Final PDF file of OPT Class 9 2077

EXERCISE : 4.10B

1. a. The co-ordinates of A, B and C are (6, 3), (– 3, 5) and (4, – 2) respectively and P is the point (x, y),

show that: ∆PBC = x + y – 2.
∆ABC 7

b. The co-ordinates of A, B and C are (3, 7), (1, – 1) and (6, 1) respectively and P is the point (x, y),

show that: ∆PAC = 2x +y– 13.
∆ABC 12

2. a. The vertices of a triangle ABC are A(2, 1), B(8, 3) and C(5, 6) and D is the mid-point of AB, then

prove that DABC = 2DADC.

b. If the co-ordinates of the vertices of ∆PQR are P(– 3, 2) Q(3, 4) and R(1, 6). A, B and C are the
mid-points of PQ, QR and RP respectively, prove that ∆PQR = 4∆ABC.

3. a. The co-ordinates of the points A, B, C and D are (6, 3), (– 3, 5), (4, – 2) and (a, 3a) respectively
b.
and ∆DBC = 21, find the value of a. Co-ordinate
∆ABC

If A and B are two points with co-ordinates (3, 4), (5, – 2) respectively, find a point P such that

PA = PB and ∆PAB = 10 sq. units.

4. P and Q are the mid-points of the sides BC and CA respectively of triangle ABC with vertices A(3, 8),
B(– 1, 2) and C(5, 4):

a. What are the co-ordinates of P and Q.

b. Find the areas of ∆ABC, ∆ABP and ∆BQC.

c. Find the ratio ∆ABC: ∆ABP: ∆BQC.

d. Prove that the area of ∆ABC: Trapezium ABPQ = 4:3.

5. a. Find the area of triangle formed by the lines x – y = 0, 2x – y = 0 and 3x – y + 4 = 0.

b. Find the area of triangle formed by the lines x + y – 6 = 0, x – 3y – 2 = 0 and 5x – 3y + 2 = 0.



Area of Triangles and Quadrilaterals ~ 147

Extra Properties of a Triangle

A A A A
R
DF O MP
B P G
O
BE O

C C BQ CB NC

The perpendicular The bisectors of angles The heights of any The medians of a

bisectors of sides of of any triangle meets triangle meets at a triangle meets at a

any triangle meets at a point and the point point and the point is point and the point is

at a point and the is known as In-centre. known as Orthocentre. known as Centroid.
point is known as Inscribed circle of

Circum-centre. the triangle can be

Circumcircle of constructed as shown.

the triangle can be

constructed as shown.

Lab Activity of Centroid of a Triangle

The centroid of a triangle divides each of the median in the ratio 2:1 from its vertex.
AD

PR C LN
G Q

BQ

Fig. (a)

A

PR C EMF
G
Fig. (c)
BQ
In fig. (a) and fig. (b) two different triangles are drawn on the graph
Fig. (b) paper. In each triangle the medians are drawn and they meet at
point G. In both of the figures, it is clear that each of the ratios
AG:GQ, BG:GR, CG:GP has the same value i.e. 2:1.

In figure (c), a triangle is constructed with DE = 4cm, EF = 5cm and DF = 6cm. Three medians DM, EN and
FL are drawn and they meet at common point Q. This is one of the beauty of any triangle that the centroid
of a triangle divides each of its medians in the ratio 2:1 from its vertex. In this experiment, if you measure
the parts of median DM, you will have DQ = 3cm and QM = 1.5cm. Hence, the ratio DQ:MQ = 3cm:1.5cm
= 2:1.
Similarly, the parts of the remaining two medians can be measured and check the ratio as mentioned in the
above statement. This is left for the students.
Also this type of activities can be performed in the case of orthocentre, circum-centre and in-centre.

148 ~ Perfect Optional Mathematics Class 9

5 TRIGONOMETRY

Curriculum Contents Included Periods

- Measurement of angle

- Introduction of trigonometric ratios sin θ
cos θ
- Trigonometric identities, e.g. sin2θ + cos2θ = 1, = tan θ problems and their
conversion
17
- Trigonometric ratios of standard angles (0°, 30°, 45°, 60°, 90°) using the concept

of unit circle

- Trigonometric ratios of (90° ± θ), (180° ± θ), (270° ± θ), (360° ± θ) and (– θ)

- Trigonometric ratios of compound angles

Objectives

On completion of this unit, students will be able to: Trigonometry
solve the problems of measurement of angles in different system as sexagesimal, centisimal and radian
measure.
solve the problems related to length of arc, radius and central angle of a circle.
solve and convert the problems of a trigonometric identities.
convert the trigonometric ratios into each other.
find the trigonometric ratios of the angle which is complement/supplement of the given angle.
solve the problems related to the trigonometric ratios of compound angles.

Materials

0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360°

sin 0 1 1 3 1 31 1 0 – 21 –  1 –  3 – 1 –  3 –  1 – 12 0
2 2 2 2 22 2 2 2 2

cos 1 31 1 0 – 21 –  1 –  3 – 1 –  3 –  1 – 12 0 1 1 3 1
2 22 2 2 2 2 2 2 2

tan 0 1 1 3 ∞ –  3 – 1 –  1 0 1 1 3 ∞ –  3 – 1 –  1 0
3 3 3 3

cosec ∞2 2 2 1 2 22 ∞ – 2 –  2 –  2 – 1 –  2 –  2 – 2 ∞
3 3 3 3

sec 1 2 22 ∞ – 2 –  2 –  2 – 1 –  2 –  2 2 ∞ 2 2 2 1
3 3 3 3

cot ∞ 31 1 0 –  1 – 1 –  3 ∞ 31 1 0 –  1 – 1 –  3 ∞
3 3 3 3

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Trigonometry 1 mark each 5 marks each Questions Marks
2 marks each 4 marks each
2 - 8 20
33

 ~ 149

5.1 Measurement of Angles

Introduction

Trigonometry is a branch of mathematics dealing with the measurement of sides and angles of a triangle.
We apply trigonometry in Engineering, Surveying, Astronomy, Geology etc.

System of Measurement of Angles

Generally the angles are measured in degrees, where 90° makes a right angle. But an angle can be measured
in following systems.
a. Sexagesimal System (Degrees/English System)
b. Centisimal System (Grades/French System)
c. Circular System (Radian Measure)

Sexagesimal System

In this system, the unit of measurement of an angle is a degree. Here a right angle is divided into 90 equal
parts and each part is known as a degree. Each degree is again divided into 60 equal parts and each part is
known as a minute. Each minute is further divided into 60 equal parts and each part is known as a second. A
degree, a minute and a second are denoted by (°), (') and (") respectively. So, we have the following relation
between degrees, minutes and seconds.

1 right angle = 90° (degrees)

1° (degree) = 60' (Sexagesimal minutes)

1' (minute) = 60" (Sexagesimal seconds)

∴ 40°26'13" means 40 degrees, 26 minutes and 13 seconds.

Centisimal System

In this system, the unit of measurement of an angle is a grade. Here a right angle is divided into 100 equal
parts and each part is known as a grade. Each grade is again divided into 100 equal parts and each part is
known as a minute. Each minute is further divided into 100 equal parts and each part is known as a second.
A grade, a minute and a second are denoted by (g), (') and (") respectively. So, we have the following
relation between grades, minutes and seconds.

1 right angle = 100g (Grades)

1g (grade) = 100' (Centisimal minutes)

1' (minute) = 100" (Centisimal seconds)

∴ 60g 20'19" means 60 grades, 20 minutes and 19 seconds B

Circular System r

In this system, the unit of measurement of an angle is a radian. Or A
Let O be the centre of the circle and OA = r be its radius. From A, cut AB = OA = r
and join OB. Then AOB = 1 radian. A radian or a circular measure is denoted by (c).
∴ 2c mean 2 radians.

150 ~ Perfect Optional Mathematics Class 9

Hence, an angle subtended by an arc equal in length to the radius of a circle at the centre of the
circle is known as one radian.
As the total length of circumference of a circle is 2pr units.

The angle subtended by cirumference of a circle at the centre of the circle is 2πr radian i.e. 2pc.
r

And 4 right angles = 2pc.

1 right angle = π c­
2

Theorem:

Radian is a constant angle.
Proof:

Let O be the centre of the circle and OP = r be the radius of the circle.

An PQ = r is taken PO and QO are joined. Produce PO to meet circle at R. Then by definition POQ = 1
radian. The diameter PR = 2r, POR = 2 right angles = 180° and

PQR = 1 × circumference = 1 × 2πr = πr.
2 2

Now, since the angles at the centre of a circle are proportional to the corresponding arcs on which they
stand.

i.e. PPOOQR = PQ Q
PQR
r Trigonometry

or, 1 radian = r R Or P
180° pr

or, 1 radian = 180 ­°
p

Since, 1 radian is independent of the radius of the circle, it is a constant angle.

Now, from the definitions of sexagesimal measure, centisimal measure and circular measure of angles

We have, 900 = 100g = πc = 1 right angle
2

Example 1:

Reduce 24g 20' 44" into centisimal seconds.

 The given angle is 24g 20' 44"

24g = (24 × 100 × 100)" = 240000"

20' = (20 × 100)" = 2000"

44" = 44"

24g 20' 44" = 242044"

Measurement of Angles ~ 151

Example 2:

Express 42° 20' 15" into the number of degrees.

 The given angle is 42° 20' 15"

42° = 42° °­ = 42°

20' = 20 = 0.333333°
60
15 °­
15" = 60 × 60 = 0.004166°

42° 20' 15" = 42.337499°

Example 3:

Express 48g 54' 68" into degrees, minutes and seconds.

 The given angle is 4­ 8g 54' 68"

= 48 + 54 + 68 g
100 100 × 100
­

= (48 + 0.54 + 0.0068)g

= 48.5468g

Now, we have the relation,

100g = 90° = 1 right angle 112° 123g

or, 1g = 9°
10
9 °
or, 48.5468g = 10 × 48.5468

= (9 × 4.85468)°

or, 48.5468g = 43.69212°

= 43° + 0.69212°

= 43° + (0.69212 × 60)'

= 43° + 41.5272'

= 43° + 41' + 0.5272'

= 43° + 41' + (0.5272 × 60)"

= 43° + 41' + 31.632"

= 43° 41' 31.63"

Hence, 48g 54' 68" = 43° 41' 31.63".

Example 4:

Express p radians into sexagesimal and centisimal measures.
6
 The p
given angle is 6 radians.

We have the relation,

pc = 90° = 1 right angle
2

152 ~ Perfect Optional Mathematics Class 9

or, 1c = 90 × 2 ° = 180 °
p p
or, p6 c = °
180 × p = 30°
p 6

p c
2
Again, = 100g

100 g 200 g
p p
or, 1c = × 2 =

or, p6 c = 200 p g
p 6
×

= 200 g = 3332 g
6

Example 5:

Reduce the following angles into radian measure.

a. 42g 75' b. 42g 15' 30"

 a. 42g 75' = 75 g
100
42 +

= (42 + 0.75)g

= 42.75g

p c
2
Now, 100g = = 1 right angle

or, 1g = pc Trigonometry
200
c
p
or, 42.75g = 200 × 42.75

= 0.21375πc

∴ 42g 75' = 0.21375πc

b. 42° 15' 30"

= 42 + 15 + 30 °
60 60 × 60

= (42 + 0.25 + 0.00833)°

= 42.25833°

Now, 90° = pc = 1 right angle
2
pc
or, 1° = 180

or, 42.25833°
= 1p80 × 42.25833 c

= 0.2348πc

∴ 42° 15' 30" = 0.2348πc

Measurement of Angles ~ 153

Example 6:

Express 40g 30' 82" as a decimal of a right angle.

