Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

or, 4x - 12y + 16 = 0
? x - 3y + 4 = 0
Taking negative sign, we get,

4x = (5y - 3) - 7y + 13
or, 4x + 2y - 10 = 0
? 2x + y - 5 = 0
Therefore, the required separate equations are x - 3y + 4 = 0 and 2x + y - 5 = 0.

Example 3. Find the equation of pair of lines represented by the equation
Solution: x2 - 2xycosecD+ y2 = 0
Given equation is

x2 - 2xy cosecD + y2 = 0
or, x2 - 2xy cosecD + y2 (cosec2D -cot2D) = 0
or, x2- 2xy cosecD + y2cosec2D - y2cot2D =0
or, (x - ycosecD)2 - (y cotD) 2 = 0.
or, (x - y cosecD + y cotD) (x - ycosecD - y cotD) = 0
or, {x - y (cosecD - cotD)} {x-y(cosecD + cotD)} = 0
Hence, the required pair of lines are

x - y (cosecD - cotD) = 0 and x - y (cosecD + cotD) = 0

Example 4. Find the angle between the pair of the lines represented by:
Solution:
(a) x2 + 7xy + 3y2 = 0 (b) x2 + xy - 6y2 + 7x + 31y- 18 = 0

(a) Given equation is 2x2 + 7xy + 3y2 = 0 ......... (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,

a = 3, h= 7 and b =3
2

Let T be the angle between the lines represented by equation (i).

Then, tanT = ± 2 h2 - ab = ± 2 49 - 2.3 =± 49 -24
a+b 4 2.5

? tanT = ± 1 2 +3

Taking positive sign,

tanT = 1 ? tanT = tan45°

? T = 45°

Taking negative sign,

tanT = - 1 or, tanT = tan135°

? T = 135°

Therefore, the required angles between the lines represented by
equation (i) are 45° and 135°

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(b) Here, x2 + xy - 6y2 + 7x + 31y - 18 = 0

Taking the homogeneous terms only, we get

x2 + xy - 6y2 = 0 ........... (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get

a = 1, h = 1 , b = -6.
2

Let T be the angle between the lines represented by equation (i), we get

=±2 h2 - ab 1 + 6 1+24
=±2 4 4
tanT a+b =2 =±2
=±1 1-6 -5
1
2 .5

Taking positive sign, -5

tanT = 1 = tan45° ? T = 45°

Taking negative sign,

tanT = - 1 = tan 135°

? T = 135°

Therefore, the required angles are 45° ans 135°.

Example 5. Find the angle between the pair of lines represented by
Solution: x2 - 2xy cosecD+ y2 = 0
Given equation is x2 - 2xy cosecD + y2 = 0
Example 6. Comparing it to ax2 + 2hxy + by2 = 0, we get
Solution:
a = 1, h = -cosecD, b- 1
Let T be the angle between the lines represented by (i),

tanT = ±2 h2 - ab = ± 2 cosec2D = ±cotD
a+b 1+1

= cot (± D) = tan(90° ± D)

? T = 90 ± D

Show that the lines represented by 3x2 + 8xy - 3y2 = 0 are perpendicular to
each other.

Equation is 3x2 + 8xy - 3y2 = 0 ........... (i)

comparing it with ax2 + 2hxy + by2 = 0,

we get a = 3, h = 4, b = -3

Now, a + b = 3 - 3 = 0

Since a + b = 0, the lines represented by equation (i) are perpendicular to
each other.

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Alternatively

Here, 3x2 + 8xy - 3y2 = 0

or, 3x2 + 9xy - xy - 3y2 = 0

or, 3x(x + 3y) - y(x + 3y) = 0

or, (x + 3y) (3x - y) = 0

Separate equation of lines are :

x + 3y = 0 ............(i)

3x - y = 0 ............. (ii)

slope of line (i), m1 = - 1
3

slope of line (ii), m2 = .3

m1.m2 = 1 × 3 = - 1
3

? m1.m2 = -1, the lines (i) and (ii) are at right angle each other.

Example 7. Find the value of k when the pair of lines represented by

(k+2) x2+8xy+ 4y2=0 are coincident.

Solution: Given equation is (k + 2) x2 + 8xy + 4y2 = 0

comparing it to ax2 + 2hxy + by2 = 0, we get,

a = k + 2, h = 4, b = 4

condition for coincidence h2 = ab

or, 16 = (k + 2). 4

or, 16 = k + 2
4

or, 4 = k + 2

? k=2

Example 8. Find the single equation passing through the point (1, 1) and parallel to the
Solution: lines represented by the equation x2 - 5xy + 4y2 = 0

Given equation is

x2 - 5xy + 4y2 = 0

or, x2 - 4xy - xy + 4y2 = 0

or, x(x - 4y) - y (x - 4y) = 0

or, (x - 4y) (x - y) = 0

Separate equations of lines are

x - 4y = 0 ........... (i) and x - y = 0 .............. (ii)

Equation of any line parallel to (i) is x - 4y + k1 = 0 ......... (iii)
Equation of any line parallel to (ii) is x - y + k2 = 0 ......... (iv)

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Above both lines pass through the point (1, 1)

1 - 4 + k1 = 0 or, k1 = 3

and 1 - 1 + k2 = 0 or, k2 = 0

putting the values of k1 and k2 in equations (iii) and (iv), we get,

x - 4y + 3 = 0 x-y =0

required single equation is given by,

(x - 4y + 3) (x - y) = 0

or, x(x - 4y + 3) - y (x - 4y + 3) = 0

or, x2 - 4xy + 3x - 3x - xy + 4y2 - 3y = 0

? x2 - 5xy + 4y2 + 3x - 3y = 0, which is the required single equation.

Example 9. Find the single equation of straight line passing through the origin and
Solution: perpendicular to the lines represented by x2 + 3xy + 2y2 = 0
Given equation is

x2 + 3xy + 2y2 = 0
or, x2 + 2xy + xy + 2y2 = 0
or, x(x+2y) + y(x + 2y) = 0
or, (x + 2y) (x + y) = 0
Separate equations of lines are

x + 2y = 0 and x + y = 0
Equations of the lines perpendicular to x + 2y = 0 and x + y = 0 and passing

through the origin are respectively 2x - y = 0 and x - y = 0
Required single equation is

(2x - y) (x - y) = 0
or, 2x2 - 3xy + y2 = 0
? 2x2 - 3xy + y2 = 0 is the required equation.

Example 10. Prove that two straight lines represented by the equation

(x2 + y2)sin2D = (xcosE - ysinE)2 makes an angle is 2D.

Solution: The given equation of two straight line is

(x2 + y2)sin2D = (xcosE - ysinE)2

or, x2sin2D + y2sin2D = x2cos2E - 2xy sinE.cosE + y2sin2E

or, x2sin2D - x2cos2E + y2sin2D - y2sin2E + 2xy sinE.cosE = 0

or, x2(sin2D - cos2E) + 2xy sinE.cosE + y2(sin2D - sin2E) = 0

By comparing the equation to ax2 + 2hxy + by2 = 0, we get

a = sin2D - cos2E, 2h = sinE.cosE i.e. h = sinE.cosE

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and b = sin2D - sin2E

we know that

tanT = ± 2 h2 - ab = ± 2 (sinE.cosE)2 - (sin2D - cos2E) (sin2D - sin2E)
a+b sin2D - cos2E + sin2D - sin2E

= ± 2sinD (1 - sin2D)
2sin2D - (sin2E + cos2E)
2sinD.cosD sin2D
= ± 2sin2D - 1 = ± cos2D = ± tan2D

or, tanT = tan2D

? T = 2D

Therefore, the angle two lines represented by given equation is 2D. Proved.

Exercise 8.2

Very Short Questions

1. (a) Write the general equation of second degree in x and y.

(b) Define a homogeneous equation.

(c) What is the angle between the pair of lines represented by ax2 + 2hxy + by2 ?

(d) Write the conditions of coincident and perpendicularity of two lines represented
by equation ax2 + 2hxy + by2 = 0

2. Find the single equation represented by the pair of straight lines:

(a) x - y = 0 and x + y = 0 (b) x + 3y = 0 and x - 3y = 0

(c) 4x + 3y = 0 and x + y = 0 (d) 3x + 4y = 0 and 2x - 3y = 0

3. Find the separate equations of straight lines represented by the following equations:

(a) x2 - y2 = 0 (b) 9x2 - 25y2 = 0

(c) x2 - y2 + x - y = 0 (d) x2 + y2 - 2xy + 2x - 2y = 0

4. Show that each pair of lines represented by equations represent a pair perpendicular lines:

(a) 3x2 + 8xy - 3y2 = 0 (b) 5x2 + 24xy - 5y2 = 0

(c) 6x2 - 5xy - 6y2 = 0

5. Show that each pair of lines represented by equations are coincidence:

(a) 16x2 + 16xy + 4y2 = 0 (b) x2 - 6xy + 9y2 = 0

(c) 4x2 - 12xy + 9y2 = 0 (d) 9x2 - 6xy + y2 = 0

Short Questions
6. Determine a single equation representing the following pair of lines.

(a) x + y + 3 = 0 and x - 2y + 3 = 0
(b) 2x + y - 3 = 0 and 2x + y + 3 = 0

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7. Determine the lines represented by each of the following equations:

(a) x2 + 6xy + 9y2 + 4x + 12y - 5 = 0

(b) x2 + 2xy + y2 - 2x - 2y - 15 = 0

(c) 2x2 + 3xy + y2 + 5x + 2y - 3 = 0

(d) 2x2 + xy - 3y2 + 10y - 8 = 0

8. Prove that two lines represented by the following equations are perpendicular to each
other:

(a) x2 - y2 + x - y = 0 (b) 9x2 - 13xy - 9y2 + 2x - 3y + 7 = 0

9. Find the value of O when the lines represented by each of the following equations are
coincident:

(a) 4x2 + 2Oxy + 9y2 = 0 (b) (O + 2) x2 + 3xy + 4y2 = 0

10. Find the value of O when the following equations represents a pair of line perpendicular
to each other:

(a) Ox2 + xy - 3y2 = 0 (b) (O - 25) x2 + 5xy = 0

(c) (3O + 4) x2 - 48xy - O2y2 = 0

11. Find the angle between the following pair of lines:

(a) x2 + 9xy + 14y2 = 0 (b) 2x2 + 7xy + 3y2 = 0

(c) x2 - 7xy + y2 = 0

(d) 9x2 - 13xy - 9y2 + 2x - 3y + 7 = 0 (e) x2 + 6xy + 9y2 + 4x + 12y - 5 = 0

12. Find the angle between the following pair of lines:

(a) x2 - 2xy cotD - y2 = 0 (b) x2 + 2xy cosecD + y2 = 0

(c) y2 + 2xy cotD - x2 = 0 (d) x2 - 2xy cot2D - y2 = 0

13. Find the two separate equation of two lines represented by the following equations.

(a) x2 + 2xy cotD - y2 = 0 (b) x2 - 2xy secD + y2 = 0

(c) x2 + 2xy cosecD + y2 = 0 (d) x2 - 2xy cot2D – y2 = 0

Long Questions

14. (a) Find the equation of two lines represented by 2x2 + 7xy + 3y2 = 0. Find the point
of intersection of the lines. Also, find the angle between them.

(b) Find the separate equations of two lines represented by x2 - 5xy + 4y2 = 0. Find
the angle between the lines and the points of intersection of the lines.

(c) Find the separate equation of two lines represented by the equation
x2 - 2xy cosecD + y2 = 0. Also find the angle between them.

15. (a) Find the pair of lines parallel to the lines x2 - 3xy + 2y2 = 0 and passing through
the origin.

