sin(A+B)sin(AB)=cos2θcos2+sin(θ+θ+)sin(θθ)=cos2θcos2sin2θsin2since, cosxcosysinxsiny=cos(x+y)=cos(2θ+2)=cos2(θ+)
Hence, the correct option is (b).


Question:46

The value of cos12+cos84+cos156+cos132 is

A. 12

B. 1
C. 12
D. 18

Answer:

The answer is the option (c)
cos12+cos84+cos156+cos132=(cos132+cos12)+(cos156+cos84)=2cos132+122cos132122+2cos156+842cos156842=2cos72cos60+2cos120cos36=2cos72×12+2×(12)cos36=cos72cos36=5145+14=24=12


Question:47

If tanA=12,tanB=13 then tan(2A+B) is equal to
A. 1
B. 2
C. 3
D. 4

Answer:

The answer is the option (c).
tanA=12tanB=13tan2A=2tanA1tan2A=43tan(2A+B)=tan2A+tanB1tan2AtanB=43+13143×13=(53)×(95)=3


Question:48

The value of sinπ10sin13π10 is
A.12
B.12
C.14
D.1

Answer:

The answer is the option (c).
sinπ10sin13π10=sinπ10sin(π+3π10)=sinπ10(sin3π10)=sin18sin54=(514)(5+14)=5116=416=14
Hence, (c) is correct option.


Question:49

The value of sin50sin70+sin10 is equal to
A. 1
B. 0
C.1/2
D. 2

Answer:

The answer is the option (b).

sin50sin70+sin10=2cos50+702sin50702+sin10=2cos60sin(10)+sin10=2×12×sin10+sin10=0


Question:50

If sinθ+cosθ=1, then the value of sin2θ is equal to
A. 1
B. 1/2
C. 0
D. –1

Answer:

The answer is the option (c).
sinθ+cosθ=1(sinθ+cosθ)2=1sin2θ+cos2θ+2sinθcosθ=1   1+2sinθcosθ=12sinθcosθ=0sin2θ=0


Question:51

If α+β=π4 then the value of (1+tanα)(1+tanβ) is
A. 1
B. 2
C. –2
D. Not defined


Answer:

tan(α+β)=tanπ4=1tan(α+β)=tanα+tanβ1tanαtanβ=1tanα+tanβ=1tanαtanβtanα+tanβ+tanαtanβ=11+tanα+tanβ+tanαtanβ=1+1(1+tanα)(1+tanβ)=2
Hence, correct option is (b).


Question:52

If sinθ=45 and θ lies in third quadrant then the value of cosθ2 is
A.15
B.110
C.15
D.110

Answer:

The answer is the option (c)
sinθ=45,θlies in third quadrant
cosθ=1(45)2=35cosθ=2cos2θ2135=2cos2θ21cos2θ2=15cosθ2=15 
[As, π2<θ2<3π4]
Hence, correct option is (c).


Question:53

Number of solutions of the equation tanx+secx=2cosx lying in the interval [0,2π] is
A. 0
B. 1
C. 2
D. 3

Answer:

The answer is the option (c).
tanx+secx=2cosxsinx+1cosx=2cosx1+sinx2cos2x=01+sinx2+2sin2x=02sin2x+sinx1=0
Since the equation is a quadratic equation in sinx. So, there will be two solutions.
Hence, correct option is (c).


Question:54

The value of sinπ18+sinπ9+sin2π9+sin5π18 is given by
\\A. \sin \frac{7 \pi}{18}+\sin \frac{4 \pi}{9}\\\\ B .1\\\\ C \cdot \cos \frac{\pi}{6}+\cos \frac{3 \pi}{7}\\\\ D \cdot \cos \frac{\pi}{9}+\sin \frac{\pi}{9}

Answer:

The answer is the option (a).
sinπ18+sinπ9+sin2π9+sin5π18=(sinπ18+sin5π18)+(sinπ9+sin2π9)=2sin5π18+π182cos5π18π182+2sinπ9+2π92cos2π9π92=2sinπ6cosπ9+2sinπ6cosπ18=2×12cosπ9+2×12cosπ18=cosπ9+cosπ18=sin(π2π9)+sin(π2π18)=sin4π9+sin7π18
Hence, correct option is (a).