 1 right angle = 100g = (100 × 100 × 100)" = 1000000"

and 40g 30' 82" = (40 × 10000 + 30 × 100 + 82)" = 403082"

Now, 1000000" = 1 right angle

or, 1" = 1 right angle
1000000
403082
or, 403082" = 1000000 right angle

Example 7:

One acute angle of a right angled triangle is three tenth of a right angle. Find the other acute angle

in grades.

 One acute angle of a right angled triangle,

= 3 of a right angle
10
3
= 10 × 100g = 30g

Another acute angle = 100g – 30g = 70g.

Example 8:

One angle of a triangle 72°. The remaining angles are in the ratio 1:2. Find all the angles in grades.

 10 g
9
One angle of a triangle = 72° = 72 × = 80g

Let the remaining two angles x and 2x in grades.

Hence, x + 2x + 80g = 200g (sum of all angles of a triangle)

or, 3x = 120g

∴ x = 40g

Hence, the remaining two angles of the triangle in grades measure are 40g and 80g.

Example 9:

7x g 9x ° px  c
2 4 50
The angles of a triangle are , and . Find the angles of the triangle in degrees.

 The first angle of the triangle = 7x g 7x 9 ° 63x °
2 2 10 20
= × =

The second angle of the triangle = 9x °
4

px c px 180 g 18x °
50 50 p 5
The third angle of the triangle = = × =

Now, the sum of all angles of a triangle is 2 right angles.

154 ~ Perfect Optional Mathematics Class 9

i.e. 6230x ° + 9x ° + 18x ° = 180°
4 5

or, 63x + 45x +72x ° = 180°
20
or, 18200x ° = 180°

∴ x° = 20°

Hence, the angles of the triangle are 63 × 20 °, 9 × 20 ° and 18 × 20 ° i.e. 63°, 45° and 72°.
20 4 5

Example 10:

Sum of the first and the second angle of a triangle is 150°. The ratio of the number of grades in the

first angle to the number of degrees in the second angle is 5:3. Find the angles of the triangle in

circular measure.

 Suppose number of grades in the first angle be 5x and the number of degrees in the second angle be 3x.

Hence, the first angle = 5xg = 5x × 9 °= 9x °
10 2

The second angle = 3x°

From question, 92x° + 3x° = 150°

or, 9x° + 6x° = 150°
2
Trigonometry
or, 15x° = (150 × 2)°

∴ x° = 20°

The first angle = 9x ° = 9 × 20° = 90°
2 2

The second angle = 3x° = 3 × 20° = 60° and

The third angle = 180° – 150° = 30°

The angles of the triangle in circular measure are

p c p c p c
180 180 180
90 × , 60 × and 30 ×

i.e. p2c, pc and p6c.
3

Exterior and Interior Angles of a Regular Polygon EF

A polygon in which all the sides and the angles (internal) are equal is DC
called a regular polygon. The exterior angles are also equal in regular
polygon. AB

We know a polygon with n sides can be divided into (n – 2) triangles No. of sides = 6
from a fixed vertex. Then the sum of all interior angles of a polygon of No. of triagles = 6 – 2 = 4
n sides = (n – 2) × 180°

or, (n – 2) × 200g

Measurement of Angles ~ 155

As all the interior angles of a regular polygon are equal and there are n angles in a polygon with n sides, if
x be the degree measure of an interior angle of a regular polygon of n sides,

Sum of all angles = (n – 2) × 180°

or, n × x = (n – 2) × 180°

∴ x = n – 2 × 180°
n

Similarly, all the exterior angles of a regular polygon are equal and there are n exterior angles in a polygon

with n sides. The sum of all exterior angles of any polygon = 360° or 400g. If q be the degree measure of an

exterior angle of a regular polygon of n sides, AF

Sum of exterior angles = 360°

or, n × θ = 360° BE

∴ θ = 360°
n
Int. angle Ext. angle

Note: CD G

a. The sum of an exterior and an interior angle of a regular polygon = 180° or 200g.

b. The size of an interior angle increases as number of sides of a regular polygon increases.

c. The size of an exterior angle decreases as number of sides of a regular polygon increases.

Example 11:

Find the interior angle of a regular pentagon in degrees.

 In a regular pentagon, number of sides (n) = 5

Interior angle (x) = ?

We know that = n – 2
n
Interior angle (x) × 180°

= 5 – 2 × 180°
5

= 3 × 36° = 108°

Example 12:

The ratio of an exterior and an interior angle of a regular polygon is 1:3. Find the number of sides

of the polygon.

 Let an exterior and an interior angle of a regular polygon are x and 3x degrees respectively. Then

x + 3x = 180°

or, 4x = 180°

or, x = 45°

∴ Exterior angle = x = 45°

Interior angle = 3x = 135°

And an exterior angle = 360°
n

156 ~ Perfect Optional Mathematics Class 9

or, 45° = 360°
n

or, n = 360°
45°

∴ n = 8

∴ The required number of sides = 8.

Example 13:

If the ratio of the number of sides of two regular polygons is 5:4 and the sum of their interior angles

is 279°. Find the number of sides of each regular polygon.

 Let the number of sides of two regular polygons be 5x and 4x respectively. The interior angle of the

first regular polygon is

x1 = n1 – 2 × 180°
n1

or, x1 = 5x – 2 × 180°
5x

or, x1 = 36(5x – 2)° and
x

The interior angle of the second regular polygon is

x2 = n2 – 2 × 180°
n2

or, x2 = 4x – 2 × 180° Trigonometry
4x

or, x2 = 90(2x – 1)° and
x

As the sum of their interior angles = 279°

or, x1 + x2 = 279°

or, 36(5xx– 2)° + 90(2x – 1)° = 279°
x

or, 180x – 72 + 180x – 90° = 279°
x

or, 360x – 162 = 279x

or, 360x – 279x = 162

or, 81x = 162

∴ x = 2

Hence, the number of sides of given two regular polygons are 5x and 4x i.e. 10 and 8.

Angle made by hands of a clock

In clock, hour hand is comparatively shorter than minute hand.
In clock, total hours is 12 and total minutes is 60 minutes.
In 12 hours, the hour hand turns through 360°.

In 60 minutes, the minute hand turns through 360°.

Measurement of Angles ~ 157

Example 14:

Find the angle between hour hand and minute hand at 4:30.

 In, 4 hours and 30 minutes

Total hours = 4 hours + 30 minutes

= 4 + 30 hours
60

= 4 + 1 hrs = (4 + 0.5) hours
2

= 4.5 hrs

In 12 hours, the hours hand turns through 360°.

In 1 hour, the hour hand turns through 31620°.

In 4.5 hours, the hour hand turns through 360° × 4.5 = 30 × 4.5 = 135°.
12

In 60 minutes, the minute hand turns through 360°.

In 1 minute, the hour hand turns through 31620°.

In 30 minutes, the hour hand turns through 6 × 30 = 180°

The angle between the hour hand and minute hand

= 180° – 135° = 45° EXERCISE : 5.1A

1. Reduce the following angles into sexagesimal seconds.

a. 45° b. 15° 20" c. 70° 20' 10"

2. Express the following angles in degrees.

a. 25° 15' 30" b. 15° 25’ 10" c. 6° 4' 8"

3. Express the following angles in grades:

a. 42g 40' 3" b. 29g 3' 8" c. 4g 80' 29"

4. Express the following angles in centisimal minutes.

a. 40g 28' b. 21g 3' 48" c. 2g 82' 4"

5. Express the following angles in degrees, minutes and seconds.

a. 28g 14' 13" b. 10g 8' 42" c. 81g 62' 28"

6. Express the following angles in degrees and grades.

a. p3c b. 34pc c. 56pc
7. Express the following angles in radian measures.

a. 45° b. 150g c. 60°

158 ~ Perfect Optional Mathematics Class 9

8. Express the following angles in the circular measures.

a. 40° 20' 15" b. 25g 32' 28" c. 6° 18' 52"

9. Express the following angles as a decimal of a right angle.

a. 24g 32' 61" b. 40° 25' 12" c. 2pc
7

10. a. One acute angle of a right angled triangle is 27°. Find the measure of another acute angle in

grades.

b. Two angles of a triangle are 70g and 80g. Find all the angles of the triangle in degrees.

c. The angles of a triangle are in the ratio 4:7:9. Find all the angles of the triangle in degrees.

d. The difference of the two acute angles of a right angled triangle is 36°. Find all the angles of the
triangle in degrees.

11. a. If D be the number of degrees and G be the number of grades of an angle, prove that:

i. D9 = 1G0 ii. G = D + D
9

b. If D be the degree measure and C be the circular measure of an angle, show that: D = 2pC.
90

c. If R and r denote the number of English and French minutes of an angle, prove that: 2R7 = 5r0.

12. Find the degree measures of the interior angles of the following regular polygons.

a. Pentagon b. Hexagon c. Heptagon Trigonometry

d. Octagon e. Nonagon f. Decagon

13. Find the number of sides of regular polygons whose exterior angles are as follows.

a. 72° b. 45° c. 36°

EXERCISE : 5.1B

1. a. One angle of a triangle is 50g. The remaining two angles are in the ratio 2:1. Find the angles of
b.
the triangle in degrees.
c.
d. One angle of a triangle is 2 of a right angle. The greater of the other two exceeds the lesser by
3
2. a.
b. 20°. Find the all angles of the triangle in grades.

One angle of a triangle is 1 of a right angle. If the difference between remaining two angles is
3
30°, find the angles of the triangle in circular measure.

The sum of two angles of a triangle is 150g and the difference of the same two angles is 25°. Find
the angles of the triangle in:

i. degrees ii. grades iii. radians

The angles of a triangle are in the ratio 3:7:10. Find the angles of the triangle in degrees, grades
and radians.

The angles of a quadrilateral are in the ratio 1:2:3:4. Find the angles of the quadrilateral in
degrees as well as grades.

Measurement of Angles ~ 159

px c 3x ° 2x g
75 2 3
3. a. The angles of a triangle are , and . Find the angles of the triangle in degrees.

b. The angles of a triangle are 13x °, 67x g and px c Find all the angles of a triangle in degrees.
4 18 75
.

c. The three angles of a triangle are 7x g 3x ° and px c find the measure of each of the angle in
grades. 3 2 75
, ,

4. a. In a triangle, the first angle is greater than the second angle by 18° and less than the third by 10g.
Express all the angles of the triangle in grades.

b. Find each of the base angles in degrees and radians of an isosceles triangle whose vertical angle

is 4 of its base angle.
7

c. The angles of a triangle are (a – d)°, a° and (a + d)°. If the number of degrees of the greatest angle

is 80, find all the angles in radian.

5. a. Divide an angle of 150g in two parts such that the ratio of the number of degrees in the first part
and the number of grades in the second part is 27:20. Find each part of the angle in circular
measure.

b. Sum of the two angles of a triangle is 150° and the ratio of the number of grades in the first part
to the number of degrees in the second part is 10:21, find the angles in circular measure.

6. a. The number of grades in an angle of a regular polygon exceeds the number of degrees in it by 16.
Find the number of sides of the polygon.

b. If an exterior angle of a regular polygon is 1 of an interior angle of a regular polygon. Find the
number of sides in the polygon. 3

7. a. The number of sides in two regular polygon are in the ratio 3:4 and the sum of their angles is
255°. Find the number of sides of the polygon.

b. If the ratio of the number of sides of two regular polygons is 1:2 and the ratio of their interior
angles is 3:4, find the number of sides of the polygons.

c. The ratio of the number of sides of two regular polygons is 3:1 and the sum of their interior
angles is 300°. Find the number of sides of the polygons.

