Vedanta Opt. Excel in Mathematics Book 8 Final (2078)

Measurement of Angles

used from the time of Babylonians. In this system, 1 right angle is divided into
90 equal parts and each 1 part is called 1 degree. It is denoted by 1° and read as
one degree.

we write, In Degree system
1 right angle = 90° 1 right angle = 90°

Similarly, 1° = 60 smaller equal part of a degree 1° = 60'
1 part = 1' (One minute) 1' = 60"

? 1° = 60'

Again, 1' = 60 smaller equal parts of a minute

1 part = 1" (One second)

? 1' = 60" A
We can write 1° = 60 × 60" = 3600"

In this system of measurement of angle, the sum of angles of

triangle is 180°

In 'ABC, ‘A + ‘B + ‘C = 180° B C

(b) Centesimal System

This system centesimal also refers to the word cent which means 100. So, right

angle here is divided into 100 equal parts, each part is called 1 grade. It is

denoted by 1g and read as one grade. In Grade system

Hence, 1 right angle = 100g 1 right angle = 100g

Similarly, 1g = 100 equal parts of angle 1g = 100'

? 1g = 100' 1' = 100"

Again, 1' = 100 equal parts of a minute A
? 1' = 100"

In this system of measurement of angle, the sum of angle of a

triangle is 200g. B

In 'ABC, ‘A + ‘B + ‘C = 200g C

Relation between Degree and Grade System

We know,

1 right angle = 1 right angle

or, 90° = 100g

1° = 100 g= 10 g
90 9

vedanta Excel in Additional Mathematics - Book 8 51

Measurement of Angles

Similarly, 100g = 90°

or, 1g = 90 °= 9°
100 10

Using these relations, we can convert degree into grades and vice versa.

Worked out Examples

Example 1. Convert into sexagesimal seconds.
Solution: 40°40'20"
Here, 40°40'20"
We know, 1° = 60'
So, 40° = (40 × 60)'
Also, 1' = 60"
? 40' = (40 × 60)"
and (40 × 60)' = (40 × 60 × 60)"
? 40° 40' 20" = (40 × 60 × 60)" + (40 × 60)" + 20"

= 144000" + 2400" + 20" = 146420"

Example 2. Convert into centesimal seconds.
Solution: 10g25'15"
Here, 10g25'15"
= (10 × 100 × 100)" + (25 × 100)" + (15)" 1g = 100'
= (100000 + 2500 + 15)" 1' = 100"
= 102515"

Example 3. Convert into grades 27°.
Solution:
We know,
Example 4.
Solution: 1° = 10 g
9

or, 27° = 10 × 27 g = 30g
9

Convert into degrees: 100g

We know,

1g = 9°
10

52 vedanta Excel in Additional Mathematics - Book 8

Measurement of Angles

or, 100g = 9 × 100 °= 90°
10

Example 5. Convert into degree : 45°30'45"

Solution: Here, 45° + 30' + 45"
Example 6.
= 45° + 30 + 60 45 60
60 ×
1° 1°
= 45° + 2 + 80

= 3600° + 40° + 1°
80
3641
= 80 = 45.5125°

Convert into grades : 25g25'25"

Solution: Here, 25g25'25"

= 25g + 25g + 25g
100 100 × 100
1g 1g
= 25g + 4 + 4 × 100

= 10000g + 100g + 1g
400
10101g
= 400 = 25.2525g

(c) Radian system/Circular Measure B

The angle made at the center of a circle by an arc whose

length is equal to the radius of the same circle is known as 1c
0
1 radian. It is written as 1c and read as 1 radian. C A

In figure, OA = AB = radius of circle

then ‘AOB = 1c

Note : The angle ‘AOC = Sc

In this system of measurement of angle, the sum of angle of a triangle is Sc.

Relation between degree and radian

Here, we have,

Sc = 2 right angles.

or, Sc = 180°

Similarly, 1° = Sc
180

? 1c = 180 °
S

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Measurement of Angles

Example 7. Convert 81° into radians.

Solution: We know,

1° = Sc
180
9S c
? 81° = S × 81 c = 20
180
Example 8.
Solution: Convert S c into degrees.
4

We know,

1c = 180°
S
180 ° = 45°
or, S c= S × S
4 4

Relation between grades and radian

We know,

Sc = 2 right angles.

or, Sc = 200g (? 1 right angle = 100g)

1c = 200 g
S

Similarly, 1g = Sc
200

Example 9. Convert into radians : 150g

Solution: We have,

1g = Sc
200
3S c
? 150g = S × 150 c= 4
200

Example 10. Convert into grades : Sc
3

Solution: We know,

1c = 200 g
S

or, S c= 200 × S g= 66 2 g
3 S 3 3

Example 11. One angle of a right-angled triangle is 709 . Find the remaining angles

in degrees. C

Solution: Here, One angle, ‘A = 90°

Second angle, ‘B = 70g

AB

54 vedanta Excel in Additional Mathematics - Book 8

Measurement of Angles

Third angle, ‘C = ?

We know,

1g = 9°
10
9
or, 70g = 10 × 70 = 63°

We know,

‘A + ‘B + ‘C = 180°

or, 90° + 63° + ‘C = 180°

or, ‘C = 180° – 153°

? ‘C = 27°

Hence, the third angle is 27°.

Example 12. The angles of a triangle are in the ratio 2:3:5. Find all the angles in
grades.

Solution: Let the angles be 2k, 3k and 5k

Now, 2k + 3k + 5k = 200g (Sum angles of a triangle = 200g)

or, 10k = 200g

k = 20g

Hence, 2k = 2 × 20g = 40g

? 3k = 3 × 20g = 60g

and 5k = 5 × 20g = 100g

Example 13. If two angles of a triangle are 72° and 60g, then find the remaining
angle in radian measure.

Solution: We have,

1st angle = 72°

2nd angle = 60g

We know,

1g = 9°
10
9
or, 60g = 10 × 60° = 54°

Let the third angle be x,

Then, sum of angle of a triangle is 180°

72° + 54° + x = 180°

vedanta Excel in Additional Mathematics - Book 8 55

Measurement of Angles

or, x = 180° – 126°

? x = 54°

Now, we know,

1° = Sc
180

? 54° = S × 54 c= 3S c
180 10

Exercise 4.1

Short Questions :

1. Convert into sexagesimal seconds.

(a) 20° (b) 60° (c) 25° 15'

(d) 40°40' 20" (e) 120° 20' 20"

2. Convert into centesimal seconds.

(a) 15g (b) 20g (c) 25g 25'

(d) 10g 20' 10" (e) 20g 20' 20"

3. Convert into grades.

(a) 63° (b) 27° (c) 54°

(d) 144° (e) 126°

4. Convert into degrees.

(a) 40g (b) 63g (c) 120g (d) 117g

5. Convert into degrees.

(a) 45° 30' (b) 60° 45' (c) 20° 30' 25" (d) 54° 12' 15"

6. Convert into grades.

(a) 40g 45' (b) 50g 15' (c) 70g 50' 25" (d) 80g 75' 75"

7. Convert into radians.

(a) 40° (b) 90° (c) 75°

(d) 120° (e) 54°

8. Convert into degrees.

(a) Sc (b) Sc (c) 3S c
4 9 4

(d) 4S c (e) 7S c
5 4

56 vedanta Excel in Additional Mathematics - Book 8

Measurement of Angles

9. Convert into radians.

(a) 50g (b) 80g (c) 75g

(d) 120g (e) 150g

10. Convert into grades.

(a) 3S c (b) Sc (c) 3S c
5 20 10

(d) 2S c (e) 7S c
5 10

11. Find the remaining in grades.

(a) One angle of a right-angled triangle is 70°. Find the remaining angle in
grades.

(b) One angle of a right-angled triangle is 72g. Find the remaining angle in
radian measure.

12. (a) The angles of a triangle are in the ratio 3 : 4 : 3. Find all the angles in

(i) degrees (ii) grades (iii) radians.

(b) The angles of a quadrilateral are in the ratio 2 : 3 : 4 : 1. Find all the angles in

(i) degrees (ii) grades (iii) radians

(c) The two angles of a triangle are in the ratio 4 : 5 and the third angle is 90°.
Find all the angles in grades.

13. (a) If two angles of a triangle are 45° and S c, find the remaining angle in
degree. 6

(b) If one angle of an right-angled triangle is 25°, find the remaining angle in
radian measure.

(c) Two acute angles of a right angled triangle are 63° and 30g. Express all
angles in radian.

14. (a) If D and G are the number of degrees and grades of the same angle, prove
G D
that 10 = 9

(b) If M and m represents the number of sexagesimal and centesimal minute
M m
of any angle respectively, prove that 27 = 50 .

vedanta Excel in Additional Mathematics - Book 8 57

Measurement of Angles

1. (a) 72000" (b) 216000" (c) 90900" (d) 146410"

2. (a) 150000" (b) 200000" (c) 252500" (d) 102010" (e) 202020"

3. (a) 70g (b) 30g (c) 60g (d) 160g (e) 140g

4. (a) 36° (b) 56.7° (c) 108° (d) 105.3°

5. (a) 45.5° (b) 60.75° (c) 20.5069° (d) 54.2042°

6. (a) 40.45g (b) 50.15g (c) 70.5025g (d) 80.7575g

7. (a) 2Sc (b) Sc (c) 5Sc (d) 2Sc (e) 3Sc
9 2 12 3 10
8. (a) 45° (b) 20° (c) 135° (d) 144° (e) 315°

9. (a) Sc (b) 2Sc (c) 3S c (d) 3S c (e) 3S c
4 5 8 5 4

10. (a) 120g (b) 10g (c) 60g (d) 80g (e) 140g

11. (a) 200 g (b) 7S c
9 50

12.(a) (i) 54°, 72°, 54° (ii) 60g, 80g, 60g (iii) 3S c, 2S c, 3S c
10 5 10

(b) (i) 72°, 108°, 144°, 36° (ii) 80g, 120g, 160g, 40g (iii) S c, 3S c, 4S c, Sc
5 5 5 5

(c) 400 g, 500 g, 100g
9 9

13. (a) 105° (b) 13S c (c) 7S c, 3S c, Sc
36 20 20 2

4.3 Measurement of Angles of Polygons

A polygon is a closed plane figure bounded by three or more than three line segments.
Triangle, quadrilateral, pentagon, etc. are some examples of polygons that may be
regular or irregular.

A polygon having all sides and angles equal is called regular polygon and otherwise
irregular.

AD
DC

EC

BC A B AB
Equilateral triangle Square Regular Pentagon

58 vedanta Excel in Additional Mathematics - Book 8

Measurement of Angles

Interior and Exterior angles of a polygon

ED

FC

AB G

Let us draw a regular hexagon ABCDEF. Each of interior angle is 120°.

Extend AB to G. Then, we get ‘CBG = 60°.

Here, ‘ABC = ‘BCD = ........... = ‘FAB = 120° are interior angles

and ‘CBG = 60° is called an enterior angle of the regular polygon.

Each interior angle of a regular polygon (x) = n – 2 × 180°
n

Sum of Interior Angles of a Polygon

In above regular hexagon, the sum of interior angles is 120° × 6 = 720°

For any polygon sum of interior angle = (n – 2) × 180°

For hexagon n = 6

Sum of interior angle = (6 – 2) × 180° = 4 × 180° = 720°

For octagon, n = 8

Sum of interior angles = (8 – 2) × 180° = 1080°

Sum of Exterior Angle of a Polygon

In case of a regular polygon, if n in the number of sides, each exterior angle is equal.

Sum of exterior angle of a regular polygon = 360°

Also sum of exterior angle of irregular polygon = 360°

Size of each exterior angle = 360°
n

Let us tabulate the number of sides and sum of interior angles

Polygon Figure No. of No. of triangles Sum of the interior
Triangle
sides (n) formed angles of polygon

31 (3 – 2) × 180° = 180°

vedanta Excel in Additional Mathematics - Book 8 59

Measurement of Angles 42 (4 – 2) × 180° = 360°
53 (5 – 2) × 180° = 540°
Quadrilateral

Pentagon

Hexagon 64 (6 – 2) × 180° = 720°
Heptagon 75 (7 – 2) × 180° = 900°

.......... ............... ........... ..................... ................................
N-gon n n–2 (n – 2) × 180°

Here, we study to find angles of different polygons in degrees and change them into
grade or radian measure.

we have

180° = 200g = