NCERT Solutions For Class 11 Maths Chapter 3

NCERT Solutions Class 11 Maths Trigonometric Functions

For students of class 11 who are looking to do their best their class 11 final exam it is important for them to practice their knowledge on questions of NCERT Solutions for Class 11 Maths Chapter 3. Hence we have provided the downloadable form of NCERT Solutions for Class 11 Maths Chapter 3 pdf for ease of access. NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions handles questions and solutions of topics such as measuring angles in radians and operations on them, definition of trigonometric functions with the help of a circle, signs of trigonometric functions, domain and range of trigonometric functions and more.

NCERT Solutions Class 11 Maths Chapter 3 Exercises

Exercise 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(i).  25

(ii).  240

(iii).  4730'

(iv).  520

 

Sol:

(i).  25

As we know that, 180 = π radian

Therefore, 25 = π180×25 radian = 5π36 radian

Hence, 25 = 5π36 radian

 

(ii).  240

As we know that 180 = π radian

Therefore, 240 = π180×240 radian = 4π3 radian

Hence, 240 = 4π3 radian

 

(iii).  4730'

= 4712

= 952 degree

As we know that, 180 = π radian

Therefore, 952 degree = π180×952 radian = 1936×2π radian = 1972π

Hence, 4730' = 1972π

 

(iv).  520

As we know that, 180 = n radian

Therefore, 520 = π180×520 radian = 26π9 radian

Hence, 520 = 26π9 radian

 

 

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = 227]

(i) 1116

 

(ii) -4

 

(iii) 5π3

 

(iv) 7π6

 

Sol:

(i).  1116:

As we know, that Π Radian = 180°

Therefore, 1116 radian = 180π×1116 degree

= 45×11π×4 degree

= 45×11×722×4 degree

= 3158 degree

= 3938 degree

= 39+3×608 minute  [1 = 60’]

= 39+22'+12 minute

= 39+22'+30  [1’ = 60’’]

 

(ii).  -4:

As we know, that Π Radian = 180°

Therefore, -4 radian = 180π×(4) degree

= 180×7(4)22 degree

= 252011 degree

= 229111 degree

= 229+1×6011 minutes     [1 = 60’]

= 229+5'+511

= 229+5'+27 [1’ = 60’’]

 

(iii).  5π3

As we know, that Π Radian = 180°

Therefore, 5π3 radian = 180π×5π3 degree

= 300

 

(iv).  7π6

As we know, that Π Radian = 180°

Therefore, 7π6 radian = 180π×7π6

= 210

 

 

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Answer:

No. of revolutions made in a minute = 360 revolutions

Therefore, no. of revolutions made in a second = 36060 = 6

In one revolution, the wheel rotates an angle of 2π radian.

Therefore, in 6 revolutions, it will turn an angle of 12π radian.

 

 

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

[Use π = 227]

Answer:

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ=lr

Given:

r = 100 m and L = 22 m

We have,

θ=22100 radian

= 180π×22100 degree

= 180×7×2222×100 degree

= 12610 degree

= 1235 degree

= 1236' [1 = 60’]

Therefore, the req angle is 1236'

 

 

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Answer:

13

Diameter of circle = 40 m

Radius of circle = 4020 m = 20 m

Let XY be the chord (length = 20 m) of the circle.

In ΔOXY, OX = OY  = radius of the circle = 20 m

XY = 20 m

Therefore,

ΔOXY is an equilateral triangle.

θ=60 = π3 radian

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ = lr

π3 = arc(AB)20

arc(AB)20 = 20π3 m

The length of the minor arc of the chord is 20π3 m.

 

 

Q.6: In two circles, arcs which has same length subtended at an angle of 60 and 75 at the center. Calculate the ratio of their radii.

 Sol:

Let, the radii of two circles be r1 and r2.

Let, an arc of length l subtend an angle of 60 at the center of the circle of radius r1, while let an arc of length l subtend an angle of 75 at the center of the circle of radius r2.

60 = π3 radian

75 = 5π12 radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ then:

θ = lr

l = r θ

l = r1π3

l = r25π12

r1π3 = r25π12

r1 = r254

r1r2 = 54

Therefore, the ratio of radii is 5: 4

 

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(i) 10 cm

 

(ii) 15 cm

 

(iii) 21 cm

 

Sol:

As we know that, in a circle of radius ‘r’ unit, if an arc of length ‘l’ unit subtends an angle θ radian at the center, then:

θ=lr

Here, r = 75 cm

(i).  10 cm:

θ = 1075 radian = 215 radian

 

(ii).  15 cm:

θ = 1575 radian = 15 radian

 

(iii).  21 cm:

θ = 2175 radian = 725 radian

 

 

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if cosy = 12 and y lies in 3rd quadrant.

Sol:

(i)  sec y :

Since, cos y = 12

Therefore, sec y = 1cosy = 1(12)

Hence, sec y = -2

 

(ii)  sin y :

Since, sin2y+cos2y=1

Therefore, sin2y=1cos2y

sin2y=1(12)2

sin2y=114

sin2y=34

siny=±32

Since, y lies in the third quadrant, the value of sin y will be negative.

Therefore, sin y = 32

 

(iii)  cosec y = 1siny = 1(32) = 23

Therefore, cosec y = 23

 

(iv)  tan y = sinycosy = tany=(32)(12) = 3

Therefore, tan y = 3

 

(v)  cot y = 1tany = 13

Therefore, cot y = 13

 

 

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = 35, where y lies in second quadrant.