8. a. Find the angle made by minute hand and hour hand of a clock at 3:30 in circular measure.

b. Find the angle made by minute hand and hour hand of a clock at quarter to 6 in degree measure.

9. a. The number of degrees in an angle of a triangle is to the number of grades in the second is to the

number of radians in the third as 72:70:p4, find the angles in degrees.

b. If G, D and C be the number of grades, degrees and radians in any angle prove that G – D = 20C
p



160 ~ Perfect Optional Mathematics Class 9

5.2 Length of Circular Arc

Introduction

There is a relationship between the arc of a circle, the circular measure of the central angle and the radius
of the circle.

The radian measure of an angle at the centre of a circle is equal to the ratio CB
of the subtending arc to the radius of the circle.
r
Let O be the centre of the circle and OA = r be the radius of the circle cut
an arc AB = r and join OB. Then by definition, AOB = 1 radian. Also let qc A
arc ABC = l and AOC = θc. 1c

Or

Since, the angles at the centre of a circle are proportional to their
corresponding arcs on which they stand,

i.e. AAOOCB = ABC
AB

or, θ1 rraaddiiaann = l
r

or, θ = l
r

Note: To apply formula θ = lr, θ must be in radian. Trigonometry

Example 1:

Find the angle subtended at the centre of a circle of radius 5.6cm by an arc of length 22cm in degrees.

 Here, radius of the circle (r) = 5.6cm

length of arc (l) = 22cm. 22cm

Angle at the centre (θ) = ? q
l
By using formula, θ = r = 22cm = 220 O 5.6cm A
5.6cm 56

or, θ = 110 radian
28

The angle of the centre (θ) B

= 110 radian
28

= 110 × 180 °
28 π

= 110 × 180 °
28 22

= 10 × 180 ° = 225º.
4 2

Length of Circular Arc ~ 161

Example 2:

The given figure is a part of the circle with centre O and the arc AB. If OA = 14cm and AOB = 3π c
4
,

calculate the length of the arc AB. Also, find the perimeter of the figure OAB.

 The radius of the given sector is = 14cm.

The angle at centre (θ) = 3πc O
4
Length of the arc AB (l) = ? 3p c
4
Perimeter of sector OAB = ?
AB
We have, θ = l
r
or, 34πc l
= 14

or, l = 3 × 22 × 14
4 7

or, l = 33

Hence, the length of arc AB = 33cm.

Perimeter of sector OAB = OA + OB + AB
= 14cm + 14cm + 33cm
= 61cm

Example 3:

A horse is tied to a stake by rope of length 50m. If the horse moves around the boundary of the

circle always keeping the rope tight, find how far will it have gone when the rope has traced out an

angle of 63º at the centre. B l=?

 Here, radius of the circle (r) = 50m qc A
1c
Length of an arc AB (l) = ?
O 50m
Central angle (q) = 63º p c 7p c
180 20
= 63 × =

By using formula,

θ = l
r

or, l = r × θ

or, l = 50 × 7p
20

or, l = 50 × 7 × 22
20 7
or, l = 55

Hence, the horse has gone about 55 metres.

162 ~ Perfect Optional Mathematics Class 9

Example 4:

If the radius of the earth is 6400km and the distance of the moon from the earth is 60 times the

radius of the earth, find the radius of the moon which subtends at the earth the angle of 16'.

 Here, O be the position of the earth and OA = OB is the distance between the earth and the moon.

As the distance between moon and earth is very large, A
diameter AB of moon can be considered as the arc of a circle
with centre O and radius OA. Moon16' O

Hence, radius of circle (r) = 60 × 6400km TrigonometryEarth

Diameter of the moon (AB = l) = ? B
Central angle (θ) = 16'

or, θ = 16 °
60
c
16 p
or, θ = 60 × 180

∴ θ= pc
675

By using formula,

θ = l
r

or, l = θ × r

or, l = p × 60 × 6400
675

or, l = 22 × 60 × 6400
7 675

or, l = 1787.937

Hence, the diameter of the moon AB = 1787.937km
and its radius = 17872.937 km = 893.968 km.

Example 5:

What is the ratio of the radii of two circles at the centre of which two arcs of the same length

subtend angles of 80° and 120°?

 Let r1 and r2 be the radii of the first and the second circles respectively. If l be the length of arc of

both circles
l

Then, r1 = θ1

= circular measure of 80°

= 80 × p
180

= 49π … … … (1)

Length of Circular Arc ~ 163

And, l = θ2
r2

= circular measure of 120°
p
= 120 × 180

= 23π … … … (2)

Now, from (1) and (2), we have EXERCISE : 5.2A

l/r2 = 2π/3
l/r1 4π/9

or, rr12 = 3
2

Hence, the required ratio is 3:2.

1. a. The radius of a circle is 14cm. An arc of the circle subtends an angle of 60° at the centre. Find
the length of the arc.

b. The radius of a circle is 21cm. An arc of the circle subtends an angle of 30º at the centre. Find the
length of the arc.

c. ACB is an arc of a circle with centre O. If AO = OB B

= 11.2cm and AOB = 3π c determine the C B l
4
, O 60°

length of the arc ACB. 3pc O 21cm A
4
d. In the given figure O is the centre of the circle and Fig.1(d)
the radius OA = 21cm. If the central angle AOB A Fig.1(c)
is 60°, find the length of arc AB.

e. A horse is tied to a pole by a rope of length 11m. It moves to trace a circle and subtends an angle

of 70°. Find the distance travelled by the horse. B

f. In the given figure, AOB is a quarter circle with

centre O. If OA = 7cm, find the perimeter of AOB. O

2. a. The radius of a circle is 7cm. Find the angle in degrees O 7cm A AB
subtended by an arc of 323 cm at the centre of the circle.
Fig.1(f) Fig.2(c)
b. What angle will be subtended by a tower 220 metres
long at the eye of an observer at a distance of 5.6km.

c. The perimeter of the adjoining sector is 64m. Find the AOB in degrees, if its radius is 21m.

d. A pendulum of 0.8m long and describes an arc of 12cm, when it swings. Through what angle
does it swing?

3. a. How long is the radius of the circle if an arc of length 11 metres subtends an angle 75º at the
centre of the circle.

b. Find the radius of a circle for which an arc 30cm long subtends an angle of 45° at the centre.

c. A cow is tethered to a stake. How long must the rope be in order to that, when the cow has moved
through 44m at the extremity of the rope, the angle traced out by the rope may be 72°?

164 ~ Perfect Optional Mathematics Class 9

EXERCISE : 5.2B

1. a. The circumference of a circle is 6.4m. Find the angle in the sexagesimal system as well as in
centisimal system subtended by an arc of length 44cm at the centre of the circle.

b. A railway train is travelling on a curve of 1500 metres radius at the rate of 66 kilometres per hour,

through what angle has it turned in 10 seconds ?
c. A train is travelling a circular track of 123 km radius at 44km per hour. Find the angle through

which it turns in 1 minute.

d. A man is running in circular track of radius 33m at 22m per 5 minutes. Find the degrees of the
angle through which it turns in 10 minutes.

2. a. A horse is tied to a stake by a rope if the horse move along a circular path of length 176m always

keeping the rope tight and it has traced out 80g at the centre, find the length of the rope.

b. The length of a pendulum is 4 metres while the pendulum swings through 1 radians, find the
2
length of the arc through which the tip of the pendulum passes.

c. A person of normal eye-sight can read print at such a distance that the letters subtend an angle 5'
at the eye. Find the height of the letters he can read at a distance of 2640 metres.

d. A man is running along a circular path of radius 700m. If he runs through the circular path which
subtends an angle 75º in 10 minutes, find his speed in km per hour.

e. The distance between the Sun and the Earth is 150000000km (approximately) and the Sun Trigonometry
subtends an angle of 32' with eye of man of the Earth, find the diameter of the Sun.

3. a. A railway curve is to be laid out on a circle. What radius should be used if the track is to change
direction by 20º at a distance of 88m?

b. Assuming that the Earth's diameter is 6336km and that subtends an angle of 56' at the centre of
the Moon, find the distance of the Moon from the Earth.

4. a. What is the ratio of the radii of two circles at the centres of which two arcs of same length
subtend angles of 150º and 75º?

b. The radius of one circle is 4 times that of a second, compare an arc subtending 45° at the centre
of the first with arc subtending 60° at the centre of the other.

5. a. In a circle of diameter 20cm, the length of a chord is 10cm. Find the length of the minor arc
divided by the chord.

b. An arc AB of a circle with radius 28cm and centre O subtends an angle AOB at the centre. If
length of the arc AB is 2931 cm, find the length of chord AB.

c. Find the measure of minor arc of a chord 21cm long in a circle having diameter 42cm

d. Find the measure of minor arc of a chord 28cm long in a circle having diameter 56cm.



Length of Circular Arc ~ 165

5.3 Trigonometric Ratios

Introduction

Trigonometry deals with the measurement of triangles, that is, it deals with the relationship between the
sides and angles of triangles.

Let q be an angle formed by rotating the initial line OA in QB
anticlockwise direction to the terminal line OB. From any arbitrary

point P on OB, draw perpendicular PM to OA. The ∆POM is a right P
angled triangle with reference to angle q, the side PM, the opposite

side of q, is called perpendicular and side OM is called the base of

right angled ∆POM. OP is the hypotenuse of right angled ∆POM. q MNA
There are six different ratios of sides of ∆POM and they are
O

PM , OOMP , OPMM, OP , OP and OPMM.
OP PM OM

These ratios of sides are called trigonometric ratios. If we draw QN⊥OA from any another point Q on OB,

∆POM~∆QON and trigonometric ratios of ∆QON are QN , ON , QN , OQ , OQ and ON .
OQ OQ ON QN ON QN
PM QN
As ∆POM ~ ∆QON, we have OP = OQ etc.

The trigonometric ratio of the perpendicular to the hypotenuse is called the sine of the angle of reference θ

and written as sin θ,

thus sin θ = Perpendicular = PM = QN i.e. sin θ = p
Hypotenuse OP OQ h

Similarly.

The ratio, Base = OM = b is known as cosine of an angle θ
Hypotenuse OP h

i.e. cos θ = b = OM
h OP

The ratio, Perpendicular = PM = p is known as tangent of an angle θ
Base OM b

i.e. tan θ = p = PM
h OM

The ratio, Hypotenuse = OP = h is known as cosecant of an angle θ
Perpendicular PM b

i.e. cosec θ = h = OP
p PM

The ratio, Hypotenuse = OP = h is known as secant of an angle θ
Base OM b

i.e. sec θ = h = OP
b OM

166 ~ Perfect Optional Mathematics Class 9

The ratio, Base = OM = b is known as cotangent of an angle θ
Perpendicuar PM p

i.e. cot θ = b = OM
p PM

Let us consider an angle θ placed in the standard position. Draw Y
circle with centre at O and radius r, let the circle intersect the terminal P(x, y)
line at some point P(x, y) as shown in the figure.
ry
Draw PN⊥OX, then ON = x, PN = y and OP = r. For PON = θ, X' q X

six trigonometric ratios can be defined in terms of x-coordinate 'x', O xN

y-coordinate 'y' and the radius 'r' by the formula,

sin θ = yr cos θ = xr tan θ = y Y'
x

cosec θ = yr sec θ = xr cot θ = x
y

Note:

a. The trigonometric ratios are independent of size of sides of triangle.

b. sin θ is a single term. It is not the product of sin and θ i.e. sin and θ can not be separated.

c (sin θ)2 is generally written as sin2θ.

Relationship Between Trigonometric Ratios

From above definitions of trigonometric ratios we can establish the following relations.

sin θ = hp, cosec θ = h Trigonometry
p
1 1
∴ sin θ . cosec θ = 1, sin θ = cosec q, cosec θ = sin q

cos θ = hb, sec θ = h
b
1 1
∴ cos θ . sec θ = 1, cos θ = sec q, sec θ = cos q

tan θ = pb, cot θ = b
p
1 1
∴ tan θ . cot θ = 1, tan θ = cot q, cot θ = tan q

tan θ = sin qq, cot θ = cos qq.
cos sin

Example 1:

In ∆ABC, AB = 5cm, BC = 12cm, find cos C and sin C. A

 We have AB = 5cm, BC = 12cm 5cm

Using pythagoras theorem in right angled triangle ABC, we have q
AC2 = AB2 + BC2
or, AC2 = (5cm)2+ (12cm)2 B 12cm C
or, AC2 = 25cm2 + 144cm2
or, AC2 = 169cm2
∴ AC = 13cm

Trigonometric Ratios ~ 167

Hence, cos C = BC = 12cm = 12 and
AC 13cm 13

sin C = AB = 5cm = 153.
AC 13cm

Example 2:

From the given triangle, find the trigonometric ratios of cos b and sin θ. 5 cmA

 From right angled triangle ABD, BD is perpendicular and AD is base b

with reference angle b, hence cos b = AADB. D
From right angled triangle BCD, BD is base and DC is perpendicular with
reference angle θ, hence sin θ = DBCC. q

Trigonometric Relations (Using Pythagoras Theorem) B 12 cm C

From right angled triangle ABC with ABC = 90°, we have the pythagoras relation, AC2 = AB2 + BC2.

If ACB = θ the angle of reference, we have h2 = p2 + b2 A
Now, sin2θ + cos2θ

= p2 + b2
h h
q
= p2 + b2
h2 h2 BC

= p2 + b2
h2

= h2
h2

∴ sin2θ + cos2θ = 1 ... ... ... ... (1)

and sin2θ = 1 – cos2θ ... ... ... ... (2)
cos2θ = 1 – sin2θ ... ... ... ... (3)
Also, sin θ = 1 – cos2θ ... ... ... ... (4)
cos θ = 1 – sin2θ ... ... ... ... (5)
Again, sec2θ – tan2θ
... ... ... ... (6)
h ­  2 p ­  2
b b
= –

= h2 – p2
b2 b2

= h2 – p2
b2

= b2
b2

= 1.
∴ sec2θ – tan2θ = 1

168 ~ Perfect Optional Mathematics Class 9

sec2θ = 1 + tan2θ ... ... ... ... (7)
tan2θ = sec2θ – 1 ... ... ... ... (8)
sec θ = 1 + tan2θ ... ... ... ... (9)
tan θ = sec2θ – 1 ... ... ... ... (10)
Lastly, cosec2q – cot2q

h ­  2 b  ­ 2
p p
= –

= h2 – b2
p2 p2

= h2 – b2
p2

= p2
p2

= 1. ... ... ... ... (11)
∴ cosec2θ – cot2θ = 1 ... ... ... .. (12)
... ... ... ... (13)
cosec2θ = 1 + cot2θ ... ... ... ... (14)
... ... ... ... (15)
cot2θ = cosec2θ – 1

cosec θ = 1 + cot2θ

cot θ = cosec2q – 1

Methods of Proving the Trigonometric Identities Trigonometry

To prove a trigonometric identity. We follow any one of the following procedures,
a. Starting from the expression on the left hand side (LHS), we bring the right hand side (RHS).

b. Starting from the expression on the right hand side (RHS), we bring the left hand side (LHS).

c. Starting from LHS and RHS, bring the common result.

d. Get the new identity by transposing or by cross multiplying and prove by using any of above
procedure.

Example 3:

Find the product of sin θ (sin θ + cos θ) cos θ (sin θ – cos θ).

 The given expression is

sin θ (sin θ + cos θ) cos θ (sin θ – cos θ)
= sin θ cos θ (sin2θ – cos2θ)
= sin3θ cos θ – sin θ cos3θ.
Example 4:

Simplify: 1 cos φ φ + 1 cos φ φ
– sin + sin

Trigonometric Ratios ~ 169

 Here, 1 cos f φ + 1 cos f φ
– sin + sin

= cos f (1 + sin φ) + cos φ( 1– sin φ)
(1 – sin φ) (1 + sin f)

= cos f + cos f sin φ + cos φ – cos f sin φ
(1 – sin φ) (1 + sin f)

= 2 cos f = 2 cos f = 2 φ = 2 sec f
1 – sin2φ cos2φ cos

Example 5:

Show that : 1 – sin4A = 2 sec2A – 1
cos4A

 LHS = 1 – sin4A = 12 – (sin2A)2
cos4A (cos2A)2

= (1 + sin2A) (1 – sin2A)
cos2A. cos2A

= (1 + sin2A) cos2A = 1 + sin2A
cos2A. cos2A cos2A

= 1 + sin2A = sec2A + tan2A
cos2A cos2A

= sec2A + sec2A – 1 = 2 sec2A – 1
= RHS (Proved)

Example 6:

Prove that: 1 cos2A + 1 sin2A = 1 + cos A sin A
– tan A – cot A

 LHS = 1 cos2A + 1 sin2A
– tan A – cot A

= cos2A + sin2A
1
– sin A 1 – cos A
cos A sin A

= cos cos2A A + sin2A
A – sin sin A – cos A

cos A sin A

= cos cos3A A – sin3A
A – sin cos A – sin A

= cos cos3A A + sin sin3A A
A – sin A – cos

= cos cos3A A – cos sin3A A
A – sin A – sin

= ccooss3AA – sin3A
– sin A

170 ~ Perfect Optional Mathematics Class 9

= (cos A – sin A) (cos2A + sin A cos A + sin2A)
cos A – sin A

= cos2A + cos A sin A + sin2A

= 1 + cos A sin A

= RHS. Proved.

Example 7:

Prove that : tan A + sec A – 1 = 1 + sin A
tan A – sec A + 1 cos A

 LHS = tan A + sec A – 1
tan A – sec A + 1

= (tan A + sec A) – (sec2A – tan2A) [ 1 = sec2A – tan2A]
tan A – sec A + 1

= (tan A + sec A) – (sec A + tan A) (sec A – tan A)
tan A – sec A + 1

= (tan A + sec A) [1 – (sec A – tan A)]
tan A – sec A + 1

= (tan A + sec A) (1 – sec A+ tan A)
(1 – sec A + tan A)

= tan A + sec A

= sin A + 1 = 1 + sin A
cos A cos cos A
A

= RHS proved. Trigonometry

Example 8:

Prove that: cosec 1 cot A – 1 A = 1 A – cosec 1 cot A
A+ sin sin A–
 By transposing, we get the new identity

cosec 1 cot A + cosec 1 cot A = 1 + 1 = 2
A+ A– sin A sin A sin A

LHS = cosec 1 cot A + cosec 1 cot A
A+ A–

= 1 1 + 1
sin A
+ cos A 1 – cos A
sin A sin A sin A

= 1 + 1 + 1 – 1 = 1 sin A + 1 sin A
cos A cos A + cos A – cos A

sin A sin A

= sin A(1 – cos A) + sin A(1 + cos A)
(1 + cos A)(1 – cos A)

= sin A – sin A cos A + sin A + sin A cos A
1 – cos2A

= 2sisnin2AA = 2 A
sin

= RHS proved.

Trigonometric Ratios ~ 171

Example 9:

Prove that : sec4A – sec2A = tan4A + tan2A

 LHS = sec4A – sec2A

= sec2A(sec2A – 1)
= (tan2A + 1) tan2A
= tan4A + tan2A
= RHS Proved

Example 10:

Prove that : (3 – 4 sin2A) (sec2A – 4 tan2A) = (3 – tan2A) (1 – 4 sin2A)

 LHS = (3 – 4 sin2A) (sec2A – 4 tan2A)

= (3 – 4 sin2A) 1 – 4 sin2A
cos2A cos2A

= (3 – 4 sin2A) 1 – 4 sin2A
cos2A

= 3 – 4 sin2A (1 – 4 sin2A)
cos2A

= co3s2A – 4 sin2A (1 – 4 sin2A)
cos2A

= (3 sec2A – 4 tan2A) (1 – 4 sin2A)
= {3(1 + tan2A) – 4 tan2A} (1 – 4 sin2A)
= (3 + 3 tan2A – 4 tan2A) (1 – 4 sin2A)
= (3 – tan2A) (1 – 4 sin2A)
= RHS (Proved)

EXERCISE : 5.3A

1. a. From the figure, find the trigonometric ratios of sin θ and P
cos α.
q

b. ABC is a right angled triangle, in which ABC = 90°, AB 5 cm a Q
= 5cm, AC = 13cm. Find the values of sin q and tan θ.
R 13 cm
A q

B C

172 ~ Perfect Optional Mathematics Class 9

c. In the given figure, AB = 3cm, BC = 4cm and ABC = 90°, A
find the values of cos α, tan α, cosec θ and sec θ.
a

6 cm 3 cm q
4.8 cm
B 4 cm C

2. a. In the given figure, PS⊥QR, PQ = 6cm, PS = 4.8cm and P
PR = 8cm,
a 8 cm
i. Prove that ∆PQR is a right angled triangle.

ii. Find the values of cos θ and sin α. QS q

R

b. In right angled ∆PQR, PQR = 90°, QR = 2 units and PQ = 2 2 units, find sin R, tan P and cot R.

3. a. Express all the trigonometric ratios in terms of cot θ.

b. Express all the trigonometric ratios in terms of sec θ.

4. Find the product of

a. (cos φ + sin φ) (cos φ – sin φ) (cos2φ + sin2φ)

b. (sin θ – cos θ) (sin2θ + sin θ cos θ + cos2θ)

c. (2 sec θ + 3 tan θ) (2 sec θ – 3 tan θ)

d. (1 – tan θ) (1 + tan θ) (1 + tan2θ) (1 + tan4θ)

5. Factorise the following expressions.

a. tan3A + cot3A b. 1 – cos4θ Trigonometry

c. cos2α – 5 cos α + 6 d. 3 tan2θ + 2 tan θ – 8

Prove the following identities.

6. a. sin2A – sin2B = cos2B – cos2A b. (sec2A – 1) cos2A = sin2A

c. cot2θ sin2θ + tan2θ cos2θ = 1 d. sin2α + tan2α sin2α = tan2α

e. sec2φ – tan2φ = sin2φ f. sec φ sin φ cosec2φ – 1 = 1
1+ cot2φ

g. 1 – 2 sin θ cos θ = cos q –  sin q h. (1 + tan2θ) (1 – cos2θ) = tan2θ

i. sec2φ – cosec2φ = tan2φ – cot2φ j. 1 – cos4φ = 1 + 2 cot2φ
sin4φ

k. 1 + sin4φ = 1 + 2 tan2f sec2f l. 1 – cot4q = 1 – 2 cos2q
cos4φ cosec4q

m. sec4φ – tan4φ = 1 + 2 tan2φ n. cos2θ – sin2θ = 1 – tan2φ
1 + tan2φ

o. sin4A + cos4A = 1 – 2 sin2A cos2A p. sec q 1 tan q = 1 – sin q
+ cos q

q. cosec 1 = 1 + cos q r. 1 + sin x = cos x
q– sin q cos x 1 – sin x
cot q

s. sin2α sec2b + tan2b cos2α = sin2α + tan2b t. sin2A cos2B – cos2A sin2B = sin2A – sin2B

Trigonometric Ratios ~ 173

7. a. 1 + 1 A + 1 – 1 A = 2 cosec2A b. 1 – 1sin θ – 1 + 1 θ = 2 tan θ sec θ
cos cos sin

c. 1 sin b + 1 + cos b = 2 cosec b d. ccooss AA + sin A = 1 + tan A
+ cos sin b – sin A 1 – tan A
b

e. tcaont θ + cot b = tan θ cot b f. 11 +– ccooss θθ = cosec θ – cot θ
θ + tan b

g. 1 – sin θ = sec θ – tan θ h. sin3α + cos3α = 1 – sin α cos α
1 + sin θ sin α + cos α

i. ssiinn3qq – cos3q = 1 + sin q cos q j. 11 +– ssiinn xx = (sec x – tan x)2
– cos q

k. 11 – cos q = (cot q – cosec q)2 l. 11 ++ csoins qq 1 – sin q = cos2q
+ cos q 1 – cos q sin2q

m. sin α sin2α – cos2α sin2α = – (cosec α + sec α)
cos2α – cos α

n. sseecc A + tan A =1 + 2 sec A tan A + 2 tan2A
A – tan A

o. (1 + cos b) + (1 – cos b )2 = 2(1 + 2 cot2b)
sin2 b

p. (cot A – 3) (3 cot A – 1) = 3 cosec2A – 10 cot A

q. (1 – tan b)2 + (1 – cot b)2 = (sec b – cosec b)2

r. tan12+q – cot2q = sin2q – cos2q
cot2q cos2q

s. 1 +tanta2nq2q. 1 + cot2q = tan2q
cot2q

t. 1 cos θ θ + 1 sin θ θ = sin q + cos q
– tan – cot

u. ssiinn3αα + cos3α + sin3α + cos3α = 2
+ cos α sin α – cos α

v. sin6a + cos6a = 1 – 3 sin2a cos2a

EXERCISE : 5.3B

Prove the following trigonometric identities.

1. a. 2 (sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0

b. sin8θ – cos8θ = (sin2θ – cos2θ)(1 – 2 sin2θ cos2θ)

c. (sin α + cosec α)2 + (cos α + sec α)2 = tan2α + cot2α + 7

d. 3(sin θ + cos θ) – (sin θ + cos θ)3 = 2(sin3θ + cos3θ)

e. 2 sec2θ – 2 cosec2θ – sec4θ + cosec4θ = cot4θ – tan4θ

f. (1 tan α + (1 cot α = sin α cos α
+ tan2α)2 + cot2α)2

g. 1 +tanta3nα2α + cot3α = 1– 2sin2α cos2 α
+ cot2α sin α cos α
1

174 ~ Perfect Optional Mathematics Class 9

h. ttaann22AA + cot2A = sin4A + cos4A
– cot2A sin2A – cos2A

i. sin2A – sec2­A + cos2A = tan4A
cos2A – cosec2A + sin2A

2. a. (3 – 4 sin2θ) (1 – 3 tan2θ) = (3 – tan2θ) (4 cos2θ – 3)

b. (3 – 4 sin2q)(sec2q – 4 tan2q) = (3 – tan2θ) (1 – 4 sin2θ)

c. (2 – cos2q)(1 + 2 cot2q) = (2 – sin2θ) (2 + cot2θ)

d. (2 + 3 sin2q)(1 – 2 tan2q) = (3 cos2q – 2) (2 + 5 tan2θ)

e. sec3α cosec3α – 3 sec α cosec α = tan3α + cot3α

f. tan α – cosec α sec α (sin2α – cos2α) = cot α

g. (sec x + cosec x)2 tan x = (tan x + 1)2 (tan x + cot x)

h. (tan q + cot q)2 = (1 + cot2q) (1 + tan2q)

i. (sin x cos y + cos x sin y)2 + (cos x cos y – sin x sin y)2 = 1

j. sin α = 1 – cos α + sin α k. ccoott q + cosec q –1 = 1 + cos q
cos α + 1+ sin α q – cosect q +1 sin q
1 + sin α

l. 11 – secA + tan A = secA + tan A –+11 m. ttaann q + sec q – 1 = 1 + sin q
+ sec A – tan A sec A – tan A q – sec q + 1 cos q

n. ssiinn q + cos q – 1 = 1 cos q q o. ssiinn qq –+ ccooss q + 1 = 1 + sin q
q – cos q + 1 + sin q – 1 cos q

p. ccoosseecc q + cot q + 1 = cosec q + cot q – 1 Trigonometry
q – cot q + 1 cot q – cosec q + 1

q. 1 + sin θ – cos θ + 1 + sin θ + cos θ = 2 cosec θ
1 + sin θ + cos θ 1 + sin θ – cos θ

r. sec θ 1 tan θ – 1 θ = 1 θ – sec θ 1 tan θ
+ cos cos –

s. 11 + tan2b 1 – tan b 2 tan2θ tan2b
+ cot2 b 1 – cot b cos2 b cos2 θ
= t. – = tan2θ – tan2b

u. sec2q 1 – sin2q = 1 – ttaann22Aq v. 1 + sec2A cot2C = 1 + tan2A cos2C
sin2A 1 + sec2B cot2C 1 + tan2B cos2C

w. 1 + 3 cos α – 4 cos3α = (1 + 2 cos a)2 x. 1 + 3 sin q – 4 sin3q = (1 + 2 sin q)2
1 – cos α 1 – sin q

y. 1 + cos y – 2 cos2y = 1 + cos y
sin2y + cos
1 y

z. 1– sin2φ + 1 + cos φ – sin φ = cot2φ (1 – cos φ)
1+ cos φ sin φ – cos
1 φ

aa. sec4q + tan4q = 1 + 2c osisn4q2q ab. cosec4q + cot4q = 1 + 2 cos2q
sin4q

ac. 11 ++ (cosec α tan b)2 = 1 + (cot α sin b)2
(cosec q tan b)2 1 + (cot q sin b)2



Trigonometric Ratios ~ 175

5.4 Conversion of Trigonometric Ratios

Introduction

We can express a given trigonometric ratio of an angle in terms of any other trigonometric ratios. Let us
study the following examples.

Example 1:

Express all other trigonometric ratios in terms of sin θ.

 We have, sin θ = sin θ

cosec θ = 1 θ
sin

cos θ = 1 – sin2θ

sec θ = 1 θ = 1
cos 1 – sin2θ

tan θ = sin q = sin q
cos θ 1 – sin2θ

cot θ = cos q = 1 – sin2θ
sin θ sin q

Example 2:

If tan θ = 152 , find the values of sin θ, sec θ and cot θ. let PRQ = θ and tan θ = 152, PQ = 5 units
right angled triangle with PQR = 90° and
 Let PQR be a

and QR = 12 units. Alternateley,

Using pythagoras theorem, Here, tan q = 5
PR2 = PQ2 + QR2 12

or, PR2 = 52 + 122 We have, sin q = 1 q = 1 = 1
cosec + cot2q
1 1 + 1
tan2θ
∴ PR = 25 + 144

= 169 = 13 tan q 5
1 + tan2q
PQ 5 = 1 = = 12
PR 13  tatna2nq2+θ 1
Now, sin θ = = 1+ 52
12

sec θ = PR = 13 55 5 12 5
QR 12 12 13 13
= 12 = 12 = × =
QR 12 144 + 25 13
cot θ = PQ = 5
12 12

Now, sec q = 1 + tan2q = 5 2 = 144 + 25 = 169 = 13
12 12 12 12
1+

Again, cot q = 1 q = 1 = 12
tan 5 5

12

176 ~ Perfect Optional Mathematics Class 9

Example 3:

If cos θ = m n2 , then prove that m sin θ = n cos θ.
m2 +

 Here, cos θ = m n2 = b
m2 + h

Hence, b = m, h = m2 + n2

Using pythagoras theorem,

h2 = p2 + b2
or, m2 + n2 = p2 + m2
or, p2 = n2

∴ p = n

Now, tan θ = p
b

or, sin θ = n
cos θ m

∴ m sin θ = n cos θ

Example 4:

If 5 cos A + 12 sin A = 13, prove that 12 cot A = 5. Trigonometry

 Here, 5 cos A + 12 sin A = 13

or, 5 cos A + 12 sin A = 13
sin A sin A sin A

or, 5 cot A + 12 = 13 cosec A

Squaring both sides, we have

(5 cot A + 12)2 = (13 cosec A)2

or, 25 cot2A + 120 cot A + 144 = 169 cosec2A

or, 25 cot2A + 120 cot A + 144 = 169(1 + cot2A)

or, 25 cot2A + 120 cot A + 144 = 169 + 169 cot2A

or, 169 cot2A + 169 – 25 cot2A – 120 cot A – 144 = 0

or, 144 cot2A – 120 cot A + 25 = 0

or, (12 cotA)2 – 2.12 cot A.5 + 52 = 0

or, (12 cot A – 5)2 = 0

or, 12 cot A – 5 = 0

∴ 12 cot A = 5 Proved.

Conversion of Trigonometric Ratios ~ 177

Example 5:

If tan q + sin q = m and tan q sin q = n then prove that 14 (m2 – n2) = mn.

 Here, m2 – n2 = (m + n) (m – n)

= {(tan q + sin q) + (tan q – sin q)} {(tan q + sin q) – (tan q – sin q)}

= (tan q + sin q + tan q – sin q) (tan q + sin q – tan q + sin q)

= 2 tan q × 2 sin q = 4 tan q sin q

Now, LHS = 41 (m2 – n2) = 1 × 4 tan q sin q = tan q sin q
4

RHS = mn = (tan q + sin q) (tan q – sin q)

= tan2q – sin2q = sin2q – sin2q = sin2q – sin2q cos2q
= cos2q cos2q

sin2q (1 – cos2q) = sin2q sin2q =  tan2q sin2q
cos2q cos2q

= tan q sin q

Hence, LHS = RHS Proved
Example 6:

If sec q+ tan q= a then show that sin q = a2 – 1  .
a2 + 1
 Here, sec q + tan q = a

or, co1s q + sin q = a
cos q

or, 1 c+ossinq q = a

or, 1 + sin q = a cos q

Squaring both sides, we have

or, (1 + sin q)2 = a2 cos2q

or, (1 + sin q)2 = a2 (1 – sin2q)

or, (1 + sin q)2 = a2 (1 + sin q) (1 – sin q)

or, 1 + sin q = a2 (1 – sin q)

or, a2 sin q + sin q = a2 – 1

or, sin q (a2 + 1) = a2 – 1

\ sin q = a2 – 1 proved.
a2 + 1

178 ~ Perfect Optional Mathematics Class 9

Example 7:

If cos4q + cos2q = 1 then show that sec4q – sec2q = 1.

 Here, cos4q + cos2q = 1

or, se1c4q + 1 = 1
sec2q

or, 1 +sesce4qc2q = 1

or, 1 + sec2q = sec4q

\ sec4q – sec2q = 1 Proved.

Example 8:

If cos q + sec q = 4 then show that cos2q + sec2q = 14. Trigonometry

 Here, cos q + sec q = 4

Squaring both sides, we have
(cos q + sec q)2 = 42
or, cos2q + 2 cos q sec q + sec2q = 16
or, cos2q + 2 + sec2q = 16
or, cos2q + sec2q = 16 – 2 = 14 Proved.

Example 9:

If 15 sin2q + 2 cos2q = 7 then find tan q.

