#### Page No 54:

#### Question 1:

Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°

#### Answer:

(i) 25°

We know that 180° = π radian

(ii) –47° 30′

–47° 30′ = degree [1° = 60′]

degree

Since 180° = π radian

(iii) 240°

We know that 180° = π radian

(iv) 520°

We know that 180° = π radian

#### Page No 55:

#### Question 2:

Find the degree measures corresponding to the following radian measures

.

(i) (ii) – 4 (iii) (iv)

#### Answer:

(i)

We know that π radian = 180°

(ii) – 4

We know that π radian = 180°

(iii)

We know that π radian = 180°

(iv)

We know that π radian = 180°

#### 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 =

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

Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e.,

12 π radian

Thus, in one second, the wheel turns an angle of 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.

#### Answer:

We know
that in a circle of radius *r* unit, if an arc of length *l* unit subtends an angle *θ* radian at the centre, then

Therefore, forr = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.

#### 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:

Diameter of the circle = 40 cm

∴Radius (*r*) of the circle =

Let AB be a chord (length = 20 cm) of the circle.

In ΔOAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ΔOAB is an equilateral triangle.

∴θ = 60° =

We know that in a circle of radius *r* unit, if an arc of length *l* unit subtends an angle *θ* radian at the centre, then.

Thus, the length of the minor arc of the chord is.

#### 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:

Let
the radii of the two circles be and.
Let an arc of length *l* subtend an angle of 60° at the centre of the circle of radius *r*_{1},
while let an arc of length *l *subtend
an angle of 75° at the centre of the circle of radius *r*_{2}.

Now, 60° =and 75° =

We
know that in a circle of radius *r* unit, if an arc of length *l* unit subtends an angle *θ* radian at the centre, then.

Thus, the ratio of the radii is 5:4.

#### Question 7:

Find the angle in radian though 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:

We
know that in a circle of radius *r* unit, if an arc of length *l* unit subtends an angle *θ* radian at the centre, then.

It
is given that *r* = 75 cm

(i) Here, *l* = 10 cm

(ii) Here, *l *=
15 cm

(iii) Here, *l *=
21 cm

#### Page No 63:

#### Question 1:

Find the values of other five trigonometric functions if , *x* lies in third quadrant.

#### Answer:

Since *x* lies in the 3^{rd} quadrant, the value of sin *x* will be negative.

#### Question 2:

Find the values of other five trigonometric functions if , *x* lies in second quadrant.

#### Answer:

Since *x* lies in the 2^{nd} quadrant, the value of cos *x* will be negative

#### Question 3:

Find the values of other five trigonometric functions if , *x* lies in third quadrant.

#### Answer:

Since *x *lies
in the 3^{rd} quadrant, the value of sec *x* will be negative.

#### Question 4:

Find the values of other five trigonometric functions if , *x* lies in fourth quadrant.

#### Answer:

Since *x* lies in the 4^{th} quadrant, the value of sin *x* will be negative.

#### Question 5:

Find the values of other five trigonometric functions if , *x* lies in second quadrant.

#### Answer:

Since *x* lies in the 2^{nd} quadrant, the value of sec *x* will be negative.

∴sec *x* =

#### Question 6:

Find the value of the trigonometric function sin 765°

#### Answer:

It
is known that the values of sin *x *repeat
after an interval of 2π or
360°.

#### Question 7:

Find the value of the trigonometric function cosec (–1410°)

#### Answer:

It
is known that the values of cosec *x *repeat
after an interval of 2π or
360°.

#### Question 8:

Find the value of the trigonometric function

#### Answer:

It
is
known that the values of tan *x *repeat
after an interval of π or
180°.

#### Question 9:

Find the value of the trigonometric function

#### Answer:

It
is known that the values of sin *x *repeat
after an interval of 2π or
360°.

#### Question 10:

Find the value of the trigonometric function

#### Answer:

It
is known that the values of cot *x *repeat
after an interval of π or
180°.

#### Page No 73:

#### Question 1:

#### Answer:

L.H.S. =

#### Question 2:

Prove that

#### Answer:

L.H.S. =

#### Question 3:

Prove that

#### Answer:

L.H.S. =

#### Question 4:

Prove that

#### Answer:

L.H.S =

#### Question 5:

Find the value of:

(i) sin 75°

(ii) tan 15°

#### Answer:

(i) sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

[sin
(*x* + *y*)
= sin *x* cos *y* + cos *x* sin *y*]

(ii) tan 15° = tan (45° – 30°)

#### Question 6:

Prove that:

#### Answer:

#### Question 7:

Prove that:

#### Answer:

It is known that

∴L.H.S. =

#### Question 8:

Prove that

#### Answer:

#### Question 9:

#### Answer:

L.H.S. =

#### Question 10:

Prove that
sin (*n* + 1)*x* sin (*n* + 2)*x* + cos (*n* + 1)*x* cos (*n* + 2)*x* = cos* x*

#### Answer:

L.H.S. =
sin (*n* + 1)*x* sin(*n* + 2)*x* + cos (*n* + 1)*x* cos(*n *+ 2)*x*

#### Question 11:

Prove that

#### Answer:

It is known that.

∴L.H.S. =

#### Question 12:

Prove that
sin^{2} 6*x* – sin^{2} 4*x* = sin 2*x* sin 10*x*

#### Answer:

It is known that

∴L.H.S.
= sin^{2}6*x* – sin^{2}4*x*

= (sin 6*x* + sin 4*x*) (sin 6*x* – sin 4*x*)

= (2 sin 5*x* cos *x*) (2 cos 5*x* sin *x*)

= (2 sin 5*x* cos 5*x*) (2 sin *x* cos *x*)

= sin 10*x* sin 2*x*

= R.H.S.

#### Question 13:

Prove that
cos^{2} 2*x* – cos^{2} 6*x* = sin 4*x *sin 8*x*

#### Answer:

It is known that

∴L.H.S.
= cos^{2} 2*x* – cos^{2} 6*x*

= (cos 2*x* + cos 6*x*) (cos 2*x *– 6*x*)

= [2 cos 4*x* cos 2*x*] [–2 sin 4*x *(–sin
2*x*)]

= (2 sin 4*x* cos 4*x*) (2 sin 2*x* cos 2*x*)

= sin 8*x* sin 4*x*

= R.H.S.

#### Question 14:

Prove that sin 2*x* + 2sin 4*x* + sin 6*x* = 4cos^{2} *x* sin 4*x*

#### Answer:

L.H.S. = sin 2*x* + 2 sin 4*x* + sin 6*x*

= [sin 2*x* + sin 6*x*] + 2 sin 4*x*

= 2 sin 4*x* cos (â€“ 2*x*) + 2 sin 4*x*

= 2 sin 4*x* cos 2*x* + 2 sin 4*x*

= 2 sin 4*x* (cos 2*x* + 1)

= 2 sin 4*x* (2 cos^{2} *x* â€“ 1 + 1)

= 2 sin 4*x* (2 cos^{2} *x*)

= 4cos^{2} *x* sin 4*x*

= R.H.S.

#### Question 15:

Prove that
cot 4*x* (sin 5*x* + sin 3*x*) = cot *x* (sin 5*x* – sin 3*x*)

#### Answer:

L.H.S = cot 4*x* (sin 5*x* + sin 3*x*)

= 2 cos 4*x* cos *x*

R.H.S. = cot *x* (sin 5*x* â€“ sin 3*x*)

= 2 cos 4*x*. cos *x*

L.H.S. = R.H.S.

#### Question 16:

Prove that

#### Answer:

It is known that

∴L.H.S =

#### Question 17:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Question 18:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Question 19:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Question 20:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Question 21:

Prove that

#### Answer:

L.H.S. =

#### Page No 74:

#### Question 22:

Prove that
cot *x* cot 2*x* – cot 2*x* cot 3*x* –
cot 3*x* cot *x* = 1

#### Answer:

L.H.S. =
cot *x* cot 2*x* – cot 2*x* cot 3*x* –
cot 3*x* cot *x*

= cot *x* cot 2*x* – cot 3*x* (cot 2*x* +
cot *x*)

= cot *x* cot 2*x* – cot (2*x *+ *x*) (cot
2*x* + cot *x*)

= cot* x *cot 2*x *– (cot 2*x *cot *x* –
1)

= 1 = R.H.S.

#### Question 23:

Prove that

#### Answer:

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

#### Question 24:

Prove that
cos 4*x* = 1 – 8sin^{2 }*x *cos^{2 }*x*

#### Answer:

L.H.S. =
cos 4*x*

= cos 2(2*x*)

= 1 – 2 sin^{2} 2*x* [cos 2*A* = 1 –
2 sin^{2} *A*]

= 1 – 2(2 sin* x *cos *x*)^{2} [sin2*A* = 2sin *A* cos*A*]

= 1 – 8 sin^{2}*x* cos^{2}*x*

= R.H.S.