Question:55

If A lies in the second quadrant and 3tanA+4=0, then the value of 2cotA5cosA+sinA is equal to

A. 5310
B. 2310
C. 3710
D. 710

Answer:

The answer is the option (b).

3tanA+4=0 [A lies in second quadrant]


tanA=43

cosA=35 [A lies in second quadrant]


sinA=45cotA=342cotA5cosA+sinA=2(34)5(35)+45=32+3+45=2310


Hence, correct option is (b).


Question:56

The value of cos248sin212 is

A.5+18B.518C.5+15D.5+122
[Hint:Usecos2Asin2B=cos(A+B)cos(AB)]

Answer:

The answer is the option (a).
cos248sin212=cos(48+12)cos(4812)=cos60cos36=12×5+14=5+18

Hence, correct option is (a).


Question:57

If tanα=17,tanβ=13then cos2α is equal to
A. sin2β
B. sin4β
C. sin2β
D. cos2β

Answer:

The answer is the option (b).
tanα=17tanβ=13cos2α=1tan2α1+tan2α=11491+149=2425tan2β=2tanβ1tan2β=2×13119=34
sin4β=2tan2β1+tan22β=2×341+(34)2=2425cos2α=sin4β=2425
Hence, correct option is (b).


Question:58

If tanθ=ab then bcos2θ+asin2θ is equal to

A.aB.bC.abD.None

Answer:

The answer is the option (b).
tanθ=abbcos2θ+asin2θ=b[1tan2θ1+tan2θ]+a[2tanθ1+tan2θ]=b[1a2b21+a2b2]+a[2ab1+a2b2]=b[b2a2b2+a2]+[2a2bb2+a2b2]
=b3a2bb2+a2+2a2bb2+a2=b(b2+a2)b2+a2=b
Hence, correct option is (b).


Question:59

If for real values of x, cosθ=x+1x then
A. θ is an acute angle
B. θ is right angle
C. θ is an obtuse angle
D. No value of θ is possible

Answer:

The answer is the option (d).
cosθ=x+1x=x2+1xx2xcosθ+1=0For real value of x, (b)24×a×c0

(cosθ)24×1×10cos2θ4cosθ±2
Hence, correct option is (d).


Question:60

The value of sin50sin130 is _______.


Answer:

sin50sin130=sin50sin(18050)=sin50sin50=1


Question:61

Fill in the blanks
If k=sin(π18)sin(5π18)sin(7π18)then the numerical value of k is

Answer:

k=sin(π18)sin(5π18)sin(7π18)k=sin10sin50sin70k=sin10cos40cos20k=sin1012×[2cos40cos20]=sin1012[cos60+cos20]k=12sin10[12+cos20]k=14sin10+12sin10cos20k=14sin10+14[sin30sin10]k=14sin30=14×12=18


Question:62

Fill in the blanks

If tanA=1cosBsinB then tan2A=......

Answer:

tanA=1cosBsinBtan2A=2tanA1+tan2A=2(1cosB)sinB1+(1cosB)2sin2B=2(2sin2B22sinB2cosB2)1(2sin2B22sinB2cosB2 )2=2(sinB2cosB2)1(sinB2cosB2)2=2tanB21tan2B2=tanB


Question:63

Fill in the blanks
If sinx+cosx=a, then
(i) sin6x+cos6x=_______
(ii) |sinxcosx|=______.

Answer:

Giventhatsinx+cosx=aOn squaring both sides,(sinx+cosx)2=a21+2sinxcosx=a2sinxcosx=a212sin6x+cos6x=(sin2x)3+(cos2x)3=(sin2x+cos2x)33sin2xcos2x(sin2x+cos2x)=13(a212)2=13(a21)24=14[43(a21)2]|sinxcosx|2=sin2x+cos2x2sinxcosx=12(a212)=1a2+1=2a2|sinxcosx|=2a2


Question:64

Fill in the blanks

In a triangle ABC with C=90 the equation whose roots are tan A and tan B is ________.
[Hint: A+B=90tanAtanB=1andtanA+tanB=2sin2A]

Answer:

x2(tanA+tanB)x+tanAtanB=0tan(A+B)=tan90tanA+tanB1tanAtanB=101tanAtanB=0tanAtanB=1tanA+tanB=sinAcosA+sinBcosB=sinAcosB+cosAsinBcosAcosB=sin(A+B)cosAcosB=1cosAsinAtanA+tanB=22sinAcosA=2sin2A x2(2sin2A)x+1=0