Sol:

sin y = 35

Therefore, cosec y = 1Siny = 135=53

Since, sin2y+cos2y=1

  cos2y=1sin2y

  cos2y=1(35)2

cos2y=1925

cos2y=1625

cos y = ±45

Since, y lies in the 2nd quadrant, the value of cos y will be negative,

Therefore, cos y = – 45

sec y = 1cosy=1(45)=54

tan y = sinycosy=(35)(45)=34

cot y = 1tany=43

 

 

 Q.3: Find the values of other five trigonometric functions if coty=34, where y lies in the third quadrant.

Sol:

cot y = 34

Since, tan y = 1coty=134=43

Since, 1+tan2y=sec2y

  1+(43)2=sec2y

  1+169=sec2y

  259=sec2y

  sec y = ±53

Since, y lies in the 3rd quadrant, the value of sec y will be negative.

Therefore, sec y = 53

cos y = 1secy=1(53)=35

Since, tan y = sinycosy

  43=siny(35)

sin y = (43)×(35)=45

  cosec y = 1siny=54

 

 

Q.4: Find the values of other five trigonometric if secy=135, where y lies in the fourth quadrant.

Sol:

sec y = 135

cos y = 1secy=1(135)=513

Since, sin2y+cos2y=1

   sin2y=1cos2y

   sin2y=1(513)2

   sin2y=125169=144169

sin y = ±1213

Since, y lies in the 4th quadrant, the value of sin y will be negative.

Therefore, sin y = 1213

cosec y = 1siny=1(1213)=1312

tan y = sinycosy=(1213)(513)=125

cot y = 1tany=1(125)=512

 

 

Q.5: Find the values of other five trigonometric function if tan y = 512 and y lies in second quadrant.

Sol:

tan  y = 512   [Given]

And, cot y = 1tany=1(512)=125

Since, 1+tan2y=sec2y

Therefore, 1+(512)2=sec2y

   sec2y=1+(25144)

  sec2y=169144

sec y = ±1312

Since, y lies in the 2nd quadrant, the value of sec y will be negative.

Therefore, sec  y = 1312

cos y = 1secy=1(1312)=(1213)

Since, tany=sinycosy

  512=siny(1213)

sin y = (512)×(1213)=513

cosec y = 1siny=1(513)=135

 

 

Q.6: Calculate the value of trigonometric function sin 765°.

Sol:

The values of sin y repeat after an interval of 360° or 2n.

Therefore, sin 765° = sin ( 2 × 360° + 45° ) = sin 45° = 12

 

 

Q.7: Calculate the value of trigonometric function cosec [-1410°]

 Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, cosec [-1410°] = cosec [-1410° + 4 × 360°] = cosec [ -1410° + 1440°] = cosec 30° = 2

 

 

Q.8: Calculate the value of the trigonometric function tan19π3.

Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, tan19π3 = tan613π = tan(6π+π3) = tanπ3 = tan 60° = 3

 

 

Q.9: Calculate the value of the trigonometric function sin(11π3).

 Answer:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, sin(11π3) = sin(11π3+2×2π) = sinπ3 = 32

 

 

Q.10: Calculate the value of the trigonometric function cot(15π4).

 Sol:

It is known that the values of tany repeat after an interval of 180 or n.

Therefore, cot(15π4) = cot(15π4+4π) = cotπ4 = 1

 

 

Exercise 3.3

Q.1: Prove:

sin2π6+cos2π3tan2π4=12

 

Sol:

Now, taking L.H.S.

sin2π6+cos2π3tan2π4:

= (12)2+(12)2(1)2

= 14+141 = 12

= R.H.S.

 

 

Q.2: Prove:

2sin2π6+cosec27π6cos2π3=32

 

Sol:

Now, taking L.H.S.

2sin2π6+cosec27π6cos2π3 :

= 2(12)2+cosec2(π+π6)(12)2

= 2×14+(cosecπ6)2(14)

= 12+(2)2(14)

= 12+44

= 12+1

= 32

= R.H.S.

 

 

Q.3: Prove:

cot2π6+cosec5π6+3tan2π6=6

 

Sol:

Taking L.H.S.

cot2π6+cosec5π6+3tan2π6 :

= (3)2+cosec(ππ6)+3(13)2

= 3+cosecπ6+3×13

= 3 + 2 + 1 = 6

= R.H.S.

 

 

Q.4: Prove:

2sin23π4+2cos2π4+2sec2π3=10

 

Sol:

Now, taking L.H.S.

2sin23π4+2cos2π4+2sec2π3 :

= 2{sin(ππ4)}2+2(12)2+2(2)2

= 2{sinπ4}2+2×12+8

= 2(12)2 + 1 + 8

= 1 + 1 + 8 = 10

= R.H.S.

 

 

Q.5: Calculate the value of:

(i).  sin75

 

(ii).  tan15

 

Sol:

(i).  sin75:

= sin(45+30)

= sin45cos30+cos45sin30

Since, [sin (x + y) = sin x cos y + cos x sin y]

= (12)(32)+(12)(12)

= 322+122

= 3+122

 

(ii).  tan15:

= tan(4530)

= tan45tan301+tan45tan30

Since, [tan (x – y) = tanxtany1+tanxtany]

= 1131+1(13)

= 3133+13

= 313+1

= (31)2(3+1)(31)

= 3+123(3)2(1)2

= 42331

= 23

 

 

Q.6:Prove:

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

 

Sol:

Now, taking L.H.S.

cos(π4x)cos(π4y)sin(π4x)sin(π4y):

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

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

Since, [2cos A cos B = cos (A + B) + cos (A – B)]

And, [2sin A sin B = cos (A + B) – cos (A – B)]

= 2×12[cos{(π4x)+(π4y)}]

= cos[π2(x+y)]

= sin (x + y)

= R.H.S.

 

 

Q.7: Prove:

tan(π4+x)tan(π4x)=(1+tanx