#### Question 25:

Prove
that: cos 6*x* = 32 cos^{6} *x* – 48 cos^{4} *x* + 18 cos^{2} *x *– 1

#### Answer:

L.H.S. =
cos 6*x*

= cos 3(2*x*)

= 4 cos^{3} 2*x* – 3 cos 2*x* [cos
3*A* = 4 cos^{3} *A* – 3 cos *A*]

= 4 [(2 cos^{2} *x *– 1)^{3} – 3 (2
cos^{2} *x* – 1) [cos 2*x* = 2 cos^{2} *x *– 1]

= 4 [(2 cos^{2} *x*)^{3} – (1)^{3} – 3 (2 cos^{2} *x*)^{2} + 3 (2 cos^{2} *x*)] – 6cos^{2} *x* + 3

= 4 [8cos^{6}*x* – 1 – 12 cos^{4}*x* + 6 cos^{2}*x*] – 6 cos^{2}*x* + 3

= 32 cos^{6}*x* – 4 – 48 cos^{4}*x* + 24 cos^{2} *x* – 6 cos^{2}*x* + 3

= 32 cos^{6}*x *– 48 cos^{4}*x* + 18
cos^{2}*x* – 1

= R.H.S.

#### Page No 78:

#### Question 1:

Find the principal and general solutions of the equation

#### Answer:

Therefore,
the principal solutions are *x* =and.

Therefore, the general solution is

#### Question 2:

Find the principal and general solutions of the equation

#### Answer:

Therefore,
the principal solutions are *x* =and.

Therefore,
the general solution is,
where *n* ∈ **Z**

#### Question 3:

Find the principal and general solutions of the equation

#### Answer:

Therefore,
the principal solutions are *x* = and.

Therefore, the general solution is

#### Question 4:

Find the general solution of cosec *x* = –2

#### Answer:

_{cosec }_{x}_{ = –2}

Therefore,
the principal solutions are *x* =.

Therefore, the general solution is

#### Question 5:

Find the general solution of the equation

#### Answer:

#### Question 6:

Find the general solution of the equation

#### Answer:

#### Question 7:

Find the general solution of the equation

#### Answer:

Therefore, the general solution is.

#### Question 8:

Find the general solution of the equation

#### Answer:

Therefore, the general solution is.

#### Question 9:

Find the general solution of the equation

#### Answer:

Therefore, the general solution is

#### Page No 81:

#### Question 1:

Prove that:

#### Answer:

L.H.S.

= 0 = R.H.S

#### Question 2:

Prove
that: (sin 3*x *+ sin *x*)
sin *x *+
(cos 3*x *–
cos *x*)
cos *x *=
0

#### Answer:

L.H.S.

= (sin
3*x *+
sin *x*)
sin *x *+
(cos 3*x *–
cos *x*)
cos *x*

= RH.S.

#### Page No 82:

#### Question 3:

Prove that:

#### Answer:

L.H.S. =

#### Question 4:

Prove that:

#### Answer:

L.H.S. =

#### Question 5:

Prove that:

#### Answer:

It is known that.

∴L.H.S. =

#### Question 6:

Prove that:

#### Answer:

It is known that

.

L.H.S. =

= tan 6*x*

= R.H.S.

#### Question 7:

Prove that:

#### Answer:

L.H.S. =

#### Question 8:

, *x* in quadrant II

#### Answer:

Here, *x* is in quadrant II.

i.e.,

Therefore, are all positive.

As *x* is in quadrant II, cos*x* is negative.

∴

Thus, the respective values of are.

#### Question 9:

Find for , *x* in quadrant III

#### Answer:

Here, *x* is in quadrant III.

Therefore, and are negative, whereasis positive.

Now,

Thus, the respective values of are.

#### Question 10:

Find for , *x* in quadrant II

#### Answer:

Here, *x* is in quadrant II.

Therefore,, and are all positive.

[cos*x* is negative in quadrant II]

Thus, the respective values of are .

**NCERT Solutions for Class 11 Chemistry Chapters**

- Chapter 1 – Sets
- Chapter 2 – Relations and Functions
- Chapter 3 – Trigonometric Functions
- Chapter 4 – Principle of Mathematical Induction
- Chapter 5 – Complex Numbers and Quadratic Equations
- Chapter 6 – Linear Inequalities
- Chapter 7 – Permutations and Combinations
- Chapter 8 – Binomial Theorem
- Chapter 9 – Sequences and Series
- Chapter 10 – Straight Lines
- Chapter 11 – Conic Sections
- Chapter 12 – Introduction to Three Dimensional Geometry
- Chapter 13 – Limits and Derivatives
- Chapter 14 – Mathematical Reasoning
- Chapter 15 – Statistics
- Chapter 16 – Probability

**NCERT Solutions for Class 11:**