Question:65

Fill in the blanks
3(sinxcosx)4+6(sinx+cosx)2+4(sin6x+cos6x)=

Answer:

3(sinxcosx)4+6(sinx+cosx)2+4(sin6x+cos6x)=3(sin2x+cos2x2sinxcosx)2+6(sin2x+cos2x+2sinxcosx)+4[(sin2x)3+(cos2x)3]=3(12sinxcosx)2+6+12sinxcosx+4[(sin2x+cos2x)33sin2xcos2x(sin2x+cos2x)]=3(1+4sin2xcos2x4sinxcosx)+6+12sinxcosx+412sin2xcos2x=3+12sin2xcos2x12sinxcosx+6+12sinxcosx+412sin2xcos2x=3+6+4=13


Question:66

Fill in the blanks
Given x > 0, the values of f(x)=3cos3+x+x2 lie in the interval _______.

Answer:

Giventhatf(x)=3cos3+x+x2Putting,y=3+x+x2f(x)=3cosy1cosy1\vspace\baselineskip33cos3+x+x23Hence the value is in [3,3]


Question:67

Fill in the blanks
The maximum distance of a point on the graph of the function y=3sinx+cosx from x-axis is _____.

Answer:

y=3sinx+cosx.(i)
The maximum distance from a point on the graph of equation (i) from x - axis
(3)2+(1)2=3+1=2


Question:68

True and False

If tanA=1cosBsinB then tan2A=tanB

Answer:

tanA=1cosBsinB=2sin2B22sinB2cosB2=tanB2tan2A=tanB

Hence, the statement is true.


Question:69

True and False
The equality sinA+sin2A+sin3A=3 holds for some real value of A.

Answer:

Given that sinA+sin2A+sin3A=3
Since the maximum value of sin A is 1 but for sin 2A and sin 3A it is not equal to 1. So, it is not possible.
Hence, the statement is ’false’.


Question:70

True and False
sin10 is greater than cos10

Answer:

If sin10>cos10Then,sin10>cos(9080)sin10>sin80
which is not possible because value of sine is in increasing order
Hence, the statement is ‘false’


Question:71

True and False
cos2π15cos4π15cos8π15cos16π15=116

Answer:

cos2π15cos4π15cos8π15cos16π15=cos24cos48cos96cos192=116sin24(2sin24cos24)(2cos48)(2cos96)(2cos192)=116sin24(2sin48cos48)(2cos96)(2cos192)=116sin24(2cos96sin96)(2cos192)=2sin192cos19216sin24=sin38416sin24=sin(360+24)16sin24=sin2416sin24=116
Hence, the statement is ‘true’.


Question:72

True and False
One value of θ which satisfies the equation sin4θ2sin2θ1 lies between 0 and 2π


Answer:

Given equation is sin4θ2sin2θ1=0

sin2θ=(2)±(2)24×1×(1)2×1=2±4+42=2±82=2±222=1±2
1sinθ1 andsin2θ1 butsin2θ=1±2
which is not possible
Hence, the given statement is ‘false’


Question:73

True and False
If cosecx=1+cotx then x=2nπ,2nπ+π2

Answer:
cosecx=1+cotx x=2nπ,2nπ+π21sinx=1+cosxsinxsinx+cosx=1
12sinx+12cosx=12cos(xπ4)=cosπ4x=2nπ+π4+π4=2nπ+π2Or,x=2nπ+π4π4=2nπ
Hence, the given statement is ‘true’


Question:74

True and False

tanθ+tan2θ+3tanθtan2θ=3 then θ=nπ3+π9

Answer:

tanθ+tan2θ=3tanθtan2θ+3tanθ+tan2θ=3(1tanθtan2θ)tanθ+tan2θ1tanθtan2θ=3tan3θ=3tan3θ=tanπ33θ=nπ+π3θ=nπ3+π9
Hence, the given statement is ‘true’ .


Question:75

True and False
If tan(πcosθ)=cot(πsinθ), then cos(θπ/4)=±122

Answer:

tan(πcosθ)=cot(πsinθ)tan(πcosθ)=tan(π2πsinθ)πcosθ=π2πsinθcosθ+sinθ=12 cosπ4 cosθ+sinπ4sinθ=12\vspace\baselineskipcos(θπ/4)=±122
Hence, the given statement is ‘true’


Question:76

In the following match each item given under the column C1 to its correct answer given under the column C2:

Answer:

sin(x+y)sin(xy)=sin2xsin2ycos(x+y)cos(xy)=cos2xcos2ycot(π4+θ)=cotπ4cotθ1cotθ+cotπ4=cotθ1cotθ+1=1tanθ1+tanθtan(π4+θ)=tanπ4tanθ1tanθ+tanπ4=1+tanθ1tanθ
Thus, (a) - (iv) , (b) - (i), (c) -(ii), (d) - (iii)

Important Notes of NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Trigonometry is an ancient concept which in ancient times was used to solve problems relating to triangles and geometry, but it has extended its use to various different fields in the present times including varied areas of studies. It was derived from Greek words meaning, “measuring the sides of a triangle” which has widened its scope to much more than the original meaning. It is basically used to measure length, height and angles of different triangles with its reach in real-life practical situations. NCERT Exemplar solutions for Class 11 Maths chapter 3 extends studying trigonometric ratios to any angle regarding or concerning radian measure and interpreting and representing it as a trigonometric ratio with the help of diagrams for a better understanding.

Students can make use of NCERT Exemplar Class 11 Maths solutions chapter 3 pdf download for further learning.

Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 3

The topics covered in the chapter are as follows:

  • 3.1 Introduction
  • 3.2 Angles
  • 3.2.1 Degree measure
  • 3.2.2 Radian measure
  • 3.2.3 Relation between radian and real numbers
  • 3.2.4 Relation between degree and radian
  • 3.3 Trigonometric Functions
  • 3.3.1Sign of trigonometric functions
  • 3.3.2 Domain and range of trigonometric functions
  • 3.4 Trigonometric Functions of Sum and Difference of Two Angles
  • 3.5 Trigonometric Equations
Background wave

What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 3?

The students will learn a variety of concepts from NCERT Exemplar solutions for Class 11 Maths chapter 3 which has a wide range of application in different fields such as engineering, sound engineers, architects, astronauts, surveyors, physicist, and much more for future references. NCERT Exemplar Class 11 Maths solutions chapter 3 is also given importance since it has various applications in real life and could be connected to routine activities that happen around us. It is even used in the gaming industry, IT sector, construction of bridges, buildings, mountains, the inclination of floors, roofs, marine biology, criminal investigations, wide use in physics for derivations and explanations, and much more. The uses of trigonometry in these many fields justify its use for inexhaustible purposes and its importance for students belonging or deciding to enter any field or subject in the future.

NCERT Solutions for Class 11 Mathematics Chapters

Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

· NCERT Exemplar Class 11 Maths chapter 3 solutions give explanation, and interpretation of different trigonometric functions for sin x, cos x, sec x, cot x, cosec x, and tan x along with values of trigonometric ratios for 0º, 30º, 45º, 60º, 90º, 180º, 270º and 360º along with the sign convention of these trigonometric functions.

· Class 11 Maths NCERT Exemplar solutions chapter 3 also covers the range and domain of trigonometric functions with the help of diagrammatic representation of the same. This chapter also extends to the trigonometric functions of sum and difference of two angles with a variety of questions and illustrations to be done for the same.

· NCERT Exemplar class 11 Maths solutions chapter 3 concludes with insight on trigonometric equations that involve equations containing trigonometric functions of any variable.

Check Chapter-Wise NCERT Solutions of Book

NCERT Exemplar Class 11 Solutions

Read more NCERT Solution subject wise -

Also, read NCERT Notes subject wise -

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. Who can use these solutions of the NCERT chapter?

Those who are prepping for their board exams and for those who are planning to appear in their JEE Main exam or any other engineering entrance exam.

2. What all topics are covered in this chapter?

This chapter covers everything related to trigonometric functions, their properties, angles, trigonometric equations, etc.

3. Who has prepared these NCERT exemplar Class 11 maths solutions chapter 3?

These solutions are prepared by other experienced maths teachers and team so as to include every detail of the solution with no mistake.

4. How to make use of these solutions?

Students can practice the questions, and while doing so, these Class 11 Maths NCERT exemplar solutions chapter 3 will act as a reference while studying.

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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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