Prime Optional Mathematics 7

Matrix

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

12
1. What is the transpose matrix of A = <3 4F ?

2. a. If A = 2 –1 2 3 , find A + B.
<3 1 F, B = <1 –1F

b. What do you mean by square matrix ? Write down with an

example.

c. If =x – 1 y + 2G = 2 15F, find the value of ‘x’ and ‘y’.
3 1 <3

3. a. If aij = 2i + j, find 2 × 2 matrix.
3 2 1 –1
b. If A + B = < F and B = < F , find the matrix A.
1 2
–2 4

21 31
4. If A = <3 –2F and B = <–1 1F, prove that (A + B)T = BT + AT.

46 PRIME Opt. Maths Book - VII

3 Co-ordinate Geometry

Specification Grid Table TM Periods

K(1) U(2) A(4) HA(5) TQ

No. of Questions – 1 1 – 2 6 10

Weight –24–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : Materials Required:
At the end of the lesson • Graph board
• Students are able to know • Graph paper
• Geo-board
concept of co-ordinate • Sheet of paper
system. • Scissors
• Students can plot the points in • Chart of sign used in
graph.
• Students can find distance quadrents
between any two points.
• Students can use section 47
formula and mid-point
formula to find the section
point.
• Students can find the centroid
of a triangle.

PRIME Opt. Maths Book - VII

3.1 Rectangular co-ordinate axis

The number line XX` drawn horizontally is taken as x-axis and the number
line YY` drawn vertically is taken as y-axis. The number line XX` and YY`
are intersected at right angle at the point ‘O’ called the origin point.

............. Y

............. 4
3
2
1
0
–1
–2
–3
–4

Y’

..X...’........ –4 –3 –2 –1 0 1 2 3 4 .........X....
The plane surface is divided by the lines XX` and YY` into four parts called
quadrants. They can be observed as shown in diagram.

Y

4

3

2nd Quadrent 2 1st Quadrent

1
0
X’ –4 –3 –2 –1 1 2 3 4 X

–1

3rd Quadrent –2 4th Quadrent

–3

–4

Y’

48 PRIME Opt. Maths Book - VII

Here, XX` and YY` are perpendicular to each other.
O is the origin point.
First quadrent → Region XOY
Second quadrent → Region X`OY
Third quadrent → Region X`OY`
Fourth quadrent → Region XOY`

Run and Rise
Y

4 P(3, 4)

3
2 4 units

1
0Q
X’ –4 –3 –2 –1 1 2 3 4 X

–1 3 units

–2

–3

–4

Y’
As shown in diagram, the position a point P is expressed as 3 units
distant from the point O horizontally and 4 units distant from the point
Q Vertically. It is written as P(3, 4) where horizontal distance is called
run and vertical distance is called rise.

i.e. Run = OQ = 3 units.
Rise = QP = 4 units.
Position of the point P is (3, 4)

For a point of reference Run is defined as the
horizontal displacement and Rise is defined as the
vertical displacement.

PRIME Opt. Maths Book - VII 49

Sign rule in the quadrant:

The two straight lines intersected at a point which are perpendicular to

each other are called axes. The sign of x-componet and y-component can

be observed as the given diagram where,
• Horizontal line is called x - axis.
• Vertical line is called y - axis.
• The intersecting point of x - axis and y - axis is called the
origin point (O).
Y

y-axis

Second First
quadrent quadrent

(– , +) (+ , +)

X` O (Orgin) x-axis X

Third Fourth
quadrent quadrent

(– , –) (+ , –)

Y`

• In first quadrant:

X - component = positive (+ve)

Y - Component = Positive(+ve)

• In second quadrant:
X - component = negative(-ve)

Y - Component = Positive(+ve)

• In third quadrant:
X - Component = negative(-ve)

y - Component = negative(-ve)

• In fourth quadrant:
X - Component = Positive (+ve)

Y - Component = negative(-ve)

50 PRIME Opt. Maths Book - VII

The pair of the X- component and Y-component of
a point written in the form of (x, y) is called the Co-
ordinate of the point.

Let us Consider a point P(3, 5) Y
X - Component = 3 (OM)
Y - Component = 5(PM)

P(3, 5)

X` O M X
Y`

Run and Rise for any two points (x1 , y1) and (x2 , y2).

Y

A(x , y ) B(x2, y2)
C
1 1

X` OM NX

Y`
Here.
Run = MN = ON – OM = x2 – x1
Rise = BC = BN – CN = BN – AM = y2 – y1

PRIME Opt. Maths Book - VII 51

Distance between any two points (x1 , y1) and (x2 , y2)

Y

A(x , y ) B(x2, y2)
C
1 1

X` OM NX

Y`
If ‘d’ be the distance between any two points A(x1 , y1) and B(x2 , y2)
Draw the perpendiculars.
AM ⊥ OX, BN ⊥ OX and AC ⊥ BN.

Then,
Run = AC = MN = ON – OM = x2 – x1
Rise = BC = BN – CN = BN – AM = y2 – y1

Distance between any two points A(x1, y1) and
B(x2, y2) is,

d = ^x2 – x1h2 + ^y2 – y1h2
d = ^Runh2 + ^Riseh2

In right angled ∆ABC,
h2 = p2 + b2

or, AB2 = AC2 + BC2
or, d2 = (x2 – x1)2 + (y2 – y1)2
` d = ^x2 – x1h2 + ^y2 – y1h2
It is the distance between any two points.
Also,
d = ^Runh2 + ^Riseh2

52 PRIME Opt. Maths Book - VII

Worked out Examples

1. Find the distance between any two points having run = 6 units and
rise = 8 units.
Solution:
Here,
Run = 6 units
Rise = 8 units
Using distance formula,
d = ^Runh2 + ^Riseh2
= 62 + 82
= 36 + 64
= 100
= 10 units

2. Find the distance between the two point A(1, 2) and B(4, 6).
Solution:
The given points are:
A(1, 2) = (x1 , y1)
B(4, 6) = (x2 , y2)
Then,
Distance formula,
d = ^x2 – x1h2 + ^y2 – y1h2
d(AB)= (4 – 1)2 + (6 – 2)2
= 32 + 42
= 9 + 16
= 25
= 5 units.

3. Find the distance of the points A(3, 4) and B(4, 3) from the origin.
Also write down the result.
Solution:
For the points A(3, 4) & origin (0, 0),
(x1 , y1) = (0, 0)
(x2 , y2) = (3, 4)
Using distance formula,

PRIME Opt. Maths Book - VII 53

d(OB)= ^x2 – x1h2 + ^y2 – y1h2
= ^3 – 0h2 + ^4 – 0h2
= 25
= 5 units

For the points B(4, 3) and origin (0, 0)
(x1 , y1) = (0, 0)
(x2 , y2) = (4, 3)
Using distance formula,
d(OB)= ^x2 – x1h2 + ^y2 – y1h2

= ^4 – 0h2 + ^3 – 0h2
= 16 + 9

= 25
= 5 units.
Here, OA = OB [They are equal].

4. Prove that the points A(1, 2), B(4, 6) and C(8, 9) are the vertices of
an isosceles triangle.
Solution:
The given points are
A(1, 2), B(4, 6) & C(8, 9)
Taking AB
(x1 , y1) = (1, 2)
(x2 , y2) = (4, 6)
d(AB)= ^x2 – x1h2 + ^y2 – y2h2
= ^4 – 1h2 + ^6 – 2h2
= 32 + 42
= 5 units.
For BC
(x1 , y1) = (4, 6)
(x2 , y2) = (8, 9)
d(BC) = ^x2 – x1h2 + ^y2 – y1h2
= ^8 – 4h2 + ^9 – 6h2
= 16 + 9

= 25
= 5 units

54 PRIME Opt. Maths Book - VII

For AC
(x1, y1) = (1, 2)
(x2, y2) = (8, 9)
d(AC) = ^x2 – x1h2 + ^y2 – y1h2

= ^8– 1h2 + ^9 – 2h2

= 72 + 72

= 98

= 7 2 units.
Here, AB = BC,
So, ∆ABC is an isosceles triangle.

5. Proved that the pionts A (2, 1), (–1, 5) and (–4, 9) are collinear
points.
Solution:
The points A & B.
A (2, 1) = (x1, y1)
B (–1, 5) = (x2, y2)
then,
d = (x2 – x1)2 + (y2 – y1)2

d = (–1 – 2)2 + (5 – 1)2

d = (–3)2 + (4)2

d = (9 + 16)

d = 25
d = 5 unites

For the points B and C
B (–1, 5) = (x1, y1)
C (–4, 9) = (x2, y2)
then,
d = (x2 – x1)2 + (y2 – y1)2

d = (–4 + 1)2 + (9 –5)2

d = 9 + 16

d = 25
d = 5 unites

PRIME Opt. Maths Book - VII 55

For the points AC,
A (2, 1) = (x1, y1)
C (–4, 9) = (x2, y2)
then,
d = (x2 – x1)2 + (y2 – y1)2
d = (–4 – 2)2 + (9 – 1)2
d = 36 + 64
d = 100
d = 10 units.

Here,
AB + BC = AC
or, 5 + 5 = 10

10 = 10
Hence, A,B & C are collinear points.

Exercise : 3.1

1. Plot the following points in graph and join one after another. Also
name the diagrams so formed.
i. (–2, 3) and (5,7)
ii. (3, 6), (1, 0) and (5, 3)
iii. (4, 3), (–3, 4), (–2, –4) and (5, –3)
iv. (4, 7), (2, 1), (2, –4), (6, –3) and (7, 2)
v. (5, 1), 6, 2), (4, 5), (–4, 5), (–5, 2), (–7, 4), (–6, 0), (–7, –3),
(–5, –2), (–4, –4), (4, –4), (6, –1) and (5, 1)

2. Find the distance between the points under the followings.

i. Run = 4 units, Rise = 3 units

ii. Run = 8 units, Rise = 6 units

iii. Run = 5 units Rise = 12 units

iv. Run = 8 units, Rise = 15 units

v. Run = 7 units, Rise = 3 units.

56 PRIME Opt. Maths Book - VII

3. Find the distance between the following pairs of points.
i. A(2, 5) and B(6, 8)
ii. P(3, 7 ) and Q(9, 15)
iii. C(1, 2) and D(6, 14)
iv. Q(4, 5) and R(–11 , –3)
v. B(0, 0) and C^ 11, 14h

4. Prove that the followings:
i. AB = BC where the points are A(3, 2), B(9, 10) and C(1, 4).
ii. Points P(3, 7), Q(–4, 3) and R(–3, –4) are equidistant from the
origin.
iii. Prove that points A (3, 2), B(6, 6) and C(7, 5) are the vertices of
an isosceles triangle.
iv. Prove that PQ = 2RS where the points are P (2, 1), Q= (6, 9),
R =(2, 1) and S (10, 7).
v. Prove that the points P(1, 2), Q (–3, 5) and R (–2, –2) are the
vertices of an isosceles triangle.

5. Prime more creative questions:
i. Prove that K (–3, 2), L(–2, 6), M (2, 7) and N(1, 3)are the
vertices of an equilateral triangle.
ii. Prove that P (0, 1), Q (2, 3) and R (3, –2) are the vertices of a
right angled triangle.
iii. Prove that A (–3, 3), B (3, –7) and C (7, 9) are the vertices of an
isosceles right angled triangle.
iv. Prove that the points (3, 4), (7, 7) and (11, 10) are collinear
points.
v. Prove that the points (0, 4), (3, –2) and (5, –6) are collinear.

Answer iv. 17 units v. 4 units
iv. 17 units v. 5 units
1. i. Show to your teacher.
2. i. 5 units ii. 10 units iii. 13 units
3. i. 5 units ii. 10 units iii. 13 units
4. Show to your teacher.
5. Show to your teacher.

PRIME Opt. Maths Book - VII 57

3.2 Mid- point of a line segment B(x2, y2)

Let us consider a point P(x, y) be a P(x, y)
point which divides the line joining

the points A(x1, y1) and B(x2, y2) in two A(x1, y1)
equal parts. The point P is called mid-

point of the line segment AB. It’s co-ordinate can be calculated by using

the concept of average value of each components of the points A and B.
x1 + x2 y1 + y2
x = 2 and y = 2

` P(x, y) = a x1 + x2 , y1 + y2 k
2 2

Examples: Mid- point of a line joining the points A(1, 2) and B(5, 6)

is calculated as, y1 + y2
x1 + x2 2
x = 2 , y =

= 1+5 , = 2+6
2 2

= 6 = 8
2 2

=3 =4

` Mid-point of AB is (3, 4) .

Centroid of a triangle: (x1A, y1)
RQ
Let us consider a triangle having vertices
G
A(x1, y1), B(x2, y2) and C(x3, y3) P, Q and R are
the mid-point of sides of the ∆ABC where
AP, BQ and CR are the medians. The medians

are intersected at point G. Where G is taken

as the centroid of ∆ABC. It’s co-ordinate can

be calculated by taking the average value of B P C
(x2, y2) (x3, y3)
the components of the points A and P. ( x1 + x2 , y1 + y2 )
x1 + x2 + x3 y1 + y2 + y3 2 2
x = 3 , y = 3

i.e. (x, y) = a x1 + x2 + x3 , y1 + y2 + y3 k
3 3

58 PRIME Opt. Maths Book - VII

Examples: The centroid of ∆ABC having vertices A(3, 1), B (5, 7) and

C (4, –2) is, y1 + y2 + y3
x1 + x2 + x3 3
x = 3 and y =

or, = 3+5+4 = 1 + 7–2
3 3

or, = 12 = 6
3 3

or, = 4 =2

` The centroid of ∆ABC is G (4, 2).

Worked out Examples

1. Find the mid- point of a line segment joining the points (3, –1) and
(5, 3).
Solution:
The givens points are,

A (3, –1) = (x1, y1)

B (5, 3) = (x2, y2)

Now, mid- point of line segment AB is,

x = x1 + x2 , y = y1 + y2
2 2

= 3+5 = –1 + 3
2 2

= 8 = 2
2 2

=4 =1

` Mid-point of AB is (4, 1)

2. If (2, 4) is the mid point of a line segment joining the points A (1, 3)
and B (a, b), find the co-ordinate of B.
Solutions:
The given points are:

A (1, 3) = (x1, y1)
B (a, b) = (x2, y2)
mid-point (2, 4) = (x, y)

PRIME Opt. Maths Book - VII 59

We have, + y1 + y2
2 2
x = x1 x2 , y=

or, 2 = 1+a , 4= 3+b
2 2

or, 1+ a = 4, 3+b=8

or, a = 4 – 1, b=8–3

or, a = 3, b=5

` Co-ordinate of point B is (3, 5).

3. Find the centroid of a triangle having vertices P(2, 1), Q (4, 5) and

R (6, 0).

Solution:

The given vertices of ∆PQR are,

P (2, 1) = (x1, y1)
Q (4, 5) = (x2, y2)
R (6, 0) = (x3, y3)
Now, Co-ordinate of centroid,
+ + y1 + y2 + y3
x = x1 x2 x3 , y = 3
3

= 2+4+6 , = 1+5+0
3 3

= 12 , = 6
3 3

= 4, =2
` The centroid is (4, 2).

4. If (1, 3) is the centroid of ∆ABC having vertices A (m, 2), B (–1, 4)
and C (2, n). find the value of ‘m’ and ‘n’.

Solution:

The given vertices of ∆ABC are:

A (m, 2) = (x1, y1)
B (–1, 4) = (x2, y2)
C (2, n) = (x3, y3)
We know,
+ + y1 + y2 + y3
x = x1 x2 x3 , y= 3
3

or, = m–1 + 2 , 3= 2+4+n
3 3

60 PRIME Opt. Maths Book - VII

or, m + 1 = 3, 6+n=9
or, m = 3 – 1, n=9–6
or, m = 2, n=3
` m=2
n =3

5. Find the length of the median drawn from (–3, –4)
�irst vertex of a triangle having vertices A A

(–3, –4), B (4, 5), and C (6, –1).

Solution:
Let, AD be the median drawn from �irst
vertex of ∆ABC,where D is mid- point of

BC. B D C
Now, (4, 5) (6, –1)
For the mid-point of BC,

B (4, 5) = (x1, y1), y1 + y2
C (6, –1) = (x2, y2) 2
We have,
x1 + x2
x = 2 , y =

= 4 + 6 , = 5–1
2 2

= 10 , = 4
2 2

= 5, = 2

Again, for the length of AD,
A (–3, –4)= (x1, y1)
D (5, 2) = (x2, y2)
then,
d = ^x2 –x1h2 + ^y2 –y1h2

= ^5 + 3h2 + ^2 + 4h2
= ^8h2 + ^6h2

= 64 + 36

= 100
= 10 units.
` Length of median AD is 10 units.

PRIME Opt. Maths Book - VII 61

Exercise : 3.2

1. Find the Co-ordinate of mid-point of the line segments from the
following points.
i. (3, 5) and (7, 3)
ii. (1, –3) and (5, 9)
iii. (–3, 4) and (1, 6)
iv. (–1, –3) and (5, –5)
v. (a + b, b – a) and (a –b, a + b)

2. Find the Co-ordinate of centroid of triangle gives below.
i. A (1, 1), B (2, 5), and C (3, 3)
ii. P (3, 2), Q (–1, 5) and R (1, 5)
iii. K (–2, 0), L (0, 6) and M (5, –3)
iv. A (5, 6), B (1, 1) and C (–3, 5)
v. D (–3, –4) E (2, –5) and C (1, 0)

3. Find the followings.
i. If (1, 2) is the mid-point of line joining the points A (a, 4) and
B (1, b), find the value of ‘a’ and ‘b’.
ii. If (2, 3) is the mid- point of line joining the points P (2, 5) and
Q (a, b), find the co-ordinate of the point Q.
iii. If the centroid of a triangle having vertices A (a, 2) B (0, b) and
C (3, –2) is (2, –1), find the value of ‘a’ and ‘b’.
iv. If (–2, 1) is the centroid of ∆PQR having vertices P(2, 1), Q(–3, 5)
and R (p, q),find the co-ordinate of R.
v. Find the mid-point of sides of ∆ABC having vertices A (3, 1),
B (–1, 5) and C (5, –3).

4. Prime more creative questions:
i. Find the mid- point of diagonals of a parallelogram having
vertices P (1, 2), Q (3, 6), R (7, 8) and S (5, 4).
ii. If B(m, n) lines in the line joining the points A(2, 5) and C(4, 7),
find the co-ordinate of B where AB = BC.
iii. If P (a, b), Q (2, 5) and R (6, 7) lines in a same straight lines,

62 PRIME Opt. Maths Book - VII

find the co-ordinate of P where PQ = QR.
iv. Find the length of the median drawn from first vertex of a

triangle having vertices A (1, 2), B (5, 3) and C (3, 9).
v. Find the length of the median drawn from first vertex of

triangle having vertices (0, 4), (5, –2) and (7, –6).

1. i. (5, 4) Answer iii. (–1, 5)
iv. (2, –4)
ii. (3, 2)
2. i. (2, 3) v. (a, b)
iv. (1, 4)
ii. (1, 4) iii. (1, 1)
3. i. a = 1, b = 0 v. (0, –3)
iv. (–5, –3)
ii. (2, 1) iii. a = 3, b = –3
4. i. (4, 5)
iv. 5 units v. (1, 3), (2, 1) and (4, –1)

ii. (3, 6) iii. (–2, 3)
v. 10 units

PRIME Opt. Maths Book - VII 63

Co-ordinate Geometry

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. Write down the formula to find distance between any two points

A(x1, y1) and B(x2, y2)
2. a. Find the distance of a point A(4, –3) from the origin point ‘O’.

b. Find the mid-point of line joining the points A(1, 2) and B(5, 4).
c. If (1, 3) is the mid-point of line joining the points P(a, 2) and

Q (3, b), find the value of a and b.
3. a. Find the co-ordinate of centroid of a triangle having vertices

A(3, 1), B(2, 5) and C(–2, 6).
b. Prove that AB = BC where the points are A(–2, 1), B(1, 5) and

C(5, 8).
4. Prove that the vertices A(5, 3), B(5, –2) and C(10, –2) are the

vertices of an isosceles triangle.

64 PRIME Opt. Maths Book - VII

4 Trigonometry

Specification Grid Table TM Periods

K(1) U(2) A(4) HA(5) TQ

No. of Questions 1 1 2 – 4 11 20

Weight 128–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : Materials Required:
At the end of the lesson • Chart paper.
• Students can solve the right • Graph paper.
• Chart of list of formulae
angled triangle.
• Students are able to know used in trigonometry.
• Chart of values of
the types of measurement of
angles. standard angles.
• Students can find the • Model of right angled
trigonometric ratios.
• Students can prove the triangle.
trigonometric identities.
• Students can find the value 65
of the ratio of the standard
angles.
• Students can find height and
distance.

PRIME Opt. Maths Book - VII

Trigonometry

The word trigonometry is defined using three words; tri → three, gones
→ angles, metron → measurement. Tri + gones + metron = trigonometry
(Measurement of three angles of a triangle.) It is said that the origin of
the trigonometry is taken from the ancient Hindu Civilization as,

Tri – lq
gono – sf0] f
metry – dfk0f

The measurement of three angles
of a triangle is called trigonometry.

It is useful to find the angles of a right angled triangle and to find the
length of the sides of it. It is used by engineers to find the heights and
distance during constraction of structures which is useful to estimate
the constructing materials and cost. It is also useful for various purposes
in physics, mathematics, statistics as well as other scientific purposes.

4.1 Measurement of angles

The angles are measured according to English system, french system and
Radian System where English system is called sexagesinal and french
system is called centesimal system. The two way of measurement of
angles sexagesinal and centesimal systems are discussing here in grade
VII.

i. Sexagesimal (English system)

1 right angle = 90° (degrees)
1° = 60’ (minutes)
1’ = 60’’ (seconds)

66 PRIME Opt. Maths Book - VII

ii. Centesimal (french system)

1 right angle = 100g (grades)
1g= 100’ (minutes)
1’ = 100’’ (seconds)

iii. Relation between Sexagesimal and Centesimal
We have,
One right angle can be defined as 90° in sexagesimal and 100g in
centesimal measurement.

So,

90° = 100g = 1 right angle

1° = a 10 g
9
k

1g = a 9 kc
10

iv. Sum of the angles of a triangle.
Sum of the interior angles of a triangle is defined as two right angle.

In DABC
∠A + ∠B + ∠C = 2 right angle.
∠A + ∠B + ∠C = 180°
∠A + ∠B + ∠C = 200g

Worked out Examples

1. Convert 15° 24’ 32’’ into seconds.
Solution:
15° 24’ 32’’ = (15 × 60 × 60 + 24 × 60 + 32)’’
= (54000 +1440 + 32)’’
= (55472)’’

PRIME Opt. Maths Book - VII 67

2. 40g 25’ 60’’ convert into minutes.

Solution: = (40 ×100 + 25 + 6 )’
40g 25’ 60’’ 100

= (4000 + 25 + 0.6)’

= (4025.6)’

3. Convert 54° 24’ 45’’ into degrees.

Solution: 24 45
60 60×60
54° 24’ 45’’ = a54 + + 0

k

= (54 + 0.4 + 0.0125)°

= (54.4125)°

4. Convert 42° 75’ 84’’ into grades.

Solution: 75 84
100 100×100
42° 75’ 84’’ = a42D e m o + + g

k

= (42 + 0.75 + 0.0084)g

= (42.7584)g

5. One angle of a triangle is 80g, second angle is 38°. Find the third

angle in degrees. A

Solution: 80g
Let, ABC be a triangle.

Where the angles are, B 38° ? C
∡A = 80g, ∡B = 38° and ∡C = ?

we have,
∡A = + ∡B + ∡C = 180°
or, 72° + 38° – ∡c = 180°
or, ∡C = 72° + 38° – 180°
or, ∡C = 110° – 180°
or, ∡c = 70°

i.e. The third angle is 70°

68 PRIME Opt. Maths Book - VII

6. The angles of a triangle are in the ratio 2:3:5, �ind the angles in
grades.

Solution: P

Let, PQR be a triangle where the angles are in the 2x
3x 5x R
ratio 2:3:5. Q
i.e. ∡P = 2x

∡Q = 3x
∡R = 5x

Where,
∡P + ∡Q + ∡R =200g
or, 2x + 3x + 5x = 200g
or, 10x = 200g
200
or, x = 2100g g
or, x =

Then,

The angles of the triangle are :
∡P = 2 × 20g = 40g
∡P = 3 × 20g = 60g
∡R = 5 × 20g = 100g

7. Find the ratio of the angles 80g and 108°.

Solution:
First angle = 80g
Second angle =108°
10 g
= a180× 9
k

= (120)g

Then, �irst angle
sec ond ange
Ratio =

= 80g
120g
= 2:3

PRIME Opt. Maths Book - VII 69

Exercise : 4.1

1. Convert the followings into seconds.

i. 12° 15’ 40” ii. 50° 25’

iii. 70° 20’ 25” iv. 42g 32’ 22”

v. 15g 35’

2. Convert the following into minutes.

i. 20° 30’ 45” ii. 80° 15”

iii. 25g 20’ 44” iv. 42g 50”

v. 72g 42’ 25”

3. Convert the followings into degrees.

i. 40g ii. 20° 18’ 54”

iii. 42° 24’ iv. 54° 30’ 36”

v. 75g 72’ 81”

4. Convert the followings into grades.

i. 72° ii. 42g 45’ 48”

iii. 80g 40’ iv. 64g 42’ 25”

v. 45° 15’ 27”

5. Find the followings:
i. Find the ratio of the angles 72° and 200g.
ii. Find the sum of the angles 80g and 27° in grades.
iii. Find the difference of the angles 90g and 54° in degrees.
iv. Find the angles of a triangle in degrees where the angles are in
the ratio 1:2:3.
v. Find the ratio of the angles 48° and 80g by converting into
grades.

6. Find the angles from the followings:

i. The two angles of a triangle are in the ratio 3:5 and the third
angle is 60°, find the angles in degrees.

ii. One angle of a triangle is 40g and second angle is 74°. Find the

third angle is degrees.

70 PRIME Opt. Maths Book - VII

iii. One angle of a triangle is 50g and second angle is 72°. Find the
third angle in grades.

iv. The two angles of a triangle are in the ratio 2:3 and the third
angle is 45°. Find the angles in grades.

v. The two angles of a triangle are 40° and 68° respectively. Find
the third angle in grades.

7. Prime more creative questions:
i. Convert 72° 36’ 45” into centesimal measurement.
ii. Convert 84g 15’ 54” into centesimal measurement.
iii. One angle of a triangle is more than the second angle by 20g.
Find the angles of the triangle in degrees. Where the third
angle is 62°.
iv. One angle of a triangle is less than the second angle by 36°.
Find the angles in grades. Where the third angle is 54°
v. If 63° is taken out from an angle of 160g what will be the
remaining part in degrees?

1. i. 44140” Answer iii. 253225”
iv. 423222” iii. 2520.44’
ii. 181500” iii. 424°
2. i. 1230.75’ v. 153500” iii. 80.4g
iv. 4200.5’ iii. 27°
ii. 4800.25’ iii. 70g
3. i. 36° v. 7242.25’ iii. 50°, 68°, 62°
iv. 54.51°
ii. 20.315°
4. i. 80g v. 68.1553°
iv. 64.4225g
ii. 42.4548g
5. i. 2:5 v. 50.2861g

iv. 30°, 60°, 90° ii. 110g

6. i. 45°, 75°, 60° v. 2:9
iv. 52g, 78g, 50g
ii. 70°
7. i. 80g, 68’ 5” v. 80g
iv. 50g, 90g, 60g
ii. 75° 44’ 23.5”
v. 81°

PRIME Opt. Maths Book - VII 71

4.2 Introduction of right angled triangle

Right angled triangle:
The triangle having one of the angle is a right angle

(90°) is called right angled triangle.

• Side opposite to right angle is called hypotenuses. It is longest side of
the triangle.

• Other two perpendiculars can be taken as perpendicular and base.

Let us consider a DABC is a right angled triangle where one of the angle

∠B = 90°. A

BC
Here,
∠B = 90° [right angle]
AC= Side opposite to right angle is hypotenuses.
AB and AC = Perpendicular to each other and can be taken as perpendicular
and base.

Relation of sides and angles in a right angled triangle:
Activity for the relation of sides:

Draw a triangle having an angle 90° and sides taking 3cm, 4cm and 5cm
respectively as shown in diagram.

A

5cm
3cm

B 4cm C

72 PRIME Opt. Maths Book - VII

Construct the squares in each sides as shown in diagram.

A 3cm 3
1 5cm

B 4cm C
2

Find the area of the squares so formed in each sides of right angled
triangle as shown below.
Area of square 1 =(AB)2 = (3 cm)2 = 9 cm2
Area of square 2 = (BC)2 =(4 cm)2 = 16 cm2
Area of square 3 = (AC)2 = (5 cm)2 = 25 cm2
Conclusion upon the area of squares as discussed above.

25 = 9 + 16
or, Square 3 = square 1 + square 2
or, AC2 = AB2 +BC2
i.e h2 = p2 + b2

The relation of sides:
• h2 = p2 + b2

w h = p2 + b2
w p = h2 – b2
w b = h2 – p2

PRIME Opt. Maths Book - VII 73

Activity for the relation of angles.

Let us consider a right angled triangle having an angle 20°as shown in

diagram. A

B 20° C

Here, In right angled ∆ABC,
∠B =90°
∠C =20°
∠A =?

We have,
Sum of the angle of a triangle is 180° so,
∠A + ∠B + ∠C =180°

or, ∠A + 90° + 20° =180°
or, ∠A + 110° = 180°
or, ∠ A =180°-110°
\ ∠ A = 70°
Conclusion for the angles.

∠B =90°
∠A= 70°
∠C=20°
Here, 90° =70°+20°
i.e ∠B = ∠A + ∠C
i.e. ∠B = 90° – ∠C

• Angles other than right angle are always acute
angles.

• Sum of the acute angles is always a right angle.
• The acute angles are called complementary angles.
• One acute angle = 90° - other.

74 PRIME Opt. Maths Book - VII

Reference angle:

In a right angled triangle one of the angle is always a right angle (90°)
angle and other reaming two angles are always acute angles. One of the
acute is taken as the reference angle.

Reference angle is used to define perpendicular side and base of a right
angled triangle.

Lets us consider a right angled DABC Where ∠C is taken as the reference

angle. A

B C

Here,
In right angled DABC
∠B =90°
∠C = reference angle.

Then
The side can be defined as,
AC = hypotenuses (h)
AB = perpendicular (p)
BC = base (b)

In a right angled triangle,
• Side opposite to right angle is hypotenuse.
• Side opposite to reference angle is perpendicular.
• Side Remaining from the others is base.

PRIME Opt. Maths Book - VII 75

Worked out Examples

1. Is the triangle having sides 8 cm,15cm and 17 cm as right angled?

Solution: A

Let, ABC be a triangle 8 cm 17cm
Where AC is taken as the longer side.
Where,

AB = 8 cm B 15cm C
BC = 15 cm
AC = 17 cm

Let us examine,

AC2 = AB2 + BC2

or, (17)2 = 82 +(15)2

or, 289 = 64 + 225

or, 289 = 289

Hence, It is right angled at B.

2. Find the unknown side from the given triangle. P

Solution

In the given triangle DPQR.

∠Q = 90° 6 cm

PQ = 6 cm = P

QR = 8 cm = b Q 8 cm R
PR = ? = h

We have,

h = p2 + b2

= 62 + 82

= 36 + 64

= 100

= 102

= 10cm

\ h = PR = 10cm

76 PRIME Opt. Maths Book - VII

3. Find the unknown angle from the given right angled triangle.

Solution: A
In the given right angled triangle ∆ABC, 50°

∠B =90°

∠A = 50°

∠C =? BC
We have,

Reaming angles of right angled triangle are complementary, so

∠A + ∠C = 90°

or, 50° + ∠C = 90°

or, ∠C = 90° – 50°

\ ∠C = 40°

4. Write down the sides base, perpendicular and hypotenuses of the

right angled triangle by taking ∠A is a reference angle in a right

angled triangle DABC from the given diagram B
Solution:
In a right angled ∆ABC

∠B = 90°(right angle)

∠A = Reference angle. AC
Then,

Side AC = hypotenuse(h)

(It is opposite to right angle)

Side BC= Perpendicular (p)
(It is opposite to reference angle)

Side AB = base (b)
(It is remaining side)

PRIME Opt. Maths Book - VII 77

Exercise : 4.2

1. Examine the triangle where sides given below are right angle or
not.
i) 3 cm, 4cm, 5cm; DABC. ii) 8cm, 10cm, 6cm; DPQR
iii) 17cm, 10cm, 8cm; DKLM iv) 7cm, 25cm, 24cm; DABC
V) 10cm, 15cm, 20cm; DPQR

2. Find the length of the unknown side from the given diagrams.

i) A ii) P

10cm
3cm
B 24cm4cmCQ 6cm R
A Q R
iii) 15cmiv)
5cm 8cm
B P
13cm
v) K C

R

25cm

Q

3. Find the unknown angle from the given triangles. C
ii) A
i) P

70°

Q 30° R B
78
PRIME Opt. Maths Book - VII

iii) K 40° L iv) P R
Q
M

v) A

CB

4. Write down the name of sides of right angled triangle according to

given reference angle from the given diagrams.

i) A B ii) P

iii) K C Q R
M A
iv) C

L B
Q
v) P

R

5. Prime more creative questions.
i) Is a side of length 4 5 a hypotenuses of right angled triangle
having other two sides 4cm and 8cm?
ii) Find the length of the perpendicular of a right angled triangle
having other two sides 5 5 cm and 10cm respectively.

PRIME Opt. Maths Book - VII 79

iii) Is an isosceles triangle having an acute angle 45° right angled
or not ? Explain with calculation?

iv) Prove that the triangle having sides 3 cm, 2 3 cm and 15
cm is a right angled triangle.

v) An electric pole of height 8 2 m is tied with a rope from 4m
away from the foot as shown in diagram. Find the length of the
rope.

rope 8 2 m

4m

Answer

1. i. Right angled ii. Right angled iii. No right angled

iv. Right angled v. No right angled

2. i. 5cm ii. 8 cm iii. 12cm iv. 17cm v. 7 cm

3. i. 60° ii. 20° iii. 50° iv. 45° v. 45°

4. Do yourself.

5. i. yes ii. 5 cm iii. Yes v. 12 m

80 PRIME Opt. Maths Book - VII

4.3 Trigonometric ratios

Let us consider DABC is a right angled triangle at ∠B and ∠A is taken as

the reference angle. C

BA

Then,
AC= Hypotenuses(h)
BC= Perpendicular(p)
AB= base(b)
Out of the three sides of a right angled triangle, ratio of any two sides can
be taken to define the trigonometric ratios.

The ration of any two sides of a right angled triangle
on the basis of reference angle is called trigonometric
ratio.

Where, p
h
Ratio of perpendicular and hypotenuses = Sin A =

Ratio of base and hypotenuse = Cos A = b
h

Ratio of perpendicular and base = Tan A= p
The receprocal ratios are: b

Cosec A = h Sec A = h Cot A = b
p b p

Note : Some person has curly brown hair to produce beauty.

Symbols used for reference angle
Different Greek letters can be used to denote the reference angles in a
right angled triangle. Some of them are as follows.

PRIME Opt. Maths Book - VII 81

a – Alpha b – Beta g – Gamma d – delta
k – Kappa p – Pai q – Theta l – Lyamda
ψ – Sai

Algebraic operations can be used for trigonometry as discussing below.

a + a = 2a [SinA + SinA = 2SinA]

3a – a = 2a [3SinA – SinA = 2SinA]

a × a = a2 [SinA × SinA = Sin2A ≠ SinA2]

a3 ÷ a = a2 [Sin3A ÷ SinA = Sin2A ≠ SinA2]

a2 + a = a(a + 1) [Sin2A + SinA = SinA(SinA +1)]

a2 – b2 = (a + b)(a – b) [Sin2A – cos2A = (SinA + CosA) (SinA -CosA)]

Worked out Examples

1. Find the trigonometric ratios of Tan A and cosec A from the given

right angled triangle. A
Solution:

In a right angled DABC

∠B = 90° (right angle)

∠A = Reference angle. BC

Then, p
b
Tan A= = BC
AB

Cocec A = h AC
p = BC

2. Find Cotθ and Secθ from the given right angled triangle. P
Q
Solution : 6cm
In a right angled ∆PQR

∠Q = 90° = right angled R θ
∠R = θ = reference angle. 8cm
PQ = 6 cm = P

QR =8 cm = b

Then,
h = p2 + b2

PR = PQ2 + QR2

82 PRIME Opt. Maths Book - VII

= 62 + 82
= 36 + 64

= 100

= 10 cm

Then, QR 8cm 4
PQ 6cm 3
Cotθ = b = = =
P

Secθ = h = PR = 10cm = 5
b QR 8cm 4

3. Add: Sin²A + SinA.CosA + Cos²A and 2sin²A – 2sinA.CosA – 3Cos²A
Solution:
= (sin²A + SinA.CosA + Cos²A) + (2sin²A – 2sinA.CosA – 3Cos²A)
= 2sin²A + Sin²A + SinA.CosA – 2sinA.CosA + Cos²A – 3cos²A
= 3Sin²A – SinA.CosA – 2Cos²A

4. Multiply: (Sin A - Tan A) (Sin² A +Sin A Tan A + Tan2A)
Solution:
(SinA – TanA) (Sin²A + SinA TanA+ Tan²A)
= SinA(Sin²A + SinA TanA + Tan²A) – TanA(Sin²A + SinA TanA + Tan²A)
= Sin³A + Sin2A.TanA + SinA.Tan2A – Sin2A.TanA – SinA.Tan2A – Tan³A
= Sin³A – Tan³A.

5. Factorise: Tan²A – Cos²A + TanA + CosA
Solution:
Tan²A – Cos²A + TanA + CosA
= (TanA + CosA) (TanA – CosA) + 1(TanA + CosA)
= (TanA + CosA) (TanA – CosA + 1)

PRIME Opt. Maths Book - VII 83

Exercise : 4.3

1. Find the given trigonometric ratios from the given right angled
triangles:

C

i) Find SinA and CosA.

BA

ii) Find TanA and secA. A
B
C

P Q
θ
iii) Find Cosθ and Cosecθ.
R
P aR

iv) Find Cosa and Sina. Q
B
A
PRIME Opt. Maths Book - VII
v) Find Cosb and Tanb.

b
C
84

2. Find the length of the unknown sides and write down the ratios
from the followings.
4 cm
P a R

i) Find Sina and Tana. 3 cm

Q

A 10cm
q

ii) Find Cosecq and Cotq.

B 8cm C

AB

iii) Find SecA and SinA. 13cm 5cm

C

P 17cm R
b
iv) Find Cosb and Cotb.

8cm

Q

K θ M
4cm
v) Find Cosecθ and Tanθ.

L

3. Add the followings:
i) 2Sin2θ +2Sinθ – Cos2θ and 3Sinθ + Sin2θ – Cos2θ
ii) 3Tan²θ + Sinθ – Cos²θ and Tan²θ – 2Sinθ + 2Cos²θ
iii) Cot2 θ – 3Cotθ + Cos2θ and 2Cos2θ – Cot θ + 3Cot2θ
iv) 3Tan2θ + 2CosθTanθ – Cos2θ and CosθTanθ – 2Tan2θ – Cos2θ.
v) 5Sec2θ – Secθ.Coseθ + 2Cosec2θ and 3Sec2θ – SecθCosecθ – 2Cosec2θ

PRIME Opt. Maths Book - VII 85

4) Subtract the trigonometric expressions given below.
i) 5Sec2θ – Secθ.Cosecθ + 2Cosec2θ and 3Sec2θ -Secθ.Cosecθ – 2Cosec2θ
ii) Cot2 θ – 3Cotθ + Cos2θ and 2Cos2θ – Cot θ + 3Cot2θ
iii) 2Sin2θ +2Sinθ – Cos2θ and 3Sinθ + Sin2θ – Cos2θ
iv) 3Tanθ + Sinθ – Cosθ and Tanθ – 2Sinθ + 3Cosθ
v) 3Tan2θ + 2Cosθ.Tanθ – 2Cos2θ and Cosθ.Tanθ – 2Tan2θ – Cos2θ.

5) Multiply the followings.
i. (Sin A + Cos A) ( Sin A – Cos A)
ii. ( Tan A + SecA) (Tan2A – TanA.SecA + Sec2A)
iii. SinA(Cos A + Cot A) – CosA(SinA + Cot A)
iv. (Sin A – CosA) (Sin2A + SinACosA + Cos2A)
v. (2TanA – 3) (3TanA + 2)

6. Factories the following:
i. Tan²A – Cot²A
ii. Sin³A – Cos³A
iii. Sin²A + Sin²A.Cot²A
iv. Sinθ (Cosθ – Sinθ) + Cosθ (Cosθ – Sinθ)
v. Sin²θ + Sinθ.Cosθ + Cos²θ + Sinθ.Cosθ

7. Prime more creative questions.
Cos²A – Sin²A
a. i. Simplify : CosA + SinA

ii. Factorise : Sin4A – Sin²A.Tan²A
iii. Expand : (CotA + SecA)²
iv. Factories : Sin4A – Tan4A
Tan³A – Cot³A
v. Simplify : Tan²A + TanA.CotA + Cot²A

b. i. Factorise :Tanα(Sinα – Cosα) + Cotα(Cosα – Sinα)
ii. Factorise : Sin8θ – Cos8θ
iii. Multiply : (Sin²θ + Cos²θ)(Sin4θ – Sin²θCos²θ + Cos4θ)
1 1
iv. Simplify : 1 + Sin – Sin – 1

v. Expand : (SinA + CosA)3

86 PRIME Opt. Maths Book - VII

Answer

1. Show to your teacher.

2. i. 5cm, 3 , 3 ii. 6cm, 5 , 3 iii. 12cm, 13 , 5
5 4 4 4 12 13

iv. 15cm, 15 , 15 v. 4 2 cm, 1 ,1
17 8 2

3. i. 3Sin²q + 5Sinq – 2Cos²q
ii. 4Tan²q – Sinq + Cos²q
iii. 4 Cot²q – 4Cotq + 3Cos²q
iv. Tan²q + 3Cosq.Tanq – 2Cos²q
v. 8Sec²q – 2Secq.Cosecq

4. i. 2Sec²q + 4Cosec²q ii. –2Cot²q – 2Cotq – Cos²q

iii. Sin²q – Sinq iv. 2Tanq + 3Sinq – 4Cosq

v. 5Tan²q + Cosq.Tanq – Cos²q

5. i. Sin²A – Cos²A ii. Tan³A + Sec³A

iii. SinA.CotA – CosA.CotA

iv. Sin³A – Cos³A v. 6Tan²A – 5TanA – 6

6. i. (TanA + CotA)(TanA – CotA)

ii. (SinA – CosA)(Sin²A + SinA.CosA + Cos²A)

iii. Sin²A(1 + Cot²A)

iv. (Cosq – Sinq) (Sinq + Cosq) v. (Sinq + Cosq)²

7. a. i. CosA – SinA ii. Sin²A(SinA + TanA)(SinA – TanA)
iii.
iv. Cot²A + 2CotA.SecA + Sec²A
v.
(Sin²A + Tan²A)(SinA + TanA)(SinA – TanA)
b. i.
ii. TanA – CotA

iii. (Sina – Cosa)(Tana – Cota)

v. (Sin4q + Cos4q)(Sin²q + Cos²q)(Sinq + Cosq)(Sinq – Cosq)

Sin6q + Cos6q iv. 2
1 – Sin2 T

Sin³A + 3Sin²ACosA + 3sinACos²A + Cos³A

PRIME Opt. Maths Book - VII 87

4.4 Trigonometric identities

1. Reciprocal relations: p h
h p
i. SinA × Cosec A = × = 1.

\ Sin A.CosecA = 1
1
SinA = CosecA

CosecA = 1
SinA

ii. CosA × SecA = b × h =1 SinA.CosecA = 1
h b

\ CosA.SecA = 1 CosA.SecA = 1
1
CosA = SecA TanA.CotA = 1

Sec A = 1 TanA = SinA
CosA CosA

iii. TanA × CotA = p × b = 1 CotA = CosA
b p SinA

\ TanA.CotA = 1

Tan A = 1
CotA
1
Cot A = TanA

iv Also, SinA
CosA
Tan A =

Cot A = CosA
SinA

2. Pythagoras Relations: p 2 b2
i. Sin2 A+ Cos2A = h h
+

= P2 + b2
h2 h2

= h2 – p2
b2

88 PRIME Opt. Maths Book - VII

= b2
b2

=1

\ Sin2A + Cos2A =1
Sin2A = 1 – Cos2A
SinA = 1– Cos2A

Cos2A = 1– Sin2A

Cos A = 1– Sin2A b 2 p j2
h b
ii. Sec2A – Tan2A = a k – `

= h2 – p2 Sin²A + Cos²A = 1
b2 b2 Sec²A – Tan²A = 1

h2 + p2 Cosec²A – Cot²A =1

= b2

= b2
b2

=1

\ Sec2A – Tan2A = 1

Sec2A = 1+ Tan2A

Sec A = 1+ Tan2A

Tan2A = Sec2A – 1

Tan A = Sec2A – 1 2 b 2
p
iii. Cosec2A – Cot2A = a h k – a k
p

= h2 – p2
b2 b2

= h2 – b2
p2

= b2
b2

=1

\ Cosec2A – Cot2A = 1

Cosec2A = 1 + Cot2A

CosecA = 1 + Cot2A

Cot2A = Cosec²A - 1

CotA = Cosec²A – 1

PRIME Opt. Maths Book - VII 89

Worked out Examples

1. Prove that : TanA.SinA . CotA. CosecA =1
Solution :
L.H.S = Tan A. Sin A. Cot A. Cosec A

= SinA . SinA . CosA . 1
CosA SinA SinA

=1

= R.H.S proved.

2. Prove that : Sec2 A – 1 .CosA = Sin A

Solution :

L.H.S = Sec2A – 1 .CosA

= TanA .CosA

= SinA .CosA
CosA
= SinA

= R.H.S Proved.

3. Prove that : 1 = Cosec A- Cot A
Solution: Co sec A + CotA

L.H.S = 1
Co sec A + CotA

= Co sec2 A – Cot2 A
Co sec A + CotA

= ^Co sec A + CotAh^Co sec A - CotAh
^Co sec A + CotAh

= CosecA - Cot A

= R.H.S Proved

5. Prove that : 1 + CosA =( Cosec A + Cot A)2
1 – CosA
Solution:
1 + CosA
L.H.S = 1 – CosA

= 1 + CosA × 1 + CosA [ consugate of 1 – Cos A is 1 + CosA]
1 – CosA 1 + CosA

90 PRIME Opt. Maths Book - VII

= ^1 + CosAh2
12 –Cos2 A
^1 + CosAh2
= Sin2 A

= ( 1 + CosA )²
SinA SinA

= (CosecA + Cot A)²

= R.H.S Proved

6. Prove that : 1 – Cos4 A = 1 + 2 Cot²A
Solution : Sin4 A

L.H.S = 1 – Cos4 A
Sin4 A
Cos4 A
= 1 – Sin4 A
Sin4 A

= Cosec4 A – Cot4 A

= ( Cosec²A + Cot² A) ( Cosec² A – Cot² A)

= ( 1+ Cot² A + Cot² A) × 1

= 1 + 2Cot²A

= R.H.S Proved.

Exercise : 4.4

1. Prove that the followings:
i. CotA. CosecA. Sin²A = Cos A
ii. CosA. TanA. Cosec A=1
iii. Co sec2 A – 1 . Sin A = Cos A
iv. CosA. Sec2 A – 1 . Cosec A= 1
v. 1 + Tan2 A . 1 – Sin2 A . 1 + Cot2 A = Cosec A

2. Prove that the followings:

i. Tan A+ Cot A = Sec A. Cosec A
1 + TanA CosA + SinA
ii. 1 – TanA = CosA – SinA

iii. SecA – 1 = 1 – CosA
SecA + 1 1 + CosA

PRIME Opt. Maths Book - VII 91

iv. Tan2A – Sin2A = Tan2A.Sin2A
v. Cot2A – Cos2A = Cot2A.Cos2A

3. Prove that the followings:
1
i. Co sec A + CotA = CosecA – CotA

ii. SecA 1 = SecA + TanA
– TanA

iii. 1 + 2SinACosA = (SinA + Cos A)2
1 – CosA
iv. 1 + CosA = (CosecA – CotA)2

v. 1 + SinA = (SecA + TanA)2
1 – SinA

4. Prove that the followings:

i. Tan2A.Sec2B – Tan2B.Sec2A = Tan2A – Tan2B

ii. Sin2A.Cos2B – Cos2A.Sin2B = Sin2A – Sin2B
1 – Sin4 A
iii. Cos4 A = 1+ 2Tan²A

iv. 1 – Cos4 A = 2Cosec2A – 1
Sin4 A

v. Cosec4 A – Cot4 A = 1 + 2Cot2A

5. Prime more creative questions.
a. Prove that the followings.
i. Sin2A.Cot2A + Cos2A.Tan2A = 1.
ii. (SinA + CosA)2 = 1 + 2SinA.CosA.
iii. Sin4A + 2Sin2A.Cos2A + Cos4A = 1
iv. Cosec2A + Sec2A = (TanA + CotA)2
v. Sin4A + Cos4A = 1 – 2Sin2A.Cos2A

b. Prove that the followings.

i. (1 + SinA)² – (1 – SinA)² = 4SinA

ii. (CosA + SinA)² + (CosA – SinA)² = 2

iii. (xCosθ + ySinθ)² + (xSinθ – yCosθ)² = x² + y²

iv. (1 + Sinθ + Cosθ) = 2(1 + Sinθ)(1 + Cosθ)

v. Tanα – Sinα = 2Cotα
Secα – 1 1 + Cosα

92 PRIME Opt. Maths Book - VII

4.5 Conversion of Trigonometric ratios

• Let us consider a trigonometric ratio Sinq= k in Right angled DABC

A

K1

B θ C

Here,
Sin θ = K
p K
or, h = 1

\ p=K

h=1

Then,

b = h2 – p2

= 12 – K2

= 1 – K2

Now,

Cosθ = b = 1 – K2 = 1 – Sin2 A
h 1
Sinq
Tanθ = p = K = 1 – Sin2 q
b 1 – K2

Secθ = h = 1 = 1
b 1 – K2 1 – Sin2 T

Cotθ = b = 1 – K2 = 1 – Sin2 q
p K Sinq

Cosecθ = h = 1 = 1
p K SinT

In Such a way all the trigonometric ratios can be expressed in terms of a
specific ratio.

• Let s consider a trigonometric ratio nTanq = m
m
i.e. Tanq = n
p m
or, b = n

PRIME Opt. Maths Book - VII 93

\ p=m

b=n

And, h = p2 + b2

= m2 + n2

Then, = m
p m2 + n2

Sinq = h

Cosq = b = n
h m2 + n2

Cotq = b = n
p m

Cosecq = h m2 + n2
p =m

Secq = h m2 + n2
b =n

Worked out Examples

1. If Sin A = 3 , Find the value of Cos A & Tan A
Solution: 5

SinA = 3
5

p = 3
h 5

Here,

p =3

h =5

\ b = h2 – p2

= 52 – 32

= 25 – 9

= 16
=4

Then, p
=b
Cos A = b = 4 Tan A = 3
h 5 4

94 PRIME Opt. Maths Book - VII

2. If n Cos A = m, Find Cosec2A – Cot2A

Solution:

Here, m
n
or, CosA =

or, b = m
h n

Here,

b=m

h=n

p = h2 + b2 = a h 2 – a b 2
= n2 – m2 p p
k k
Then,

Cosec2A – Cot2A

=a n m2 2 – a m m2 2
n2 – n2 –
k k

= n2 – m2
n2 – m2 n2 – m2

n2 – m2 1
= n2 – m2

=1

3. Express all the trigonemetrical ratios in terms of TanA.

Solution

Let, Tan A = K
p
or, b = K
1

p=K

b=1

h = p2 + b2 = 1 + K2

Then, = K = TanA
p 1 + K2 1 + Tan2 A

SinA = h

CosA = b = 1 = 1
h 1 + K2 1 + Tan2 A

CosecA = h 1 + K2 1 + Tan2 A
p =K = TanA

PRIME Opt. Maths Book - VII 95