(b) Find the pair of lines parallel to the lines 2x2 - 7xy + 5y2 = 0 and passing through
the point (1, 2).

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16. (a) Find the pair of lines perpendicular to the lines x2 - 5xy + 4y2 = 0 and passing
through the origin.

(b) Find the equation of two straight lines which pass through the point (2,3) and
perpendicular to the lines x2 - 6xy + 8y2 = 0

17. (a) Find the two separate equations when lines represented by kx2 + 8xy - 3y2 = 0 are
perpendicular to each other.

(b) Find the two separate equations when the lines represented by 6x2 + 5xy - ky2 = 0 are
perpendicular to each other.

18. (a) Show that the pair of lines 3x2 - 2xy - y2 = 0 are parallel to the lines
3x2 - 2xy - y2 - 5x + y + 2 = 0

(b) Show that the pair of lines, 4x2 - 9y2 = 0 and 9x2 - 4y2 = 0 are perpendicular to
each other.

Project Work

19. Write short notes with examples on the following

(a) linear equations (b) quadratic equations (c) homogeneous equations

2. (a) x2 - y2 = 0 (b) x2 - 9y2 = 0 (c) 4x2 + 7xy + 3y2 = 0

(d) 6x2 - xy - 12y2 = 0 3.(a) x + y = 0, x - y = 0

(b) 3x + 5y = 0, 3x - 5y = 0 (c) (x + y + 1) = 0, x - y = 0

(d) x - y = 0, x - y + 2 = 0 6.(a) x2 - xy - 2y2 + 6x - 3y + 9 = 0

(b) 4x2 + 4xy + y2 - 9 = 0 7.(a) x + 3y + 5 = 0, x + 3y - 1 = 0

(b) x + y - 5 = 0, x + y + 3 = 0 (c) 2x + y = 1, x + y + 3 = 0

(d) x - y + 2 = 0, 2x + 3y = 4 9.(a) ± 6 (b) - 23
16

10. (a) 3 (b) 25 (c) -1, 4

11. (a) T = tan–1(±13) (b) 45°, 135° (c) tan-1 ± 35 (d) 90° (e) 0°
12. (a) 90° (b) 90° ± D 2 (d) 90°
(c) 90°

13. (a) x + y (cosecD + cotD) = 0, x - y (cosecD - cotD) = 0

(b) x + y(secD + tanD) = 0, x + (secD - tanD) = 0

(c) x + y(cotD + cosecD) = 0, x - y (cotD - cosecD) = 0

(d) x + y (cosec2D - cot2D) = 0, x - y (cosec2D + cot2D) = 0

14. (a) x + 3y = 0, 2x + y = 0, (0, 0), 45°, 135°

(b) x - 4y = 0, x - y = 0, tan-1 ± 3 , (0, 0)
5

(c) x - y(cosecD + cotD) = 0, x + y(cotD - cosecD) = 0, 90° ± D

15. (a) x - 2y = 0, x - y = 0 (b) x - y + 1 = 0, 2x - 5y + 8 = 0
16. (a) 4x + y = 0, x + y = 0 (b) 4x + y = 11, 2x + y = 7
17. (a) 3x - y = 0, x + 3y = 0 (b) (3x – 2y) = 0, 2x + 3y = 0

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8.3 Conic Section

Conic sections are special types of locus which can be obtained geometrically as the intersection
of a plane and a right circular cone. The following are basic types of conic sections.

(a) Parabola (b) Ellipse (c) Circle (d) Hyperbola

Conic sections are also called conics. They play vital roles in the field of mathematics,
engineering, aeronautics etc. In present days, conics are very important in the study of basic
level mathematics.

The path of missile in the form of projectile and path of cables of suspension bridges are
parabolic in nature. In present age of science and technology, parabolic metal surfaces are
used in television dishes, telescopes, headlights, etc.

The planets have elliptical paths while moving around the sun. A smooth elliptical surface
can be used to reflect sound waves and light rays from one focus to another.

Definition : A conic section is the intersection of a plane and a right circular cone.

Before defining different types of conic section, let us define a right circular cone.

Conic means "cone" and section means "to cut" or "to divide".

8.3 Cone

A cone has a closed curved surface as its base Upper A axis AG
and the lateral surface is generated by a line Nappie G T
segment that rotates around the perimeter of Lower Semi vertical V
the closed curved surface keeping one end Nappie
fixed as the vertex. If the base is circle then D angle
the cone is a circular cone. Generator

From above figure we have the following V Vertex
terms :

Semi-vertical angle

The angle D is called semi-vertical angle.

If we rotate the line VG about the line AV so that the angle D does not change.

The surface generated is double napped right circular hollow cone.

Vertex
The point V is called the vertex.

Axis
The fixed line AV is called the axis of the cone.

Generator
The rotating line VG is called the generator of the cone. It is also slant height of the cone.

Nappies
The vertex divides the cone into two parts called nappies.

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8.5 Conic section as a plane section of a right circular cone

On intersecting a right circular cone by a plane in different positions, different sections so
obtained are called conic sections or conics. The shape of conic section depends on the
position of the intersecting plane with respect to cone and by the angle made by it with the
axis of the cone.

(a) Circle V Circle
O
The section of a right circular cone cut by a plane Plane
parallel to its base represents a circle. In other words,
when a cutting plane is at right angles to axis of the
cone, then the section is called circle.

(b) Parabola V Plane

The section of a right circular cone by a plane parallel D
to a generator of the cone represents a parabola. In T
other words, when a cutting plane is inclined to the
axis such that it is parallel to the generator of the cone, Parabola
then the section so obtained is called a parabola. If
T is a angle made by the plane with the axis of cone, O
then D = T.

(c) Ellipse D
T
Section of a right circular cone T
cut by a plane not parallel to any
generator and not parallel or not D
perpendicular to the axis of the cone
is called ellipse. In other words, 183
when a cutting plane is inclined at
some angle to the axis of the cone,
then the section so formed is called
a ellipse. In this case D < T < 90°,
where D is the semi-vertical angle
and T is the angle made by the plan
ze with the axis of the cone.

(d) Hyperbola

When the double cone is cut with
a plane parallel to the axis, we get
hyperbola.

In other words, when the cutting
plane is inclined to the axis such
that it also cuts the other part of

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the double cone, the section is called hyperbola with two branches.
In this case T < D, where T is the angle made by the plane with the axis of the cone.

Exercise 8.3

1. (a) Name conic section formed by a plane with right circular cone parallel to its base?
(b) Name conic section formed by a plane with right circular cone parallel to its
generator.
(c) Name conic section formed by plane with right circular cone inclined at some
angle to axis (but not 90°)

2. Name the following geometrical shapes:

(a) (b) (c) (d)
3. Name the curves which is intersection of plane and cone in the following figures:

(a) (b)

V V

(c) V (d)

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Project Work

4. Cutting a potato or cucumber, folding a paper how a parabola or an ellipse or a hyperbola
can be formed. Discuss in the classes.

5. Take a potato or radish or carrot, make a base of right circular cone. Draw an axis of
the cone on a paper. Then cut the slices of potato or radish or carrot perpendicular to
the base, parallel to the base at different angles to axis. Identity the slices so formed
parabola or ellipse or hyperbola or circle.

1. (a) circle (b) parabola (c) ellipse (d) hyperbola
2. (a) circle (b) ellipse (c) Parabola (d) hyperbola
3. (a) parabola (b) circle (d) ellipse

8.6 Circle

Introduction :

Let us draw a circle with centre at the Y
origin and radius 5 units as shown in the
graph. Discuss the answers of the following C (–3, 4)
questions :

(a) Take two points A(5,0) and B(-5, 0) on the O A (5, 0) X
circumference find the mid-point of AB. Y'

(b) Take point C(-3, 4) on the circumference, X' B (–5, 0)
join OC. Find the length of OA, OB
and OC, Are they equal ? Is there any
common name for them ?

(c) Is AB2 = AC2 + BC2 ?

(d) Find the slopes of AC and BC, is there any relations of slopes of AC and BC?

(e) What is equation of locus of points on above circle ?

Discuss the answers of above questions and draw conclusions from the discussion.

Definition of Circle

The locus of a moving point which moves in such a way that its distance from a fixed point
is always constant is said to be a circle. The fixed point is called centre of the circle and the
constant distance is called the radius of the circle.

In the above figure in graph, O(0,0) is the centre, OA, OB or OC are radii of the circle. Then,
we write OA = OB = OC = r, radius of the circle.

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Equations of Circles Y

(a) Equation of a circle with centre at the P(x, y)
origin (standard form) r
O
Let us draw a circle with centre at the X' X
origin. Let us take P(x, y) on the circle.
Then OP = r, is the radius of the circle.

Then by using distance formula.

OP = r

Squaring on both sides, we get

OP2 = r2 Y'

or, (x - 0)2 + (y - 0)2 = 0

? x2 + y2 = r2 which is the required equation.

Example 1. Find the equation of a circle with centre at the origin and radius 6 units.
Solution: Here, radius of the circle, r = 6 units centre = 0(0,0)
Equation of required circle is,

x2 + y2 = r2
or, x2 + y2 = 62
? x2 + y2 = 36

(b) Equation of a circle with centre at other than origin (Central Form)

Let C(h, k) be the centre of circle and radius CP = r. Let Y
P(x, y) be any point on the circle. Then, by using distance
formula, P(x, y)

We get C(h, k)

CP = r X' X
or, CP2 = r2 [squaring on both sides, we get] O
or, (x - h)2 + (y - k)2 = r2
which is the equation of circle called central form. Y'

Example 2. Find the equation of the circle whose centre is at (4, 5) and radius 6 units.
Solution: Here, centre (h, k) = (4, 5)

Radius (r) = 6 units
Equation of required circle is

(x - h)2 + (y - k)2 = r2
or, (x - 4)2 + (y - 5)2 = 62
or, x2 - 8x + 16 + y2 - 10y + 25 = 36

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or, x2 + y2 - 8x - 10y + 5 = 0
The required equation of circle is x2 + y2 - 8x - 10y + 5 = 0

Particular Cases Y C(h, k) X
r=k
(i) Equation of a circle touching the X - axis. X' O
Let a circle with centre at C(h, k) touch the X-axis at A. Y' A
Then radius r = AC = k.
Equation of the circle on given by
(x - h)2 + (y - k)2 = k2

Example 3. Find the equation of a circle with centre (4, 5) and touches the X - axis.

Solution: Here, centre (h, k) = (4, 5) Y
Since the circle touches X- axis.
Then r = k = 5 X' O C(4, 5)
Now, equation of the circle is, Y' r=5

(x - h)2 + (y - k)2 = k2 X
or, (x - 4)2 + (y - 5)2 = 52
or, x2 - 8x + 16 + y2 - 10y + 25 = 25

? x2 + y2 - 8x - 10y + 16 = 0 Y
which is the required equation of the circle.

(ii) Equation of a circle touching the Y-axis. B hC
Let a circle with centre c(h, k) touch the Y-axes at B. k

Then radius r = BC = h. X' O A X
Now, equation of the circle is (x - h)2 + (y - k)2 = h2 Y'

Example 4. Find the equation of the circle with centre (-2, -4) and touching the Y-axis.
Solution:
Here, centre (h, k) = (-2 -4) Y
Since the circle touches Y-axis r = h = -2 X' OX
Now, equation of the circle is,

(x - h)2 + (y - k)2 = h2 (-2, -4)
or, (x + 2) + (y + 4)2 = (-2)2

or, x2 + 4x + 4 + y2 + 8y + 16 = 4 Y'
? x2 + y2 + 4x + 8y + 16 = 0 is the required equation.

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(iii) Equation of a circle touching both of the axes X and Y. Y
Let a circle with centre (h, k) touch

both of the axes X and Y at A and B. B h
Then AC = BC = h = k = r C
Then equation of the circle is
X' k X
(x - h)2 + (y - h)2 = h2 OA
or, (x - k)2 + (y - k)2 = k2.
Y'

Example 5. Find the equation of the circle with centre (3,3) Y
Solution: touching both of the axes.
B 3 C(3, 3)
Here, centre (h, k) = (3, 3) 3

Since the circle touch both of the axis, X' OA X
h=k=r=3 Y'

Now, equation of the circle is

(x - h)2 + (y -h)2 =h2

or, (x - 3)2 + (y - 3)2 = 32

or, x2 - 6x + 9 + y2 - 6y + 9 =9

? x2 + y2 - 6x - 6y + 9 = 0 is the required equation of the circle.

(C) Equation of a circle in diameter form

Let A(x1, y1) and B(x2, y2) be the P(x,y) ends of a diameter P(x, y)
AB of a circle and O is the its centre.

Let P(x,y) be any point on the circle. Join AP and BP A B
Then by plane geometry, (x1, y1) (x2, y2)

‘APB = 90° [ Angle at semi-circle is right angle.]

Now, slope of AP = (m1) = y - y1
slope of BP = m2 = y -xy-2x2

x- x2
Since AP is perpendicular to PB,

we write, m1.m2 = - 1

or, y - y1 . y - y2 =-1
x- x1 x- x2
or, (y - y1) (y - y2) = - (x - y1) (x - x2)

or, (x - x1) (x - x2) + (y - y1) (y - y2) = 0

which is the required equation circle in diameter form.

The following properties of circles are to be noted

(i) The straight line joining the centre of the circle and middle point of a chord is
perpendicular to the chord.

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(ii) The perpendicular drawn from the centre of the circle on the chord bisects it.

(iii) The angle at semi-circle is a right angle.

(iv) A straight line is tangent to the circle if the length of the perpendicular drawn from the
centre to the line is equal to the radius of the circle.

(v) Two circles touch externally if the distance between the centres of two circles is equal
to the sum of radii of the circles. Two circles touch internally if the distance between
the centres of two circles is equal to difference of their radii.

r1 r2 P C1
C1 C2 C2

Externally touched circles Internally touched circles
r1 + r2 = C1 C2 PC1 = r1, PC2 = r2
r1 – r2 = c1 c2

Example 6. Find the equation of a circle whose ends of diameter are (-2, -3) and (2,3).
Solution: The ends of a diameter are (-2. -3) and (2,3).
i.e. (x1, y1) = (-2, -3), (x2, y2) = (2, 3)
Now, equation of circle is given by,

(x - x1) (x - x2) + (y - y1) (y - y2) = 0
or, (x + 2) (x - 2) + (y + 3) (y - 3) = 0
or, x2 - 4 + y2 - 9 = 0
? x2 + y2 = 13 is the required equation of circle.

(d) General Equation of the circle
Consider the equation
x2 + y2 + 2gx + 2fy + c = 0.......... (i)
In which the coefficients of x2 and y2 are equal each having unity and there is no term
containing xy.
The above equation (i) can be written as
x2 + 2gx + y2 + 2fy = -c
or, x2 + 2gx + g2 + y2 + 2fy + f2 = g2 + f2 - c
or, (x + g)2 + (y + f)2 = ( g2 + f2 - c )2 .......... (ii)
which is in the form of
(x - h)2 + (y - k)2 = r2 ........... (iii)
Hence, the above equation of (i) is equation of circle. Comparing equation (ii) to (iii)
Centre = (h, k) = (-g, -f)

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radius (r) = g2 + f2 - c
The above equation (i) is called general equation of circle.

Worked out Examples

Example 1. Find the equation of circle with centre (4,1) and radius 5.
Solution: Here, centre (h, k) = (4, 1)

Example 2. radius (r) = 5
Solution: Now, equation of the circle is

Example 3. (x -h)2 + (y - k)2 = r2
Solution: or, (x - 4)2 + (y - 1)2 = 52
or, x2 - 8x + 16 + y2 - 2y + 1 = 25
190 ? x2 + y2 - 8x - 2y = 8 is the required equation.

Find the centre and radius of the circle x2 + y2 + 4x - 6y + 4 = 0
The given equation is

x2 + y2 + 4x - 6y + 4 = 0
or, x2 + 4x + 4 + y2 - 6y + 9 = 4 + 9 - 4
or, (x + 2)2 + (y - 3)2 = 32
which is in the form of (x-h)2 + (y - k)2 = r2,
? centre (h, k) = (-2, 3), radius (r) = 3 units.

If the equation of circle 4x2 + 4y2 - 24x - 20y - 28 = 0, find the coordinates of
the centre and diameter of the circle.

Equation of given circle is

4x2 + 4y2 - 24x - 20y - 28 = 0

or, x2 + y2 - 6x - 5y - 7 = 0

or, 5 52 (= 7 + 32 + 5 (2
2 2 2
( (or, 25
x2 - 2.3.x + 32 + y2 - 2.y.( + +
(x - 3)2 + 9+ 4
y- 5 ((2
2 =7
(
(or, y- 5 ( 2 89
(x - 3)2 + 2 (=4

( (or, y- 5 2 89 2
(x - 3)2 + 2 = 2

which is in the form of

(x - h)2 + (y - k)2 = r2

(centre (h, k) = 3, 5 and radius (r) = 89 units.
2 2

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Example 4. Find the equation of a circle whose ends of diameter are (2,4) and (4, 5)
Solution: The ends of diameter are (x1y1) = (2, 4) and (x2,y2) = (4, 5)
Now, equation of circle in diameter form is

(x - x1) (x - x2) + (y - y1) (y - y2) = 0
or, (x - 2)(x - 4) + (y -4)(y - 5) = 0
or, x2 - 6x + 8 + y2 - 9y + 20 = 0
? x2 + y2 - 6x - 9y + 28 = 0 is the required equation.

Example 5. Find the equation of the circle whose centre is the point of intersection of
Solution: x + 2y - 1 = 0 and 2x - y - 7 =0 and which passes through the point (3, 1).

Given equation are

x + 2y - 1 = 0 ...... (i)

2x - y - 7 = 0 ....... (ii)

We get from equation (i), (3, 1) r= 2x–y–7=0
x = 1 - 2y .......... (iii) P
C x+y–1=0
Putting the value of x in equation (ii),

2(1 - 2y) - y - 7 = 0

or, 2 - 4y - y - 7 = 0

y=-1

Putting the value of y in equation (iii), we get,

x = 1 - 2(-1) = 3

? centre of circle is C (3, – 1)

Radius = CP = (3 – 3)2 + (1 + 1)2 = 2 units.

Now, equation of circle is,

(x – h)2 + (y – k)2 = r2

or, (x – 3)2 + (y + 1)2 = 22

or, x2 – 6x + 9 + y2 + 2y + 1 = 4

? x2 + y2 – 6x + 2y + 6 = 0 is the required equation.

Example 6. Find the equation of a circle with centre (6, 5) and touching the x-axis.
Solution:
Here, centre of the circle = (h, k) = (6, 5) Y

Since the circle touches the X-axis C(6, 5)

Radius = r = k = 5 r=5

Now, the equation of circle is X' O X
(x – h)2 + (y – k)2 = k2

or, (x – 6)2 + (y – 5)2 = 52 Y'

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or, x2 – 12x + 36 + y2 – 10y + 25 = 25
Example 7. or, x2 + y2 – 12x – 10x + 36 = 0
Solution: ? x2 + y2 – 12x – 10y + 36 = 0 is the required equation of the circle.

Find the equation of a circle passing through the origin and making
Example 8. intercepts 4 and 6 on positive x and y axes.
Solution:
Let X-intercept OA=4 and Y-intercept OB=6 units of given circle. Let C be the
centre of the circle, draw CM perpendicular to OA and CN perpendicular to
OB. Y

Then,ON = 1 OB = 1 × 6 = 3 B
2 2
1 1
OM = 2 ×OA = 2 ×4 = 2 N C (2, 3)

Then centre of circle has coordinates (2,3) 3
X' 2
Now, radius of circle (r) = OC = 22 + 32 A X
OM
= 4 + 9 = 13
Y'

Equation of required circle is

(x - h)2 + (y - k)2 = r2

or, (x - 2)2 + (y - 3)2 = ( 13 )2

or, x2 - 4x + 4 + y2 - 6y + 9 = 13

or, x2 + y2 - 4x - 6y = 0

? x2 + y2 - 4x - 6y = 0 is the required equation.

Find the equation of a circle passing through the points (1,2), (3,4) and (5,2).

Let, C(h,k) be the centre of circle which passes through the points M(1,2),

N(5,2) and P (3,4) P (3, 4)

Then CM = CN = CP

CM2 = CN2 = CP2 (radii of same circle) M (1, 2) C (h, k) N (5, 2)
CM2 = (1 - h)2 + (2 - k)2

= 1 - 2h + h2 + 4 - 4k + k2

= h2 + k2 - 2h - 4k + 5 ...... (i)

CN2 = (5-h)2 + (2 - k)2

= 25 - 10h + h2 + 4 - 4k + k2

= h2 + k2 - 10h - 4k + 29 ..... (ii)

CP2 = (3 - h)2 + (4 - k)2

= 9 - 6h + h2 + 16 - 8k + k2

= h2 + k2 - 6h - 8k + 25 ...... (iii)

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Taking CM2 = CN2 [from (i) and (ii)]
h2 + k2 - 2h - 4k + 5 = h2 + k2 - 10h - 4k + 29

or, 8h = 24
? h=3
Again, taking CN2 = CP2
or, h2 + k2 - 10h - 4k + 29 = h2 + k2 - 6h - 8k + 25
or, - 4h + 4k = - 4
or, h - k = 1
putting the value of h in this equation, k = 2

centre (h, k) = (3, 2)
and radius (r) = Distance between C(3,2) and M(1,2)

= (1-3)2 + (2-2)2 = 2 units.
Now, equation of the required circle is (x-h)2 + (y-k)2 = r2
or, (x - 3)2 + (y -2)2 = 22
or, x2 - 6x + 9 + y2 - 4y + 4 = 4
? x2 + y2 - 6x - 4y + 9 =0 which is required equation.

Example 9. Find the equation of circle touching coordinate axes at (6,0) and (0,6)
Solution:
The circle touches the X-axis at A(6,0). Y-axis at B(0,6). Draw CA and CB
perpendicular on X and Y-axis respectively. Then OACB will be a square.

OA = BC = 6 Y
OB = AC = 6

centre of the circle is C(6,6)

and radius r = 6. B (0, 6) C (6, 6)
Now, equation of the circle is given by

(x - h)2 + (y - k)2 = r2 X' O X

or, (x - 6)2 + (y - 6)2 = 62 A (6, 0)

or, x2 - 12x + 36 + y2 - 12y + 36 = 36 Y'

? x2 + y2 - 12x - 12y + 36 = 0 is the required equation.

Example 10. Find the equation of a circle which passes through the points (6,5) and (4,1)
and with centre on the line 4x + y = 16

Solution: Let C(h, k)be the centre of the circle. The centre (h,k) lies on the line

4x + y = 16 4h + k = 16 ...... (i)

Also, let equation of circle be (x-h)2 + (y-k)2 = r2 ....... (ii)

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The circle (ii) passes through the points

P(6,5) and Q (4,1) Q (4, 1) P (6, 5)
or, (h - 6)2 + (k - 5)2 = (h - 4)2 + (k + 1)2

or, -4h - 8k = -44

or, h + 2k = 11.........(iii) C (h, k)
solving equations (i) and (iii), we get

h = 3 and k = 4

centre = C (3,4)

Radius (r) = CP = (6-3)2 + (5-4)2 = 32 + 12 = 10 units

Now, equation of the circle is

(x -h)2 + (y - k)2 = r2

or, (x-3)2 + (y - 4)2 = 10

or, x2 - 6x + 9 + y2 - 8y + 16 = 10

? x2 + y2 - 6x - 8y + 15 = 0 is the required equation.

Example 11. A line y - x = 2 cuts a circle x2 + y2 + 2x = 0 at two distinct points. Find the
coordinates of the points. Also, find the length of the chord.

Solution: Given equation of the line is y - x = 2 ......... (i)

The line (i) cuts the circle at two distinct points A and B.

We solve equations (i) and (ii) to find the coordinates of A and B.

From equation (i), y = x + 2

Put the value of y in equation (ii), we get

x2 + (x+2)2 + 2x = 0 x2 + y2 + 2x = 0
or, x2 + x2 + 4x + 4 + 2x = 0

or, 2x2 + 6x + 4 = 0

or, x2 + 3x + 2 = 0

or, x2 + 2x + x + 2 =0 y–x=2 B
or, x(x + 2) + 1(x + 2) = 0 A
or, (x + 2) (x + 1) = 0

x = -1, -2

when x = -1, y = -1 + 2 = 1

when x = -2, y = -2 + 2 = 0

The coordinates of A and B are (-1, 1) and (-2, 0) respectively.

Now, the length of chord AB is given by

AB = (-2+1)2+(0 -1)2 = 1 + 1= 2 units.

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Example 12. Find the equation of a circle which touches the X-axis at the point (6,0) and
passes through the points (3,3).

Solution: Let C(h,k) be the centre of the circle. Since it touches X-axis at D(6,0).

radius (r) = k and h = 6. Y
Now, the equation of circle is

(x - h)2 + (y - k)2 = k2

or, (x - h)2 + (y - k)2 = k2

It passes through the point (6,6), we get C = (h, k)

(3 - 6)2 + (3 - k)2 = k2

or, 9 + 9 - 6k + k2 = k2 (3, 3)

? k=3 X' O D (6, 0) X
Therefore, the required equation is Y'

(x - 6)2 + (y - 3)2 = 32

or, x2 - 12x + 36 + y2 - 6y + 9 = 9

? x2 + y2 - 12x - 6y + 36 = 0 is the required equation of circle.

Example 13. Show that two circles x2 + y2 = 100 and x2 + y2 - 24x - 10y + 160 = 0 touch
each other externally.

Solution: The equation of given circles are

x2 + y2 = 100.......... (i)

x2 + y2 - 24x - 10y + 160 = 0 .......... (ii)

From equation (i), radius (r1) = 10 units r1 = 13 r2 = 3
centre C1 = (0, 0) C1 = (0, 0) C2 = (12, 5)

From equation (ii),

radius = g2 + f2 - c
(r2) = (-12)2 + (-5)2 - 160

= 144 + 25 - 160

= 9 = 3 units

Centre (C2) = (12, 5)
Distance between the centres = (12 - 0)2 + (5 - 0)2

= 144 + 25

= 169 = 13 units

Sum of the radii of the circle = 10 + 3 = 13

Since the distance between the centres of circles is equal to sum of the radii
of circles, two circles touch externally.

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Example 14. Find the equation of the circle which touches the positive Y-axis at a distance of
4 units from the origin and cuts of an intercept 6 from the X-axis.

Solution : Let C (h, k) be the centre of the circle touching Y-axis at N and ON = 4.

From C, draw perpendiculars CM and CN on X-axis and Y-axis respectively.

Then, ON = CM = 4 =k

k=4 Y

Also, AM = MB = 1 AB = 3,
2

From right angled triangle CMA,

AC2 = AM2 + CM2 N C (h, k)
= 9 + 16 = 25 4
X' O A 4 BX
AC = radius (r) = 5 units 3M
(h, k) = (5, 4)

Now, equation of the circle is Y'
(x - h)2 + (y - k)2 = r2

or, (x - 5)2 + (y - 4)2 = 52

or, x2 - 10x + 25 + y2 - 8y + 16 = 25

or, x2 + y2 - 10x - 8y + 16 = 0

? x2 + y2 - 10x - 8y + 16 = 0 is the required equation.

Exercise 8.4

Very Short Questions

1. (a) Write down the equation of a circle with centre at O(o,o) and radius R.

(b) Write down the equation of a circle with centre (h,k) and radius r.

(c) Write radius and centre of general circle x2 + y2 + 2gx + 2fy + c = 0

(d) Write the equation of a circle whose ends of a diameter are (x1, y1) and (x2, y2) .
2. Write the equation of the circle under the following conditions.

(a) centre at O(0,0) and radium 5 units.

(b) centre at (-2, -3) and radius 6 units.

(c) ends of a diameter are (1,2) and (3,4).

(d) ends of a diameter are (4,5) and (-2, -3).

(e) ends of a diameter are (-a,0) and (a,0).

3. Write the centre and radius of the following circles.

(a) (x - 2)2 + (y - 4)2 = 25 (b) (x + 3)2 + (y + 5)2 = 81

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(c) (x - 2)2 + (y + 7)2 = 36 (d) (x - a)2 + (y - b)2 = c2

4. Find the centre and radius of the following circles :

(a) x2 + y2 = 81 (b) x2 + y2 + 6x + 4y - 12 = 0

(c) x2 + y2 + 2px - 2qy + p2 + q2 = m2 + n2

(d) 4x2 + 4y2 + 16x - 24y - 52 = 0

(e) 9x2 + 9y2 - 36x + 6y = 107 (f) x2 + y2 - 2axcosT -2aysinT = 0

Short Questions

5. (a) Find the equation a circle whose centre is the (4,5) and touches x-axis.

(b) Find the equation of a circle with circle with centre (-3, 5) and touching x-axis.

(c) Centre (1,3) and touching the y-axis.

6. Find the equations of circles under the following condition.

(a) Centre at (4, -1) and through (-2, -3)

(b) radius 5 units and touching positive x-axis and y-axis.

(c) touching the coordinate axes at (a,0) and (a,0).

Long Questions

7. (a) Find the equation of a circle whose centre is the point of intersection of
x + 2y - 1 = 0 and 2x - y - 7 = 0 and passing through the point (6,4)

(b) Find the equation of a circle with equations of two diameters are x + y = 14 and
2x - y = 4 and passing through the point (4, 5).

8. Determine the points of intersections of a straight line and circle in each of the following
cases:

(a) x + y = 3, x2 + y2 - 2x - 3 = 0 (b) 2x - y + 1 = 0, x2 + y2 = 2

(c) x + y = 2, x2 + y2 = 4

Also find the lengths of the intercepts (chords) in each case.

9. Find the centre and radius of circles.

(a) passing through the points P(-4, -2), Q(2,6) and R(2, -2)

(b) passing through the points P(2, -1), Q(2, 3) and R(4, 1)

10. Find the equations of circle.

(a) passing through the points (5,7),(6,6) and (2, -2).

(b) passing through the points (-6,5), (-3,-4) and (2,1)

11. (a) Find the equation of circle passing through the origin and the point (4,2) and
centre lies on the lines x + y = 1.

(b) Find the equation of the circle passing through the points (3,2) and (5,4) and
centre lies on the line 3x - 2y = 1.

12. (a) Find the equation of the circle which touches the x-axis at (4,0) and cuts of an
intercept of 6 units from the y-axis positively.

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(b) Find the equation of the circle which touches the y-axis at (0, 4) and cuts of an
intercept 6 units from the positive part of x-axis.

13. (a) Prove that the points A(2, -4), B(3, -1), C(3, -3) and D(0,0) are concyclic.
(b) Prove that the points P(3,3), Q(7,1), R(4,0) and S(6,4) are concyclic.

14. (a) Show that the two circles x2 + y2 = 36 and x2 + y2 - 12x - 16y + 84 = 0 touch
externally.

(b) Show that the two circles x2 + y2 = 81 and x2 + y2 - 6x - 8y + 9 = 0 touch
internally.

Project Work

15. Draw a circle of radius 5 units taking centre at (2,3). On a graph, find equation of
the circle. Take any four points on the circle and show that the points satisfy the
equation of the circle. Discuss the conclusion.

2. (a) x2 + y2 = 25 (b) x2 + y2 + 4x + 6y = 23 (c) x2 + y2 - 4x - 6y + 11 = 0

(d) x2 + y2 - 2x - 2y - 23 = 0 (e) x2 + y2 = a2

3. (a) (2, 4), 5 (b) (-3, -5), 9 (c) (2, -7), 6 (d) (a, b), c

4. (a) (0, 0), 9 (b) (-3, -2), 5 (c) (-p, q), m2 + n2

(d) (-2, 3), 26 (e) 2, - 1 , 4 (f) (acosT, asinT), a
3 (b) x2 + y2 + 6x - 10y + 9 = 0
5. (a) x2 + y2 - 8x - 10y + 16 = 0

(c) x2 + y2 - 2x - 6y + 9 = 0 6.(a) x2 + y2 - 8x + 2y = 23

(b) x2 + y2 - 10x - 10y + 25 = 0 (c) x2 + y2 - 2ax - 2ay + a2 = 0

7. (a) x2 + y2 - 6x + 2y = 24 (b) x2 + y2 - 12x - 16y + 87 = 0

8. (a) (1, 2), (3, 0), 2 2 (b) (-1, -1), 1 , 7 , 65 (c) (0, 2), (2, 0), 2 2
5 5 2

9. (a) (-1, 2), 5 (b) (2, 1), 2

10. (a) x2 + y2 - 4x - 6y - 12 = 0 (b) x2 + y2 + 6x - 2y = 15

11. (a) x2 + y2 - 8x + 6y = 0 (b) x2 + y2 - 6x - 8y + 21 = 0

12. (a) x2 + y2 - 8x - 10y + 16 = 0 (b) x2 + y2 - 10x - 8y + 16 = 0

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Trigonometry

9.0 Review of Trigonometricy Ratios of Compound Angle

Let A and B be any two angles, the sum (A+ B) and the difference (A - B) are known as
compound angles. The following are the list of formula of compound angles.

(i) sin (A + B) = sinA.cosB + cosA. sinB

(ii) cos (A + B) = cosA. cosB - sinA. sinB

(iii) tan (A + B) = tanA + tanB
1 - tanA tanB
cotA.cotB-1
(iv) cot (A + B) = cotB + cotA

(v) sin (A -B) = sinA. cosB - cosA. sinB

(vi) cos (A - B) = cosA.cosB + cosA. sinB

(vii) tan (A -B) = tanA - tanB
1 + tanA.tanB
cotA.cotB + 1
(viii)cot (A-B) = cotB - cotA

(ix) sin (A + B). sin (A - B) = sin2A - sin2B = cos2B - cos2A

(x) cos(A + B).cos (A - B) = cos2A - sin2B = cos2B - sin2A

(xi) tan(A + B). tan(A - B) = tan2A – tan2B
1 - tan2A.tan2B
cot2A.cot2B-1
(xii) cot(A + B). cot (A - B) = cot2B - cot2A

9.1 Trigonometric Ratios of Multiple Angles

Let A be any angle. Then, 2A, 3A, 4A, ... etc. are called multiple angles of A. In this
sub-units, we discuss the trigonometric ratios of angles 2A and 3A.

I. Trigonometric Ratios of 2A
We know that sin (A + B) = sinA.cosB - cosA.sinB.

(a) sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA
+ tan2A + cot2A

Here, sin2A = sin(A + A) = sinA.cosA + cosA.sinA

? sin2A = 2sinA.cosA............ (i)

sin2A 2sinA.cosA = 2sinA.cosA
1 sin2A + cos2A

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2sinA.cosA
cos2A
= sinA +cosA = 2tanA
tan2A + 1
cos2A
2tanA
? sin2A = 1 + tan2A ...............(ii)

2tanA 2 1 1 2cot2A 2cotA
1 + tan2A cotA cotA cot2A + 1 + cot2A
Also, sin2A = = = . =
1 1
1 + cot2A
2cotA
? sin2A = 1 + cot2A ......... (iii)

Combining above (i), (ii) and (iii), we get,

sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA
+ tan2A + cot2A
1-tan2A cot2A -1
(b) cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1 + tan2A = 1 + cot2A

Here, cos(A + B) = cosA.cosB - sinA.sinB

Now, cos2A = cos(A + A) = cosA.cosA - sinA.sinA

? cos2A = cos2A - sin2A ............ (i)

cos2A = cos2A - sin2A = 1 - sin2A - sin2A

? cos2A = 1 - 2sin2A ............... (ii)

or, 2sin2A = 1 - cos2A.

? cos2A = 2cos2A - 1 .............(iii)

2cos2A = 1 + cos2A

Again, cos2A = cos2A - sin2A

cos2A-sin2A
cos2A
= cos2A - sin2A = sin2A + cos2A = 1- tan2A
sin2A + cos2A 1 + tan2A
cos2A
1-tan2A
? cos2A = 1+tan2A ........... (iv)

1 - 1 cot2A - 1
cot2A cot2A + 1
and cos2A = = .......... (v)
1
1 + cot2A

Combining (i), (ii), (iii), (iv), and (v), we get

cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1- tan2A = cot2A -1
1 + tan2A cot2A + 1
2tanA
(c) tan2A = 1 - tan2A

Here, tan(A + A) = tanA + tanA = 2tanA
1 - tanA.tanA 1 - tan2A
2 tanA
? tan2A = 1 – tan2A

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(d) cot2A = cot2A - 1
2cotA
cotA.cotB - 1
Here, cot(A + B) = cotB + cotA

Now, cot2A = cot(A + A) = cotA.cotA -1 = cot2A - 1
cotA + cotA 2cotA
cot2A - 1
? cot2A = 2cotA

II. Trigonometric Ratios of 3A.

(a) sin3A = 3sinA - 4sin3A
Now, sin3A = sin(A + 2A)

= sinA.cos2A + cosA.sin2A

= sinA (1 - 2sin2A) + cosA. 2sinA.cosA

= sinA - 2sin3A + 2sinA.cos2A

= sinA - 2 sin3A + 2sinA(1 - sin2A)

= sinA - 2sin3A + 2sinA - 2sin3A = 3sinA - 4sin3A

? sin3A = 3sinA - 4sin3A

Also, 4sin3A = 3sinA - sin3A

(b) cos3A = 4cos3A - 3cosA

Here, cos3A = cos(A + 2A)

= cosA.cos2A - sinA.sin2A

= cosA. (2cos2A - 1) - sinA.2sinA.cosA

= 2cos3A - cosA -2sin2A.cosA

= 2cos3A - cosA - 2(1 - cos2A) cosA

= 2cos3A - cosA - 2cosA + 2cos3A = 4cos3A - 3cosA.

Also, 4cos3A = 3cosA + cos3A

(c) tan3A = 3tanA - tan3A
1 - 3tan2A

Here, tan3A = tan (A +2A)

= tanA + tan2A = tanA + 2tanA
1-tanA.tan2A 1-tan2A
1- tanA.
tanA-tan3A +2tanA 2tanA
1-tan2A

= 1 - tan2A = 3tanA - tan3A
1 - tan2A -2tan2A 1 - 3tan2A
1-tan2A

? tan3A = 3tanA - tan3A
1- 3tan2A

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(d) cot3A = cot3A - 3cotA
3cot2A - 1

Here, cot3A = cot(A + 2A)

= cotA.cot2A - 1 = cotA. cot2A -1 - 1
cot2A + cotA 2cotA

cot2A -1 + cotA
2cotA

= cot3A - cotA - 2cotA . 2cotA
2cotA cot2A - 1+2cot2A
cot3A- 3cotA
= 3cot2A - 1

? cot3A = cot3A - 3cotA
3cot2A - 1

III. Geometrical Interpretation of Trigonometric Ratios of 2A.

Let a revolving line OP make an angle ‘XOP = 2A with an initial line OX.

Now, taking O as the centre and OP as radius of a circle, circle PMN is drawn. MN is the
diameter of the circle. OP, MP, PN are joined. PR is drawn perpendicular to MN. MPN
is semi-circle with a diameter MN, ‘MPN = 90° (angle at semi-circle is right angle).

‘MNP = 1 ‘NOP = 1 × 2A = A
2 2
(Circumference angle is half central angle standing on the same arc PN)

‘OPM = ‘OMP = A (? ∆ MOP is an isosceles base angles are equal).

‘MPR = 90° - A, ‘MPN = 90°, ‘RPN = 90° - ‘MPR = 90° - 90° + A = A

? ‘RPN = A Y

In right angled ∆ PRM, P

sinA = PR , cosA = RM , tanA = PR X' A 2A A X
PM PM MR M RN
In right angled ∆MPN, O

sinA = PN , cosA = PM
MN MN
Now, in right angled ∆ ORP, Y'

(i) sin2A = PR = PR = 2PR = 2PR . PM = 2sinA.cosA
OP MN PM MN
1 MN
2
OR 2OR OR + OR
(ii) cos2A = OP = 2OP = 2OP

= (ON - RN) + (RM - OM)
2OP
OM - RN + RM - OM
= 2OP

= RM - RN
MN
RM RN RM PM RN PN
= MN - MN = PM . MN – PN . MN

= cosA.cosA - sinA.sinA = cos2A - sin2A

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(iii) tan2A = PR
OR
PR PR
= 2PR = 2PR = 2. RM = 2. RM
2OR RM-RN
1- RN 1- RN . PR
RM PR MR
2tanA 2tanA
= 1 - tanA.tanA = 1 - tan2A

Worked Out Examples

Example 1. If sinA = 3 , find sin2A, cos2A and tan2A.
Solution: 5
3
Here, sinA = 5

cosA = 1 - sin2A = 1 - 9 = 4
25 5
3
5 3
tanA = sinA = 4 = 4
cosA
5
Now, sin2A = 2sinA. cosA = 2. 53 . 4 = 24
5 25

cos2A = cos2A - sin2A

= 16 - 9 = 16 - 9 = 7
25 25 25 25

24
25
tan2A = sin2A = 7 = 24
cos2A 7

25

Example 2. Prove that, cotT = ± 1+cos2T
Solution: 1-cos2T
We have 2cos2T = 1 + cos2T............ (i)
Example 3.
Solution and 2sin2T = 1 - cos2T ........... (ii)

Dividing (i) by (ii), we get,

2cos2T = 1 + cos2T
2sin2T 1-cos2T
1 + cos2T
or, cot2T = 1-cos2T

? cotT = ± 1+cos2T = RHS Proved.
1-cos2T
4
If sinA = 5 , then, find the values of sin3A, cos3A and tan3A.

Here, sinA = 4
5
16
Now, cosA = 1 - sin2A = 1 - 25 = 3
5

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4
5
tanA = sinA = 3 = 4
cosA 3

5
sin3A = 3sinA - 4sin3A

= 3. 4 - 4. 64
5 125

= 12 - 256 = 300 - 256 = 44
5 125 125 125

cos3A= 4cos3A - 3cosA

=4 27 - 3. 3 = 108 - 9 = 108-225 =- 117
125 5 125 5 125 125
3tanA - tan3A
tan3A = 1 - 3tan2A

3. 4 - 64 4– 64 108 - 64 44
3 27 1- 27 7(3 - 16) 117
= = 16 = = –
16 3
1- 3. 9

Example 4. (If = 1 m + 1 (, then, show that :
Solution: sinA 2 m ((

((a) 1 m2 (+1 ((b) = - 1 m3 + 1
cos2A = - 2 (m2 sin3A 2 m3
((
(Here, sinA = 1 m + ((1
2 m
(
(a) cos2A = 1 - 2sin2A
(
(= 1 - 2 . ( 1 m + 1 2
4 m
(= ( 1 1 1
((1 - 2. 4 m2 + 2.m. m + m2

=( 1- 1 m2 - 1 - 1 . 1
2 2 m2
(=
- 1 m2 + 1 = LHS. Proved.
2 m2

(b) sin3A = 3sinA - 4sin3A

(= 3 1 m+ 1 (- 4 .1 m + 1 3
2 m 8 m
[ ( ]3 -
(= 1 m+ 1 m + 1 2
2 m m
[ ]3
(= 1 m+ 1 - m2 - 2.m. 1 - 1
2 m m m2
[ ]-
(= 1 m+ 1 (m2 - 1 + 1
2 m m2
( (=
– 1 m + 1 m2 - m. 1 + 1
2 m m m2
(= –
1 m3 + 1 = RHS Proved.
2 m3

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Example 5. Prove the following:
Solution:
(a) sin2A - sinA = tanA (b) 1- cosA = tan2A
1-cosA + cos2A 1+ cos2A
(d) 1 + sec2A = cotA
(c) tanA + cotA = 2cosec2A tan2A

(a) LHS = sin2A - sinA
1-cosA + cos2A
2sinA.cosA - sinA sinA(2cosA - 1)
= 1 - cosA + 2cos2A - 1 = cosA(2cosA -1)

= sinA = tanA = RHS. proved
cosA
1 - cos2A 2sin2A sin2A
(b) LHS = 1 + cos2A = 2cos2A = cos2A

= tan2A = RHS Proved.

(c) LHS = tanA + cotA = sinA + cosA
cosA sinA
sin2A + cos2A 1 2
= sinA.cosA = 2. 2sinA.cosA = sin2A

= 2cosec2A = RHS Proved.

(d) LHS = 1 +sec2A = 1+ = cos2A + 1 × cos2A
tan2A cos2A cos2A sin2A

sin2A
cos2A
2cos2A cosA
= 2sinA.cosA = sinA = cotA = RHS Proved.

Example 6. Proved that:
Solution:
(a) sin6T + cos6T = 1 (1 + 3cos22T) (b) cos4T = 1 - 8sin2T + 8sin4T
4
(a) LHS = sin6T + cos6T

= (sin2T)3 + (cos2T)3

= (sin2T + cos2T) (sin4T - sin2T.cos2T + cos4T)

= 1.(sin4T + cos4T - sin2T.cos2T)

= (sin2T + cos2T)2 - 2sin2T.cos2T - sin2T.cos2T

= 12 - 3sin2T.cos2T

=1- 3 (2sinT . cosT)2
4
3
=1- 4 sin22T

= 1 (4 - 3sin22T)
4
1
= 4 (1 + 3 - 3sin22T)

= 1 {1 + 3 (1 - sin22T)} = 1 (1 + 3cos22T) = RHS. Proved.
4 4
(b) LHS = cos4T

= cos2(2T)

= 1 -2sin22T

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= 1 - 2[2sinT .cosT]2 = 1 – 8sin2T (1 - sin2T)]
= 1 - 8sin2T + 8sin4T = RHS. Proved.

Example 7. PLHroSved=thsaitn110s°in1-10ct°aons-6100c°°os130° =4
Solution:
= 1 - sin60°
Example 8. sin10° cos60°.cos10°
Solution: cos60°.cos10° - sin60°.sin10°
= sin10°.cos60°.cos10°
Example 9.
Solution: = cos (60° + 10°) = 2 . 2cos70° = 4cos70°
2 sin10°.cos10° sin20°
206 sin10° . 1 cos10°
2
4cos70° 4cos70°
= sin(90° -70°) = cos70° = 4 = RHS. Proved.

Proved that (a) cosec2T + cot4T = cotT - cosec4T

(b) 4cosec2T . cot2T =cosec2T - sec2T

(a) LHS = cosec2T + cot4T

= 2ssiicnn11o22sTT2T+++cs2coisnosi4sn4c2TT2o(T2s.4TcT)os2T
= 2sin2T . cos2T
=

= 2cos2T + 2cos22T - 1
2sin2T.cos2T
2cos2T(1 + cos2T) - 1
= 2sin2T.cos2T

= 2cos2T.2cos2T - 1 = 2cos2T – 1
= 2sin2T.cos2T sin2(2T) 2sinT.cosT sin4T
cosT 1
sinT - sin4T = cotT - cosec4T = RHS. Proved.

(b) LHS = 4cosec2T.cot2T

= 4. 1 . cos2T
sin2T sin2T
4cos2T
= 2sinT.cosT.2sinT.cosT

= cos2T - sin2T = cos2T - sin2T
sin2T.cos2T sin2T.cos2T sin2T.cos2T
1
= 1 - cos2T = cosec2T - sec2T = RHS. Proved.
sin2T

Proved that (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1) = 2cos8T + 1

LHS = (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1)

= {(2cosT)2 - 12} (2cos2T - 1) (2cos4T -1)

= (4cos2T –1) (2cos2T - 1) (2cos4T - 1)

= {2(2cos2T - 1) + 1} (2cos2T - 1) (2cos4T - 1)

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vedanta Excel In Opt. Mathematics - Book 10

= (2cos2T + 1) (2cos2T - 1) (2cos4T - 1)
= (4cos22T - 1) (2cos4T - 1)
= {2 (2cos22T - 1) + 1} (2cos4T - 1)
= (2cos4T + 1) (2cos4T – 1)
= 4cos24T - 1
= 2(2cos24T - 1) + 1
= 2cos8T + 1 = RHS. Proved.

Example 10 Prove that cos2D + sin2D . cos2E = cos2E + sin2E . cos2D
Solution: LHS = cos2D + sin2D . cos2E
= cos2D + sin2D(1 - 2sin2E)

= cos2D + sin2D - 2sin2E . sin2D
= 1 - 2sin2E . sin2D = cos2E + sin2E - 2sin2E . sin2D
= cos2E + sin2E(1 - 2sin2D)
= cos2E + sin2E . cos2D = RHS Proved.

Example 11. Prove that 2+ 2+ 2+ 2 + 2cos16T = 2cosT

Solution: LHS = 2+ 2+ 2+ 2 + 2cos16T

= 2+ 2+ 2+ 2(1+cos16T)

= 2+ 2+ 2+ 2.2cos28T

= 2+ 2+ 2 + 2cos8T

= 2+ 2+ 2(1+cos8T)

= 2+ 2+ 2 . 2cos24T

= 2+ 2+2cos4T

= 2+ 2(1+cos4T) = 2+ 2.2cos22T)
= 2(1+cos2T) = 2.2cos2T = 2cosT = RHS. Proved.

Example 12. Prove that 1 - 1 cotA = cot2A
tan3A - tanA cot3A -
1 1
Solution: LHS = tan3A - tanA - cot3A - cotA

= 1 - 1
tan3A -
tanA 1 - 1
tan3A tanA

= 1 tanA - tanA.tan3A
tan3A - tanA - tan3A

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= 1 + tanA.tan3A = 1= 1
tan3A - tanA tan3A - tanA tan(3A - A)

= 1 = cot2A 1 +tanA.tan3A
tan2A = RHS. Proved.

Example 13. Proved that 4(cos320° + cos340°) = 3(cos20° + cos40°)

Solution: LHS = 4 (cos320 + cos340°)

= 4cos320 + 4cos340°

= 3cos20° + cos(3.20°) + 3cos40° + cos(3.40°)

= 3cos20° + cos60° +3cos40° + cos120°

= 3(cos20° + cos40°) + 1 - 1
2 2
= 3(cos20° + cos40°) = RHS. Proved.

Example 14. Prove that sin3T + sin3T = cotT
Solution: cos3T - cos3T
sin3T + sin3T 3sinT - 4sin3T + sin3T
LHS = cos3T - cos3T = cos3T - 4cos2T + 3cosT

= 3sinT - 3sin3T = 3sinT(1-sin2T) = sinT . cos2T
3cosT - 3cos3T 3cosT(1-cos2T) cosT . sin2T
cosT
= sinT = cotT = RHS. Proved

Example 15. Prove that : tanT + 2tan2T + 4tan4T + 8cot8T = cotT

Solution: LHS = tanT + 2tan2T + 4tan4T + 8cot8T

= tanT + 2tan2T + 4tan4T + 8
tan(2.4T)
8
= tanT + 2tan2T + 4tan4T + 2tan4T

1 - tan24T

(= tanT + 2tan2T + 8(1-tan24T)
2tan4T
4tan4T + (

(= tanT + 2tan2T + 4 tan4T + 1-tan24T(
tan4T(
(
(= tanT + 2tan2T + 4 tan24T + 1-tan24T
tan4T
4
= tanT + 2tan2T + tan4T

= tanT + 2tanT 4
2tan2T

(= tanT + 2 1 - tan22T
1-tan22T
tan2T + tan2T

= tanT + 2 tan22T +1- tan22T
tan2T

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= tanT + 2
2tanT

1 - tan2T

= tanT + 2(1 - tan2T)
2tanT
tan2T + 1 - tan2T 1
= tanT = tanT = cotT = RHS. Proved.

Example 16. Prove that

(a) 4sin3A.cos3A + 4cos3A.sin3A = 3sin4A

(b) cos3A.cos3A + sin3A. sin3A = cos32A

Solution: (a) We have, 4sin3A = 3sinA - sin3A

4cos3A = 3cosA + cos3A

Now, LHS = 4sin3A.cos3A + 4cos3A.sin3A

= (3sinA – sin3A). cos3A + (3cosA + cos3A).sin3A

=3sinA.cos3A - sin3A.cos3A + 3sin3A.cos3A + cos3A.sin3A

= 3(sinA.cos3A + sin3A.cosA)

= 3.sin(A + 3A) = 3sin4A = RHS Proved.

(b) LHS = cos3A cos3A + sin3A.sin3A

= 1 (3cosA + cos3A).cos3A + 1 (3sinA - sin3A).sin3A)
4 4
1
= 4 [3cosA.cos3A + cos23A + 3sinA.sin3A - sin23A]

= 1 [3(cosA.cos3A + sinA.sin3A) + (cos23A - sin23A)]
4
1
= 4 [3cos(A - 3A) + cos6A]

= 1 (3cos2A + cos6A)
4
1 1
= 4 [3cos2A + cos(3.2A)] = 4 [3cos2A + 4cos32A – 3cos2A]

= 4cos32A
4
= cos32A =RHS. Proved.

Example 17. Prove that cot(45° + T) + tan(45° - T) = 2cos2T
1 + sin2T

Solution: LHS = cot(45° + T) + tan(45 - T)

= cot45°.cotT - 1 + tan45° - tanT
cot45° + cotT 1 +tan45°.tanT
cotT - 1 1-tanT
= 1 + cotT + 1 + tanT

1 -1 1 - tanT
tanT 1 + tanT
= +
1 + 1
tanT

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= 1-tanT + 1-tanT
1 + tanT 1 + tanT
sinT
2(1-tanT) 1- cosT
1+tanT
= = 2. 1 + sinT
cosT
(cosT - sinT) cosT - sinT cosT + sinT
=2 cosT + sinT = 2. (cosT + sinT) × cosT + sinT

= 2. cos2T = 2cos2T = RHS. Proved.
sin2T + cos2T + 2sinT.cosT 1 + sin2T

Example 18. Prove that 2sinT + 2sin3T + 2sin9T = tan27T - tanT
cos3T cos9T cos27T
2sinT 2sin3T 2sin9T
Solution: LHS = cos3T + cos9T + cos27T

= 2sinT.cosT + 2sin3T.cos3T + 2sin9T.cos9T
cos3T.cosT cos9T.cos3T cos27T.cos9T

= sin2T + sin6T + sin18T
cos3T.cosT cos9T cos3T cos27T.cos9T

= sin(3T - T) + sin(9T - 3T) + sin(27T - 9T)
cos3T.cosT cos9T.cos3T cos27T.cos9T
sin3T.ccoossT3T-.ccoossT3T.sinT+ sin9T.cos3T - cos9T.sin3T
= cos9T.cos3T

+sin27T.ccooss99TT.-cocos2s277TT.sin9T

= tan3T - tanT + tan9T - tan3T + tan27T - tan9T

= tan27T - tanT = RHS. Proved.

Example 19. Without using table or calculator. Show that sin18° = 5-1 .
Solution: Let T = 18° 4

Then 5T = 90°
or, 2T + 3T = 90°
or, 3T = 90° - 2T
Taking cosine on both sides, we get

cos3T = cos(90° - 2T)
or, cos3T = sin2T
or, 4cos3T - 3cosT = 2sinT cosT
since sinT = sin18°≠ 0, dividing both sides by cosT.

4cos2T - 3 = 2sinT
or, 4 - 4sin2T - 2sinT - 3 = 0
or, -4sin2T - 2sinT + 1 = 0
or, 4sin2T + 2sinT - 1 = 0

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This equation is quadratic is sinT in the form of ax2 + bx + c = 0, where
a = 4, b = 2, c = -1

x= b± b2 - 4ac
2a

or, sinT = - 2 ± 22 - 4.4(-1) = 2± 4 + 16
2.4 8

= 2± 20 = 2±2 5 = -1 ± 5
8 8 4

since value of sin18° is positive. We take only positive sign.

Hence, sinT = sin18° = 5 -1 . Proved.
4

( ( (Prove that cosSc 2Sc 3Sc 1
Example 20. 7 (, cos 7 , cos 7 = 8
Solution:
( ( (LHS = cos Sc ( (( 2Sc 3Sc
7 ( (( , cos 7 , cos 7

2sin Sc . cos Sc . cos 2Sc . cos 3Sc
= 7 7 7 7
Sc
2sin 7

2.sin 2Sc . cos 2Sc . cos 2Sc
7 7 7
=
2.2sin Sc
7
4Sc 3Sc
sin 7 . cos 7

= 4sin Sc
7
3Sc 3Sc
2sin 7 . cos 7 [ ( ) ]
sin 4Sc Sc = sin 3Sc
= 4. 2sin Sc 7 = sin S - 7 7
7
(=
sin 6Sc sin Sc - Sc sin Sc 1
7 7 7 8
= = = = RHS. Proved.
8sin Sc 8sin Sc 8sin Sc
7 7 7

Exercise 9.1 211

Very Short Questions
1. (a) Define multiple angles with an example.

(b) Express sin2T in terms of sinT, cosT and tanT.
(c) Express cos2T in terms of sinT, cosT and tanT.
(d) Express cos2T in terms of cotT.

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vedanta Excel In Opt. Mathematics - Book 10

2. (a) Express sin3T in terms of sinT (b) Express cos3T in terms of cosT

(c) Express tan3T in terms of tanT. (d) Express tanT in terms of sin2T and cos2T

Short Questions

3. (a) If sinT = 1 , find the value of cos2T.
2
1
(b) If tanT 3 , find the value of tan3T

(c) If tanT = 1 , find the value of tan3T.
2
4
4. (a) If sinT = 5 , find the value of sin2T, cos2T, tan2T.

(b) If sinT = 1 , find the values of sin2T, cos2T, tan2T.
2
3
(c) If tanT = 4 , find the values of sin2T, cos2T and tan2T

(d) If sinT = 3 , find the value of sin3T and cos3T.
2
1 7
5. (a) If sinT = 4 , show that cos2T = 8

(b) If cosT = 6 , show that cos2T = 11
52 25

(c) If tanT = 5 , show that tan2T = 120
12 119

6. By using the formula of cos2T, establish the following :

(a) sinT = ± 1-cos2T (b) cosT = ± 1+cos2T
2 2

(c) tanT = ± 1-cos2$
1+cos2A
(7. 1 1
(a) If sinT = 2 p + P ((, show that :
(

((
(
( ((i) 1 1 (ii) 1 1
cos2T = – 2 p2 + P2 sin3T = – 2 p3 + P3

((b) 1 1 , show that :
If cosT = 2 p + P

((i) 1 1 ((ii) 1 1
cos2T = 2 p2 + P2 cos3T = 2 p3 + P3

8. Prove that :

(a) 1 - cos2T = tanT (b) sin2T = cotT
sin2T 1-cos2T
sin2T 1 - cos2T
(c) tanT = 1 + cos2T (d) 1 + cos2T = tan2T

(e) sin2T = 2cotT 1 (f) 1 - sin2T = 1 - tanT
cot2T + cos2T 1 +tanT
cos2T 1 - tanT cosT cosT
(g) 1 + sin2T = 1 + tanT (h) cosT - sinT - cosT + sinT = tan2T

(i) 1 + sin2A = sinA + cosA (j) cotT - tanT = cos2T
cos2A cosA - sinA cotT + tanT
4tanT cosT - sinT
(k) tan2T + sin2T = 1 - tan4T (l) cosT + sinT = sec2T - tan2T

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sin5T cos5T sin3T + cos3T 1
sinT cosT cosT + sinT 2
(m) - = 4 cos2T ((n) = 1 - sin2T

1 - sin2T Sc
cos2T 4
(o) cos4T - sin4T = cos2T ( (p) = tan - T
(
((q) Sc
cos2T = tan 4 - T
1 +sin2T
9. Prove that

(a) sin2T - cosT = cotT (b) 1 + sin2T - cos2T = tanT
1 - sinT - cos2T 1 + sin2T + cos2T

(c) sinT + sin2T = tanT (d) 1-cos2T + sin2T = tanT
1 + cosT + cos2T 1 + sin2T + cos2T

(e) (sinT + cosT)2 - (sinT - cosT)2 = 2 sin2T

(f) sinT + cosT + cosT - sinT = 2 sec2T
cosT - sinT cosT + sinT

(g) (1 + sin2T + cos2T)2 = 4cos2T(1 + sin2T)

(h) 1 - 1 cotT = cotT
tan2T - tanT cot2T -

10. Prove that following :

((a) cotT -1 2 (((b) 1 + tan2 Sc - T(
cotT+1 4 (
1-sin2T = ( = cosec2T
1+sin2T ( Sc -
((c) sin2 1 - tan2 4 T
Sc
4 - T = 1 (1 - sin2T)
2
b
11. If tanT = a , prove that a cos2T + b sin2T = a.

Long Questions

12. Prove that

(a) cos4T + sin4T = 1 (3 + cos4T) (b) cos6T - sin6T = cos2T (1 - 1 sin22T)
4 4
1 1
(c) sin4T = 8 (3 - 4 cos2T + cos4T) (d) cos8T + sin8T = 1 - sin22T + 8 sin42T.

(e) cos5T =16cos5T – 20cos3T + 5cosT (f) sin5T = 16sin5T - 20sin3T + 5sinT

13. Prove that

(a) 3 + 1 =4 (b) cosec10° - 3 sec10° = 4
sin40° cos40°
14. Prove that

(a) (2cosT + 1) (2cosT - 1) = 2cos2T + 1)

(b) sec8T - 1 = tan8T
sec4T - 1 tan2T
(c) tanT + 2tan2T + 4tan4T + 8cot8T = cotT

(d) sin2D - cos2D.cos2E = sin2E - cos2E.cos2D

(e) 2+ 2+ 2 + 2cos8T = 2cosT

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(f) sin2D - sin2E = tan (D + E)
sinD.cosD - sinE.cosE
15. Prove that

(a) 4(cos310° + sin320°) = 3(cos10° + sin20°)

(b) sin310° + cos320° = 3 (cos20° + sin10°)
4
(c) 4(cos320° + sin350°) = 3(cos20° + sin50°)

( ((d) Sc Sc
tanA + tan ( ( ((3 + $ – tan 3 - A = 3tan3A
(( ( (
16. Prove that ((( (

(a) cot (A + 45°) - tan (A - 45°) =(( 2cos2A
1 + sin2A
(b) tan(A + 45°) - tan (A - 45°) = 2sec2A.

(c) tan(A + 45°) + tan (A - 45°) = 2tan2A.

17. (a) If 2tanD = 3tanE, then prove that,

(i) tan (D - E) = sin2E (ii) tan(D + E) = 5sin2E 1
5 - cos2E 5cos2E -
1 1
(b) If tanT = 7 and tanE = 3 , prove that : cos2T = sin4E

18. Prove that

(a) cosA - 1 + sin2A = tanA (b) 1 - 1 = cot4T
sinA - 1 + sin2A tan3T + tanT cot3T + cotT

(c) cotA - tanA = 1.
cotA - cot3A tan3A - tanA
19. Prove that
( ( ( ((a)
8 1 + sin Sc 1+sin 3Sc 1- sin 5Sc 1-sin 7Sc =1
8 8 8 8
( ( ( ((b)
1-cos Sc 1- cos 3Sc 1- cos 5Sc 1- cos 7Sc = 1
8 8 8 8 8
( ( ( ((c)
sin4 Sc + sin4 3Sc + sin4 5Sc + sin4 7Sc = 3
8 8 8 8 2

20. Prove that cos233° – cos257° = 2
sin210.5° – sin234.5°
3
21. (a) Prove that cos3A + cos3(120° + A) + cos3(240° + A) = 4 cos3A

(b) Prove that sin3A + sin3(60° + A) + sin3(240° + A) = – 3 sin3A
4
Project Work

22. Prepare a report by calculating the values of sin18°, cos18°, sin36°, cos36°, cos54°,
tan18°, and tan54° by using relations of multiple angle ratios in trigonometry. For this
project work, the students of class can be divided in groups as required.

3. (a) 1 (b) f (c) 11 4.(a) 24 , - 7 , - 24
2 2 25 25 7
3 1 24 7 24
(b) 2 , 2 , 3 (c) 25 , 25 , 7 (d) 0, -1

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9.2 Trigonometric Ratios of Sub-multiple Angles

Let A be any angles. Then A , A , A , ....... etc. are called sub-multiple angles of A. In this
sub-unit, we discuss 2 3 4 ratios of sub-multiple angles $2 , $
the trigonometric 3 , etc.

(I) Trigonometric Ratios of Half Angles :

sinA = 2sin A A 2tan A2 = 2cot A
((a) 2 cos 2 = 2
= A A
1 + tan2 2 1 + cot2 2
A
Here, sinA A + 2 (
2 (
A A A A A A
= sin 2 cos 2 + cos 2 . sin 2 = 2sin 2 cos 2 .

⸫ sinA = 2sin A cos A ............ (i)
2 2
A A
sinA = 2sin 2 cos 2

2sin A . cos A
2 2
= A A
sin2 2 + cos2 2

Dividing numerator and denominator by cos2 A
2
A A
2tan 2 2tan 2

= 1sh+owtanth2 aA2t ⸫ sinA = 1+ tan2 A ........... (ii)
can 2
Similarly we

2cot A
2
sinA = A .......... (iii)
1+ cot2 2

Combing (i), (ii) and (iii), we get,

A A 2tan A 2cot A
2 2 2 2
⸫ sinA = 2sin . cos = A = A
1+ tan2 2 1+ cot2 2

A A 1-tan2 A cot2 A - 1
2 2 2 cot2 2
cosA - sin2
((b) = cos2 = 1+ tan2 A = A +
2 21

Here, cosA = cos A + A
2 2

= cos A . cos A - sin A . sin A
2 2 2 2
A A
= cos2 2 - sin2 2

cosA = cos2 A - sin2 A ........... (i)
2 2

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= cos2 A - 1 + cos2 A
2 2
A
= 2cos2 2 - 1

Also, 2cos2 A = 1 + cosA
2
A A
Again, cosA = cos2 2 - sin2 2

= 1 - 2sin2 A
2
A
2.sin2 2 = 1 - cosA

cosA = cos2 A - sin2 A
2 2
A A
cos2 2 - sin2 2

= sin2 A + cos2 A
2 2
Dividing numerator and denominator by cos2 A .
2
A
1-tan2 2

⸫ cosA = 1+ tan2 A ....... (ii)
2 cot2
A - 1
Similarly, we can show cotA = 2 ............ (iii)
A
cot2 2 + 1

Combining (i), (ii) and (iii), we get

A A 1-tan2 A cot2 A -1
2 2 2 2
⸫ cosA = cos2 - sin2 = A = .
1+ tan2 2 A
cot2 2 + 1

2tan2 A
2
tanA =
((c) 1- tan2 A
2 A A
tan 2 + tan 2
A( + A
Here, tanA = tan 2 ( 2 = 1- tan A . tan A
2 2
2tan A
2
⸫ tanA = A
1- tan2 2

((d) A + A
cotA = cot 2 2

cot A . cot A - 1 cot2 A - 1
2 2 2
Here, cotA = A A = A
cot 2 + cot 2 2cot 2

cot2 A - 1
2
⸫ cotA = A
2cot 2

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II. Trigonometric Ratios of A in terms of A .
3
((a) A 2A ( A A
sinA = sin 3 + 3 = 3sin 3 - 4sin3 3
(
(Here, sinA ( 2A A
= sin 3 + 3

= sin 2A . cos A + cos 2A . sin A
3 3 3 3
(= A A A A ( A
2sin 3 . cos 3 cos 3 + 1- 2sin2 3 . sin 3

(= A 1- sin2 A A A
2sin 3 3 + sin 3 - 2sin3 3

= 2sin A - 2sin3 A + sin A - 2sin3 A
3 3 3 3
A A
? sinA = 3sin 3 - 4 sin3 3

= 4cos3 A - 3cos A
((b) cosA 3 3

Here, cosA = cos (2A+ A
( 3 3

= cos 2A (.cos A - sin 2A . sin A
3 ( 3 3 3
(= A A A A A
2cos2 3 -1 . cos 3 - 2sin 3 . cos 3 . sin 3

(= A - cos A - A 1 - cos2 A (
2cos3 3 3 2cos 3 3

= 2cos3 A - cos A - 2cos A + 2cos3 A = 4cos3 A - 3cos A
3 3 3 3 3 3
A A
? cosA = 4cos3 3 - 3cos 3

((c) tanA = tan A 3tan A - tan2 A
3 3 3
3 . = A
1 - 3tan2 3

(Here, tanA 2A A
= tan 3 + 3

2tan A A
3 3
+ 2tan
tan 2A + tan A 1 - tan2 A
3 3 3
= =
1 - tan 2A . 3tan A A
3 3 1 - 2tan 3 . A
3
1 - tan2 A tan
3
A A A A A
2tan 3 + tan 3 - tan3 3 3tan 3 - tan3 3

= 1 - tan2 A - 2tan2 A = 1 - 3. tan2 A
3 3 3
A A
3tan 3 - tan3 3

? tanA = 1 - 3. tan2 A
3

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((d) cotA = cot A ( cot3 A - 3cot A
3 3 3
3 . = A (For practice do yourself.)
3cot2 3 - 1

Worked out Examples

Example 1. If sin T = 3 find the values of.
Solution: 2 5
(a) sinT (b) cosT (c) tanT
Example 2.
Solution: Here, sin T = 3
2 5

cos T = 1 - sin2 T = 1 - 9
2 2 25

= 25-9 = 16 = 4
25 25 5
T 3
T sin 2 5 3
2 4 4
tan = cos T = 5 =
2
3 4 24
(a) sinT = 2 sin T . cos T = 2. 5 . 5 = 25
2 2
T T 16 9 16-9 7
(b) cosT = cos2 2 - sin2 2 = 25 - 25 = 25 = 25

2tan T 2. 3 3 16 3 16 24
2 4 2 16-9 2 7 7
(c) tanT = = 9 = × = × =
1 - tan2 T 1- 16
2

Alternative 24

tanT sinT = 25 = 24
cosT 7 7
25

If cos45° = 1 , show that cos2212 °= 1 . 2+ 2
2 2

Here, cos45° = 1 2

(or, 45° =1
2 2
cos2 (

or, 224cc25oo°ss22is442255p°°osi-=t1iv=e12[( 1 ( cosT = 2cos2 T - 1)
or, 2 quadrant] 2
of cos +
45° 1
2 lies
( (Value in 1st

? cos 45° = 1 + 1 = 2+1 × 2 (
2 22 2 22 2

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= 2+ 2 = 1 . 2+ 2
2 2

? cos 45° = 1 . 2 + 2 Proved.
2 2

Example 3. Prove that
Solution:
(a) tan T = sinT (b) 1 + sinT - cosT = tan T
Example 4. 2 1 + cosT 1 + sinT + cosT 2
Solution: sinT
(a) RHS = 1 + cosT

2sin T . cos T sin T T
2 2 2 2
= = = tan = LHS. Proved.
2cos2 T cos T
2 2
1 + sinT - cosT
1 + sinT + cosT
((b) LHS =

1 + (2sin T . cos T - 1- 2sin2 T
(( 2 2 2
=
( T T T
1 + 2sin 2 . cos 2 + 2cos2 2 -1
(
1 + (2sinT . cos T - 1+ 2sin2 T
(= ( 2 2 2

((2cosT sin T + cos T
2 2 2

2sin T cos T + sin T
(= 2 2 2
= RHS. Proved.
T T T
2cos 2 cos 2 + sin 2

Prove that

((a) tan Sc + A = secA + tanA
4 2
( ((b) cos2 Sc D Sc D D
4 - 4 - sin2 4 - 4 = sin 2

((a) Sc + A
LHS = 4 2

tan Sc + tan A 1+ sin A
4 2 2
= =
Sc $ $
1 - tan 4 . tan 2 1 - tan 2

1 + sin A
2

cos A cos A + sin A cos $
2 2 2 2
= = ×
$ $ A A
1 - sin 2 cos 2 cos 2 + sin 2

cos A
2

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cos $ + sin $ cos $ + sin $
2 2 2 2
= ×
A A A A
cos 2 - sin 2 cos 2 + sin 2

cos2 $ + sin2 $ + 2sin $ .cos $ 1 + sinA
2 2 2 2 cosA
= =
A A
cos2 2 - sin2 2

= 1 + sinA = secA + tanA = RHS. Proved.
cosA cosA
( ((b)
LHS = cos2 Sc - D (- sin2Sc-D
( (= cos2 4 4 (( 4 4
((
Sc - D = cos Sc - D = sin D = RHS Proved.
(Prove that (cosD + cosE)2 + (sinD + sinE)2 = 4cos244 2 2 2

Example 5: D-E
Solution: 2

LHS = (cosD + cosE)2 + (sinD + sinE)2

= cos2D + 2cosD cosE + cos2E + sin2D + 2sinD.cosD + sin2E

= (cos2D + sin2D) + (sin2E + cos2E) + 2(cosD.cosE + sinD.sin2E)

(= 2[1 + cos(D - E)] = 2.2cos2
= 1 + 1 + 2cos (D - E)

(= 4cos2 D-E ((
(( D-E 2 ((
( 2
= RHS. Proved. (
((
Example 6: (Proved that (( ( ( (1+ cos3Sc
1 + cos Sc ((
1+ cos5Sc 1+ cos7Sc = 1
(( 8
Solution: Here,
( ( { ( )}{ ( )}LHS. = (
( ( ( (= 1+ cos Sc 1+ cos3Sc 1 + cos S - 3Sc 1 + cos Sc - Sc
( (= 8 8
( { ( )}=
( (= 1+ cos Sc 1+ cos3Sc 1- cos3Sc 1- cosSc


1- cos2 Sc 1- cos23Sc


1- cos2 Sc 1 - cos2 Sc - Sc
2 8

1- cos2Sc 1- sin2Sc


= sin2 Sc. cos2 Sc

( (= 1
2
2sinSc. cos Sc

( (=1
2
sin2Sc


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( ( (= 1 sin2Sc = 1 1 = 1 = RHS Proved.
(

( (Example 7.
( ( ( (Solution: D 1 1 1 a2 +a12
If cos 2 = 2 a + a , prove that cosD = 2

Here, cos D = 1 a + 1 ,
a

cosD = 2cos2 D - 1
[ ( (]= 2

1 a + 1 2 -1
( (= a

2. 1 a2 + 2.a 1 +a12 -1
( (= 1 a
( (
a2 + 1 +1-1= 1 a2 + 1 = RHS. Proved.
a2 a2

Exercise 9.2

Very Short Questions

1. (a) Define sub-multiple angle with an example.

(b) Express sinT in terms of sin T and cos T .

T T T
(c) Express cosT in terms of cos , sin and tan .

(d) Express tanT in terms of tan T .

T
2. (a) Express sinT in terms of sin .

(b) Express cosT in terms of cos T .

T
(c) Express tanT in terms of tan .

3. (a) If sin T = , find the value of sinT.

T 3
(b) If cos = 2 , find the value of cosT.

(c) If tan T = 3, find the value of tanT.
(d) If = find the value of cosT.
T 4
cos 5 ,

Short Questions

4. (a) If cos30° = 23, find the values sin15°, cos15° and tan15°.
1
(b) If cos45° = 2 , then prove that :

(i) sin2212° = 1 2- 2 (ii) cos2212° = 1 2+ 2 (iii) tan2212° = 3-2 2
2 2

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(c) If cos330° = 3 prove the following :
2
3-1 3-1
(i) cos165° = - 22 (ii) sin165° = 22 (iii) tan165° = - (2 - 3)

5. Prove that:
( ( ( ((a)
1 - tan2 x (b) sinx = 3 sin x - 4sin3 x
2
cosx = x
1 + tan2 2
( ( ( ( ( ( ( (3tan x - tan3 x
cot3 x - 3cot x

1 - 3tan2 x 3cot2 x - 1
( ( ( ((c) tanx =
(d) cotx =

( ( ( (6.

(a) If cos T = 1 p + 1 , prove that cosT = 1 p2 + 1 .
( ( ( ((b) 2 p 2 p2

If cos T = 1 p + 1 , prove that cosT = 1 p3 + 1 .
( ( ( ((c) 2 p 2 p3

If sin T = 1 p+ 1 , prove that cosT = - 1 p2 + 1 .
2 p 2 p2
(7. Prove that:
((b) 1 + sinT =
(a) 1 - sinT = sin T - cos T ((2 sin T +cos T 2
2 2 2 2

(c) 1 + cosT = cot T (d) 1 + secT = cot T
sinT 2 tanT 2
1 + cosT
(e) cosecT - cotT =tan T (f) 1 - cosT = cot2 (2T)
2
A
1 - sinA 1 - tan 2 sin3 T - cos3 T 1
cosA 2 2 2
(g) = A (h) = 1 + sinT
1 + tan 2 sin T - cos T
2 2

1+ secT T 1+ sinT cos T + sin T
secT 2 cosT 2 2
(i) = 2cos2 (j) =
cos T - sin T
(((l) 2 2

((k) (( Sc T 1 - tan2 Sc - T ( T
(4 2 4 4 2
1 - 2sin2 - = sinT ( = sin
Sc T (
1 +tan2 4 – 4

Long Questions

( (8. Prove that:T T 2sinT - sin2T = tan2 T
(a) cos4 2 - sin4 2 = cosT 2sinT + sin2T 2
((b)
(c) 1 sin2T . cosT = tan T (d) 1 +sinT = tan2 Sc + T
+ cos2T 1 + cosT 2 1 - sinT 4 2

cos T 1+sinT T sin T 1+sinT T
2 2 2 2
(e) = tan (f) = cot
T T
sin 2 1+sinT cos 2 1+sinT

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9. Prove that
((a)
tan Sc + T(((( = secT + tanT
((b) 4 2( (((
Sc 1 - sinT
tan 4 – T (=1 + sinT
( ((c) 2 (

sec Sc + T(( . sec Sc – T = 2secT
((d) 4 2(( 4 2

tan Sc – T( = cosT
( ((e) 4 2(( 1 +sinT
2cosT
cot T + Sc ( – tan T – Sc = 1 +sinT
( ((f) 2 4 ( 2 4

tan Sc + T + tan Sc - T = 2secT
4 2 4 2

10. Prove that
((a) (cosD - cosE)2 + (sinD - sinE)2 = 4sin2
((b) (sinD + sinE)2 + (cosD + cosE)2 = 4cos2 D-E
2
D-E
2

11. Prove that
( ( ( ((a)
cos 2Sc . cos 4Sc . cos 8Sc . cos 16Sc = 1
( ( ( ((b) 15 15 15 15 16

1+cos Sc 1+cos 3Sc 1+cos 5Sc 1+cos 7Sc = 1
8 8 8 9 8

Project Work

12. Discuss how to find the value of tan 7 1° . Then show that
2
1°
tan 7 2 = 6– 3+ 2 – 2.

3. (a) 3 (b) 1 (c) - 3 (d) - 44
2 2 125
3 -1 3 + 1
4. (a) 2 2 , 2 2 , 2 – 3

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9.3 Transformation of Trigonometric Formulae

The sum or difference forms of trigonometric ratios can be transformed into the product
forms and the product forms can be transformed into the sum or difference forms.

(a) Transformation of product into sum or difference form formula of compound angles;
we have
sin (A + B) = sinA cosB + cosA. sinB ................ (i)
sin (A - B) = sinA. cosB - cosA.sinB ................. (ii)
cos (A + B) =cosA.cosB - sinA.sinB ................... (iii)
cos (A -B) = cosA. cosB - sinA.sinB ................... (iv)
Adding identities (i) and (ii), we get,
sin (A + B) + sin (A - B) = 2cosA.sinB.
Again, adding identities (iii) and (iv)
cos (A + B) + cos(A - B) = 2cosA.cosB
Subtracting identity (iii) from(iv), we get,
cos(A - B) - cos(A + B) = 2 sinA. sinB
Hence, we have the following formulae:
(I) 2sinA. cosB = sin(A + B) + sin(A - B)
(II) 2cosA. sinB = sin(A + B) - sin(A - B)
(III) 2cosA. cosB = cos(A + B) + cos(A - B)
(IV) 2sinA.sinB = cos(A -B) – cos(A + B)

(b) Transformation of Sum or difference into product
Let, A + B = C ..................... (v)

and A - B = D ...................... (vi)

Adding (v) and (vi), we get,

2A = C + D

or, A= C+D
2
Again subtraction (vi) from (v), we get,

2B = C - D

or, B= C–D
2
Now, express above four formulae (i), (ii), (iii) and (iv) in terms of C and D. We get,

(V) sinC + sinD = 2 sin C + D. cos C-D
2 2
C + D C-D
(VI) sinC - sinD = 2cos 2 . sin 2

(VII) cosC + cosD = 2cos C + D . cos C-D
2 2

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