 Here, 15 sin2q + 2 cos2q = 7

or, 15 sin2q + 2 (1 – sin2q) = 7

or, 15 sin2q + 2 – 2 sin2q = 7

or, 15 sin2q – 2 sin2q = 7 – 2

or, 13 sin2q = 5

or, sin2q = 5
13
5
Again, 1 – cos2q = 13

or, cos2q = 1 – 5 = 13 – 5
13 13
8
or, cos2q = 13

Now, tan2q = sin2q = 5/13 = 5 × 13
cos2q 8/13 13 8
5
or, tan2q = 8

\ tan q = 5 = 2 5
8 2

Conversion of Trigonometric Ratios ~ 179

EXERCISE : 5.4

1. a. If tan θ = 152, find sin θ and cos θ.

b. If sec A = mn , find the value of cot A.

c. If sin θ = 3 , find the values of tan θ and cos θ.
5

d. If sin φ = 12, find the value of the value of tan φ + cot φ.

2. a. If cos x = 110, prove that sec2x – tan2x = 1.

b. If tan θ = 5 , show that 3 cos θ – 2 sin θ = 27.
6 3 cos θ + 2 sin θ

c. If sin α = cos b = 23 then show that tan α – tan b = 13.
1 + tan α tan b

d. If cot θ = m , then prove that m cos θ – n sin θ = m2 – n2 .
n m cos θ + n sin θ m2 + n2

3. a. If cos θ = m2 – n2 , prove that 2mn cosec θ = m2 + n2.
m2 + n2

b. If sin θ = m , show that n2 – m2 t an θ = m.
n

c. If 1 – cos A = 21, show that tan2A + 4 sin2A = 6.

d. If tan α = 3 – 11, prove that sin α = 3 –21.
3 + 2 

4. a. If 4 cos2θ + 4 sin θ – 5 = 0, show that sin θ = 12.

b. If 3 sin θ + 4 cos θ = 5, show that sin θ = 53.

c. If 5 cos θ + 12 sin θ = 13, then show that 5 tan θ = 12.

d. If tan A + cot A = 3, prove that tan2A + cot2A = 7.

e. If cos α + sin α = 2cos α, show that cos α – sin α = 2 sin α.

f. If sin q + cosec q = 4, then prove that sin2q + cosec2q = 14.

5. a. In a triangle ABC, right angled at C, if tan A = 1 and tan B = 3, show that
3

sin A cos B + cos A sin B = 1.

b. If tan A = 1 , tan B = 1 , then show that
2 3

180 ~ Perfect Optional Mathematics Class 9

cos A sin B + sin A cos B = 1
2

c. If sin α = 3 and sin β = 153, show that :
5

sin α cos β + cos α sin β = 5665.

d. If sin α = 12 and cos β = 35 , prove that tan α + tan β = – 3536.
13 1 – tan α tan β

e. If x sin3α + y cos3α = sin α cos α and x sin α – y cos α = 0, prove that: x2 + y2 = 1.

6. a. If tan4q + tan2q = 1, then prove that cot4q – cot2q = 1

b. If sin4q + sin2q = 1, then prove that cot4q + cot2q = 1

c. If cos4q + cos2q = 1, then prove that tan4q + tan2q = 1

7. a. If 8 sin2q + 7 cos2q = 8, then find sin q.

b. If 3 sec2q + 4 tan2q – 3 = 0, then find sec q.

c. If sec2q = 2 (1 + tan q), find the value of tan q.

d. If 15 sin2q + 2 cos2q = 7, find the value of tan q.

e. If 4 cos q + 3 sin q = 5, then what is the vlaue of cos q?

f. If 9 cos q + 40 sin q = 41, then what is the value tan q? Trigonometry

g. If 2 cos2q + 4 sin2q = 3, then what is the value of sin q?

h. If 9 cos2q + 8 sin2q = 11, then what is the value of cos q?

i. If 7 sin2q + 3 cos2q = 4, what is the value of sin q?

j. If 3sin2q – cos2q = 1, then what is the value of sin q?

k. If 5 sec2q – tan2q = 9, then what is the value of tan2q?

l. If 7 cosec2q – 9 cot2q = 7, then what is the value of tan q?

8. a. If sec q – tan q = k, prove that cosec q = 1 + kk22 .
1 –

b. If tan a = n and sin a = m, then prove that sec2q = 1 + mn22 .
tan b sin b 1 –



Conversion of Trigonometric Ratios ~ 181

5.5 Trigonometric Ratios of Any Angles

Sign of Trigonometric Ratios of Angles of any Magnitude

Let OP, a revolving line revolves about O in the anticlockwise P2(– a2, b2) r Y P1(a1, b1)
direction and describes a circle with centre O and radius
OP = r. r

Let OP1, OP2, OP3 and OP4 be the positions of OP in the first, b
the second, the third and the fourth quadrants respectively.
q
Let the co-ordinates of P1, P2, P3 and P4 are (a1, b1), X' a PX
O
(– a2, b2), (– a3, – b3) and (a4, – b4) respectively where r
fr
a1, b1, a2, b2, a3, b3, a4 and b4 are all positive numbers P3(– a3, – b2)
P4(a4, – b4)
OP = OP1 = OP2 = OP3 = OP4 = r is always taken positive. Y'
POP1 = θ, POP2 = b, reflex POP3 = α and reflex
POP4 = φ, then for angle θ,

sin θ = b1 = +ve, cosec θ = r = +ve
r b1

cos θ = a1 = +ve, sec θ = r = +ve,
r a1

tan θ = b1 = +ve and cot θ = a1 = +ve
a1 b1

For angle b,

sin b = b2 r
r = +ve, cosec b = b2 = +ve

cos b = – a2 r
r = – ve, sec b = – a2 = – ve

tan b = b2 = – ve, cot b = – a2 = – ve
– a2 b2

For angle α,

sin α = – b3 = – ve, cosec α = r = – ve
r – b3

cos α = – a3 = – ve, sec α = r = – ve
r – a3

tan α = – b3 = +ve, cot α = – a3 = +ve
– a3 – b1

And for angle φ,

sin φ = – b4 = – ve, cosec φ = r = – ve
r – b4

182 ~ Perfect Optional Mathematics Class 9

cos φ = a4 = + ve, sec φ = r = + ve
r a4

tan φ = – b1 = – ve, cot φ = a4 = – ve
a4 – b4

The following table shows the signs of all the trigonometric ratios of angles in all quadrants.

Trigonometric I Quadrants IV
ratios of angles II III

sin q + + – –

cos q + – – +

tan q + – + –

cosec q + + – –

sec q + – – +

cot q + – + –

The sign of the trigonometric ratios of an angle of any magnitude can be summarized in the following
diagram as well

Y 90°

Sin +ve All +ve Y 90° + q 90° – q
cosec +ve I quadrant SA
180° – q + q
II quadrant O
TC 360° + q Trigonometry
X' O X or, X' X 180°
Y' O 0° or 360°
Tan +ve Cos +ve

cot +ve sec +ve 180° + q 270° + q
270° – q 360° – q
III quadrant IV quadrant

Y'

270°

Trigonometric Ratios of Negative Angles

Let OA be a revolving line revolves about O in the anticlockwise Y

direction and describes a circle with centre O and radius OA = r. Let P(a, b)

P(a, b) be any point on the circle so that POA = θ. X' q AX
P'(a, – b)
Now, sin θ = br , cos θ = a and tan θ = ba. O – q
r

Let OP' be the image of OP under the reflection about X– axis.

Then P(a, b)  P'(a, – b) and POP' = – θ. Y'

Now, sin (– θ) = –  b = – sin θ, cos (– θ) = a = cos θ,
r r

tan (– θ) = –  b = – tan θ
a

Similarly, we can have cosec (– θ) = – cosec θ, sec (– θ) = sec θ and cot (– θ) = – cot θ.

Trigonometric Ratios of Any Angles ~ 183

Hence, we have the following relations. cosec (– q) = – cosec q
sec (– q) = sec q
sin (– q) = – sin q cot (– q) = – cot q
cos (– q) = cos q
tan (– q) = – tan q

Trigonometric Ratios of Complementary Angles (i.e. 90° – q)

Let OA be the revolving line which describes a circle with centre Y
P'(b, a)
O and radius OA = r by revolving in anticlockwise direction. Let
P(a, b) be a point on the circle such that POA = θ q P(a, b)

Now, sin θ = br , cos θ = a and tan θ = ba. X' q AX
r
O

Let OP' be the image of OP under the reflection on the line y = x.

Then P(a, b) → P'(b, a) and P'OB = θ and P'OA = 90° – θ. Y'

Now, sin (90° – θ) = a = cos θ, cos (90° – θ) = b = sin θ
r r
a a
tan (90° – θ) = b = cot θ As cot θ = b

Similarly, we can prove that cosec (90° – θ) = sec θ,

sec (90° – θ) = cosec θ and cot (90° – θ) = tan θ

Hence, we have the following relations. cosec (90° – q) = sec q
sec (90° – q) = cosec q
sin (90° – q) = cos q
cos (90° – q) = sin q cot (90° – q) = tan q
tan (90° – q) = cot q

Trigonometric Ratios of the Angle (90° + θ)

Let OA be the revolving line which describes a circle with centre Y
O and radius OA = r by revolving in anticlockwise direction. Let
P(a, b) be a point on the circle so that POA = θ P'(– b, a) 90°+q P(a, b)
X'
Now, sin θ = br , cos θ = a and tan θ = ba. q AX
r
O

Let OP' be the image of OP under the rotation through 90° about O.

Then P(a, b) → P'(– b, a) and P'OA = 90° + θ . Y'

So, sin (90° + θ) = a = cos θ, cos (90° + θ) = –  b = – sin θ
r r

tan (90° + θ) = b = – cot θ As cot θ = b
– a a

Similarly, we can show, cosec (90° + θ) = sec θ, sec (90° + θ) = – cosec θ and cot (90° + θ) = – tan θ

184 ~ Perfect Optional Mathematics Class 9

Hence, we have the following relations. cosec (90° + q) = – sec q
sin (90° + q) = cos q sec (90° + q) = – cosec q
cos (90° + q) = sin q
tan (90° + q) = – cot q cot (90° + q) = – tan q

Trigonometric Ratios of Supplementary Angle (180° – θ) Y

Let OA be a revolving line which makes a circle with radius OA = r P'(– a, b) P(a, b)

and centre O by revolving in anticlockwise direction. Let P(a, b) X' C qq AX
be a point on the circle so that POA = θ.
O
br , a ba.
Then sin θ = cos θ = r and tan θ =

Let OP’ be the image of OP under the reflection on Y– axis then Y'
P(a, b) → P'(– a, b) and P'OC = θ and P'OA = 180° – θ.

Now, sin (180° – θ) = –  b = sin θ, cos (180° – θ) = –  a = – cos θ
r r
b
tan(180° – θ) = – a = – tan θ

Similarly, we can have cosec (180° – θ) = cosec θ, sec (180° – θ) = – sec θ and cot (180° – θ) = – cot θ.

Hence, we have the following relations. cosec (180° – θ) = cosec θ Trigonometry
sin (180° – θ) = sin θ sec (180° – θ) = – sec θ
cos (180° – θ) = – cos θ cot (180° – θ) = – cot θ
tan (180° – θ) = – tan θ

Trigonometric Ratios of Angle (180° + θ)

Let P(a, b) be a point on the circle so that POA = θ, then sin θ = br , Y
cos θ= tan θ = ba.
a and P(a, b)
r
Let OP' be the image of OP under the rotation through 180°
X' C q AX
about O, then P(a, b) → P'(– a,– b) and P'OC = θ and reflex
qO

P'OA = 180° + θ. P'(– a, – b)

Now, sin (180° + θ) = – b = – sin θ, cos (180° + θ) = – a = – cos θ Y'
r r
– b
tan (180° + θ) = – a = tan θ.

Similarly, we can have cosec (180° + θ) = – cosec θ, sec (180° + θ) = – sec θ and cot (180° + θ) = cot θ

Hence, we have the following relations.

sin (180° + θ) = – sin θ cosec (180° + θ) = – cosec θ

cos (180° + θ) = – cos θ sec (180° + θ) = – sec θ

tan (180° + θ) = tan θ cot (180° + θ) = cot θ

Trigonometric Ratios of Any Angles ~ 185

Trigonometric Ratios of the Angle (270° – q) Y

Let P(a, b) be any point on the circle so that POA = θ then sin θ = br,
cos OP' be the image of OP under the reflection
θ = a and tan θ = ba. Let P'(– a, b) P(a, b)
r X' C AX
q
on the Y– axis, then
O
P(a, b) → P'(– a, b) and let OP" be the image of OP' under the rotation
through 90° about origin, then q

P'(– a, b) → P"(– b, – a) and P"OD = q and reflex P"OA = 270° – θ. P"(– b, – a)
Y'

Now, sin (270° – θ) = – a = – cos q, cos (270° – θ) = – b = – sin θ,
r r
– b b
tan (270° – θ) = – a = cot θ As cot θ = a

Similarly, cosec (270° – θ) = – sec θ, sec (270° – θ) = – cosec θ and cot (270° – θ) = tan θ.

Hence, we have the following relations.

sin (270° – θ) = – cos θ cosec (270° – θ) = – sec θ

cos (270° – θ) = – sin θ sec (270° – θ) = – cosec θ

tan (270° – θ) = cot θ cot (270° – θ) = tan θ

Trigonometric Ratios of the Angle (270° + θ)

Let P(a, b) be a point on the circle such that POA = θ, then sin θ = br , Y
cos tan θ = ba.
θ = a and P(a, b)
r AX

Let OP' be the image of OP under the rotation through 270° about O, X' q

then P(a, b) → P'(b, – a) and P'OD = θ and reflex P'OA = 270° + θ. O
q

Now, sin (270° + θ) = – a = – cos θ, cos (270° + θ) = b = sin θ, P'(b, – a)
r r Y'
– a a
tan (270° + θ) = b = – cot θ As cot θ = b

Similarly, cosec (270° + θ) = – sec θ, sec (270° + θ) = cosec θ and cot (270° + θ) = – tan θ.

Hence, we have the following relations.

sin (270° + θ) = – cos θ cosec (270° – θ) = – sec θ

cos (270° + θ) = sin θ sec (270° + θ) = cosec θ

tan (270° + θ) = – cot θ cot (270° + θ) = – tan θ

Trigonometric Ratios of the Angle (360° – θ) Y

Let P(a, b) be a point on the circle so that POA = θ, then sin θ = br, P'(– a, b) q P(a, b)
cos under the reflection X'
θ = a and tan θ = ba. Let OP' be the image of OP Oq AX
r q P'(a, – b)

on Y– axis, then P(a, b) → P'(– a, b).

Again, let OP" be the image of OP' under the rotation through 180° Y'

186 ~ Perfect Optional Mathematics Class 9

about O, then

P'(– a, b) → P"(a, – b) and P"OA = θ and reflex P"OA = 360° – θ.

Now, sin (360° – θ) = –  b = – sin θ, cos (360° – θ) = a = cos θ,
r r
–  b
tan (360° – θ) = a = – tan θ.

Similarly, cosec (360° – θ) = – cosec θ, sec (360° – θ) = sec θ and tan (360° – θ) = – tan θ.

Hence, we have the following relations.

sin (360° – θ) = – sin θ cosec (360° – θ) = – cosec θ

cos (360° – θ) = cos θ sec (360° – θ) = sec θ

tan (360° – θ) = – tan θ cot (360° – θ) = – cot θ

As (360° + θ) represents the same angle as θ

sin (360° + θ) = sin θ cosec (360° + θ) = cosec θ

cos (360° + θ) = cos θ sec (360° + θ) = sec θ

tan (360° + θ) = tan θ cot (360° + θ) = cot θ

Here, we conclude the following information from above trigonometric ratios.

sin (90° – q) = cos q = sin (1 × 90° – q); 90° – q lies on the 1st quadrant

sin (90° + q) = cos q = sin (1 × 90° + q); 90° + q lies on the 2nd quadrant

sin (180° – q) = sin q = sin (2 × 90° – q); 180° – q lies on the 2nd quadrant Trigonometry

sin (180° + q) = – sin q = sin (2 × 90° + q); 180° + q lies on the 3rd quadrant

sin (270° – q) = – cos q = sin (3 × 90° – q); 270° – q lies on the 3rd quadrant

sin (270° + q) = cos q = sin (3 × 90° + q); 270° + q lies on the 4th quadrant

sin (360° – q) = – sin q = sin (4 × 90° – q); 360° – q lies on the 4th quadrant

sin (360° + q) = sin q = sin (4 × 90° + q); 360° + q lies on the 1st quadrant

sin (450° – q) = cos q = sin (5 × 90° – q); 450° – q lies on the 1st quadrant

sin (450° + q) = cos q = sin (5 × 90° + q); 450° + q lies on the 2nd quadrant

sin (540° + q) = – sin q = sin (6 × 90° + q); 540° + q lies on the 3rd quadrant

... ... ... ... ... ...

sin (900° – q) = sin q = sin (10 × 90° – q); 900° – q lies on the 2rd quadrant

sin → cos, when n is odd.
Hence, sin (n × 90° – q) = sin → sin, when n is even.

cos → sin, when n is odd.
Similarly, cos (n × 90° – q) = cos → cos, when n is even.

tan → cot, when n is odd.
tan (n × 90° – q) = tan → tan, when n is even.

Trigonometric Ratios of Any Angles ~ 187

And, + ve, when (n × 90° ± q) lies in the 1st or 2nd quadrant.
sin (n × 90° ± q) = – ve, when (n × 90° ± q) lies in the 3rd or 4th quadrant.

+ ve, when (n × 90° ± q) lies in the 1st or 4th quadrant.
Similarly, cos (n × 90° ± q) = – ve, when (n × 90° ± q) lies in the 2nd or 3rd quadrant.

+ ve, when (n × 90° ± q) lies in the 1st or 3rd quadrant.
tan (n × 90° ± q) = – ve, when (n × 90° ± q) lies in the 2nd or 4th quadrant.

Example 1:

Find the value of sin (4500° + q) and tan 855°.

 Here, sin (4500°+ q)

= sin (50 × 90° + q)
= – sin q [As 50 is even and (50 × 90° + q) lies in the 3rd quadrant.]
tan 855°
= tan (9 × 90° + 45°)
= – cot 45° [As 9 is even and (9 × 90° + q) lies in the 2nd quadrant.]
= – 1

Example 2:

Prove that : tan 25° + sin 35° + sec 40° = cot 65° + cos 55° + cosec 50°

 LHS = tan 25° + sin 35° + sec 40°

= tan (90° – 65°) + sin (90° – 55°) + sec (90° – 50°)
= cot 65° + cos 55° + cosec 50°
= RHS (Proved)

Example 3:

Solve for x: x cot α tan (90° + α) = tan (90° + α) cot (180° – α) + x sec (90° + α) cosec α

 x cot a tan (90° + a)

= tan (90° + α) cot (180° – α) + x sec (90° + α) cosec α

or, x cot α × (– cot α) = – cot α × (– cot α) + x × (– cosec α) × cosec α

or, – x cot2α = cot2α – x cosec2α

or, x (cosec2α – cot2α) = cot2α

∴ x = cot2α ( cosec2α – cot2α = 1)

Example 4:

a. Prove that : cot2 pc + cosec 5pc + 3 tan2 pc = 6
6 6 6

b. Find the value of x : cosec (90° – q) – tan x – q) = 1
sin (90° – q) (90°

188 ~ Perfect Optional Mathematics Class 9

 a. LHS = cot2 pc + cosec 5pc + 3 tan2 pc b. Here, cosec (90° – q) – tan x – q) = 1
6 6 6 sin (90° – q) (90°

= cot2 180° + cosec 5 × 180° + 3 tan2 180° or, sec q – x = 1
6 6 6 cos q cot
q

= cot2 30° + cos 150° + 3 tan2 30° or, 1 q × 1 q – 1 = x q
cos cos cot

2 12 or, 1 – cos2q = x
3 cos2q cot
= 5 +2+3× q

1 or, sin2q = x q
3 cos2q cot
= 5 + 2 + 3 ×
x
or, tan2q = cot q

=3+2+1 or, x = tan2q cot q

= 6 Proved. or, x = tan q

Example 5:

Simplify : sin (180° – q) × sin (90° + q) × sin – cot2(270° + q) – q)
cot (90° – q) tan (180° – q) (90° + q) cos (90°

 Here, sin (180° – q) × sin (90° + q) × sin – cot2(270° + q) – q)
cot (90° – q) tan (180° – q) (90° + q) cos (90°
Trigonometry
= sin q × cos q × – tan2q q
tan q – tan q cos q sin

= sin q × cos q × – tan2q
– tan2q cos q sin
q

= sin q cos q
cos q sin q

=1

Example 6:

Prove that : sin2 pc + sin2 6pc + sin2 5pc + sin2 2pc = 2
14 14 14 14

 LHS = sin2 pc + sin2 6pc + sin2 5pc + sin2 2pc
14 14 14 14

= sin2  pc – 6pc + sin2 6pc + sin2  pc – 2pc + sin2 21p4c
2 14 14 2 14

= cos2 6pc + sin2 6pc + cos2 21p4c + sin2 21p4c
14 14

=1+1

= 2 Proved.

Trigonometric Ratios of Any Angles ~ 189

EXERCISE : 5.5A

Find the trigonometric ratio of the following.

1. a. sin (– q) b. cos (– q) c. tan (– q)
f. cos (90° – q)
d. sin (90° – a) e. tan (90° – b) c. tan (180° + q)
f. tan (360° – a)
2. a. sin (180° – q) b. cos (180° + q)

d. sin (270° + b) e. cos (360° + q)

3. Find the value of:

a. sin 120° b. cos 150° c. tan pc
4
d. sin p2c e. cos 23pc 3pc
4. Find the value of: f. tan 4

a. sin 840° b. cos 750° c. cos 780°
c. tan (– 1020°)
d. sin (– 1140°) b. sec (– 1380°)

Evaluate without using table or calculator.

5. a. sin 36° – cos 54° b. tan 80° – cot 10° c. cos 50° – sin 40°
f. cos 110° – sin 200°
d. sin 110° – sin 70° e. tan 80° + tan 100°

6. a. sin 1800°° b. scions12100°° c. 35 ctaont 2655°°
cos

7. a. cos 390° sin 420° + sin (– 330°) cos (– 660°)

b. cos 315° sin 450° – cos (– 420°) cos (– 570°)
8. a. sin (90° – q) cos (90° – q) – sin q cos q b. tan 9° cot 27° – cos 81° tan 63°

c. cot 16° cot 32° cot 58° cot 74° d. tan 7° tan 39° tan 90° tan 51° tan 83°

9. a. – sin (180° – q) cos (90° – q) – cos2(180° – q) b. sin q cos (90° – q) + cos q sin (90° – q)

c. cosec2(90° – q) – cot2(90° – q) d. sec2(90° – a) – tan2(90° + a)

10. a. sin2135° + cos2150° – tan2225° b. sin2120° – cos2120° – sin2135° – tan2150°

c. sin2135° + cos2120° – sin2120° + tan2150° d. cos2135° + tan2120° – cos2120° + sin2150°

e. sin2135° – cos2135° + 2 tan 135° – sec2135° f. 3 tan2135° + 4 cos2150° – 1 sec2135° – 13 sin2120°
g. 3 tan2 34pc – sin2 23pc – 21 cot2 56pc + 81 sec2 34pc h. 3 2

2 cos2 34pc + sin  5pc  – 21 cos pc + tan2 23pc
6

Prove that:

11. a. cos 50° + cos 75° = sin 15° + sin 40° b. sin 9° – sin 81° = cos 81° – cos 9°

c. cot 62° – tan 48° = tan 28° – cot 42° d. sin 75° – sin 15° = cos 105° + cos 15°

190 ~ Perfect Optional Mathematics Class 9

12. a. sin 112° + cos 74° – sin 68° + cos 106° = 0

b. cos 20° + cos 40° + cos 140° + cos 160° = 0
c. tan 9° tan 27° tan 45° tan 63° tan 81° = 1
d. tan 32° + cot 53° – cosec 80° = tan 37° + cot 58° – sec 10°
e. sin 9° sin 27° sin 63° sin 81° = cos 9° cos 27° cos 63° cos 81°

13. a. scsioinns2 1p8600c °°++cocssoi2ns p327c00–°° cs=oins222 pp44cc = – 12 b. cot 40° ++sc isconions1s262575p563p°c°c = 2
14. a. b. tan 50° =
3
2 sin2 p6c 2

EXERCISE : 5.5B

1. Prove the following:

a. sisnin((9108°0°+ – A) sec (90° + A) tan (90° + A) = – sec2A
A) cos (180° – A) cot (180° + A)

b. ssiinn((514800°°–+qq))ccooss(2(9700°° + q) tan (270° + q) = cosec q
+ q) cos (360° + q)

c. sins(i3n6(09°0°– + q) cos (– q) cot (180° – q) q) = cot q
q) cos (180° + q) tan (90° +

2. Prove that:

a. sin2 p4c + sin2 34pc + sin2 54pc + sin2 7pc = 2 b. sin2 pc + sin2 3pc + sin2 5pc + sin2 7pc = 2 Trigonometry
4 8 8 8 8

c. cos2 p4c + cos2 34pc + cos2 54pc + cos2 74pc = 2 d. cos2 pc + cos2 3pc + cos2 5pc + cos2 7pc = 2
8 8 8 8

e. sin2 1p6c + sin2 31p6c + sin2 51p6c + sin2 71p6c = 2 f. cos2 pc + cos2 3pc + cos2 5pc + cos2 7pc = 2
16 16 16 16

3. Find the vaue of x from the following equations:

a. sin (180° – q) cos (90º + q) + x tan (90° – q) + sin (90° + q) cos (180° – q) = 0

b. cosec (90° + q) + x cos q cos (90° + q) = cos q

c. x cot q tan (90° + q) = tan (90° + q) cot (180° – q) + x sec (90° + q) cosec q

4. Find the value of x:

a. sin (90° – q) + x cos (90° – q) = 1 b. x sin q + cos q = cosec q
cos q sin q sin (90° – (90° – cos q
q) cos q)

c. x cos (90° + a) cos (90° – a) + tan (180° – a) cot (90° + a) = sin a sin (180° – a)

d. x sec (90° + a) cosec a + tan (90° + a) cot (180° – a) = x cot a tan (90° + a)

e. sin (270° – q) sin (450° – q)) – cos (630° – q) cos (90° + q) + x cos 90° = 0

f. cos (180° – q) + tan (180° + q) cos2 (270° – q) cosec (– q) – x sin (90° – q) = 0



Trigonometric Ratios of Any Angles ~ 191

5.6 Trigonometric Ratios of Standard Angles

Introduction

We have to remember the values of trigonometric ratios of the standard angles like 0°, 30°, 45°, 60°, 90°
and so on. Similarly, we can establish the relations of the trigonometric ratios of the allied angles like
90° + θ, 90° – θ, 180° + θ, 180° – θ, 270° + θ and so on.

Trigonometric Ratios of 45º Y

Let OP be the revolving line starting from the initial line OX. P
Trace out an angle POX = 45º in anticlockwise direction.
From P, draw PM^OX. Then POM = 45º = OPM (why?) 2a
a
Let OM = a units then PM = a units and by using pythagoras theorem,
OM
OP2 = OM2 + PM2 X' X
or, OP2 = a2 + a2

or, OP2 = 2a2

∴ OP = 2 a

Now, in right angled DPOM,

sin 45º = PM = a = 1
OP 2a 2

cos 45º = OM = a = 1 and
OP 2a 2

tan 45º = PM = a = 1
OM a

Similarly, cosec 45º = OP = 2a = 2,
PM a

sec 45º = OP = 2a = 2 and
OM a

cot 45º = OM = a = 1.
PM a
Trigonometric Ratios of 30° and 60° Y
P
Let OP be the revolving line starting from the initial line OX. Trace out 3a
an angle POX = 60º in anticlockwise direction. 2a

From P, draw PM^OX. Then POM = 60º and

OPM = 90º – 60º = 30º. X' O aMa N X

Join PN then OP = ON and OPN = ONP = 60º.

Hence, DOPN is an equilateral triangle.

Let OP = 2a units then PN = ON = 2a units. Y'

OM = MN = a units (OM^OX)

192 ~ Perfect Optional Mathematics Class 9

Now in right angled DOPM,
OP2 = OM2 + PM2
or, (2a)2 = a2 + PM2
or, PM2 = 3a2
\ PM = 3a
From in DOPM, taking POM as the angle of reference.

sin 60º = PM = 3a = 3 ,
OP 2a 2

cos 60º = OM = a = 1 and
OP 2a 2

tan 60º = PM = 3a = 3
OM a

Similarly, cosec 60º = OP = 2a = 23,
PM 3a

sec 60º = OP = a = 2 and cot 60º = OM = a = 1
OM 2a PM 3a 3

Again from DPOM, taking OPM = 30º as angle of reference,

sin 30º = OM = a = 21,
OP 2a

cos 30º = PM = 3a = 3 and Trigonometry
OP 2a 2

tan 30º = OM = a = 1
PM 3a 3

Similarly, cosec 30º = OP = 2a = 2,
OM a

sec 30º = POMP = 2a = 2 and
3a 3

tan 30º = OM = 3a = 3
PM a

Trigonometric Ratio of 0º and 90º Y
QP
Let OP be the revolving line starting from the initial line OX. Trace out
an angle POX in anticlockwise direction. From P, draw PM^OX. As OP
approaches to OX, POM → 0º,

PM → 0 and OM → OP.

\ sin 0º = PM = 0 = 0, X' O NM X
OP OP

cos 0° = OM = OP = 1 and
OP OP

Y'

Trigonometric Ratios of Standard Angles ~ 193

tan 0° = PM = 0 =0
OM OP

Similarly, cosec 0º = OP = OP = undefined (∞),
PM 0

sec 0° = OP = OP = 1 and
OM OP

cot 0° = OM = OP = undefined (∞)
PM 0

Again, the revolving line reach at OQ making angle QOX with OX in anticlockwise direction.

From Q, draw QN^OX
As OQ approaches OY, QON → 90º , QN → OQ and ON → 0.

\ sin 90º = QN = OQ = 1,
OQ OQ

cos 90° = ON = 0 = 0 and
OQ OQ

tan 90° = QN = OQ = undefined (∞)
ON 0

Similarly, cosec 90º = OQ = OQ = 1,
QN OQ

sec 90° = OQ = OQ = undefined (∞) and
ON 0

cot 90° = ON = 0 = 0
QN OQ

From above relations, the values of various trigonometric ratios of standard angles can be summarised as
follows.

0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360°

sin 0 1 1 3 1 31 1 0 – 21 –  1 –  3 – 1 –  3 –  1 – 12 0
2 2 2 2 22 2 2 2 2

cos 1 31 1 0 – 12 –  1 –  3 – 1 –  3 –  1 – 21 0 1 1 3 1
2 22 2 2 2 2 2 2 2

tan 0 1 1 3 ∞ –  3 – 1 –  1 0 1 1 3 ∞ –  3 – 1 –  1 0
3 3 3 3

cosec ∞ 2 2 2 1 2 22 ∞ – 2 –  2 –  2 – 1 –  2 –  2 – 2 ∞
3 3 3 3

sec 1 2 22 ∞ – 2 –  2 –  2 – 1 –  2 –  2 – 2 ∞ 2 2 2 1
3 3 3 3

cot ∞ 31 1 0 –  1 – 1 –  3 ∞ 31 1 0 –  1 – 1 –  3 ∞
3 3 3 3

194 ~ Perfect Optional Mathematics Class 9

Example 1:

Find the value of sin230º + sin245º + sin260º.

 sin230º + sin245º + sin260º

1 2 1 2 32
2 2 2
= + +

= 1 + 1 + 3
4 2 4

= 1 + 2 + 3 = 6 = 32.
4 4

Example 2:

Find the value of : tan pc sin pc + sin pc cos pc + cos pc – sin pc
3 3 4 2 2 3

 Here, the given expression is

tan pc sin pc + sin pc cos pc + cos pc sin pc
3 3 4 2 2 3

= tan 60° sin 60° + sin 45° cos 90° + cos 90° sin 60°

= 3 × 3 + 1 × 0 + 0 × 3
2 2 2
= 32 + 0 + 0
Trigonometry
= 23.
Example 3:

Prove that: 1 + tan 30° = cos 30° + sin 30°
1 – tan 30° cos 30° – sin 30°

 LHS = 1 + tan 30° RHS = cos 30° + sin 30°
1 – tan 30° cos 30° – sin 30°

1+ 1 3 + 1
13 = 2 2
=
1– 3 1
3 2 – 2

3+1 3+1

= 3 = 2
3–1 3–1

32

= 3 + 11 = 33 +– 11
3 –

\ LHS = RHS Proved.

Trigonometric Ratios of Standard Angles ~ 195

Graphs of Trigonometric Functions

A function can be represented in various ways. One of the important way of representing a function is
the graph. In trigonometric function, for different values of the angle, the values of trigonometric ratio
will be different. Taking different values of the angle as the x co-ordinates and corresponding values of
the trigonometric ratio as the y co-ordinates, we can plot the points on the graph and join them with the
freehand curve to get the trigonometric function.

Example 4:

Draw the graph of y = sin x.

 As y = sin x,

We list the values of the angle x ranging from 0º to 360º (i.e. from 0 to 2pc) and
the corresponding values of y i.e. sin x.

x 0° 30° 45° 60° 90° 120° 135° 150° 180°
y = sin x 0 0.5 0.71 0.87 1 0.87 0.71 0.5 0

x 210° 225° 240° 270° 300° 315° 330° 360°
y = sin x –0.5 –0.71 –0.87 –1 –0.87 –0.71 –0.5 0

Y 90° 180° 270° 360°
90°
1.0 90° 180° 270° 360° X
0.8
0.6 180° 270° 360°
0.4
Example 5: 0.2

X'

-0.2
-0.4
-0.6
-0.8
-1.0

Y'

Draw the graph of y = cos x.

 We have y = cos x.

Firstly list the values of the anlge x ranging from 0º to 360º (i.e. from 0 to 2pc) and the
corresponding values of y i.e. cos x.

x 0° 30° 45° 60° 90° 120° 135° 150° 180°
y = cos x
1 0.87 0.71 0.5 0 –0.5 –0.71 –0.87 –1

196 ~ Perfect Optional Mathematics Class 9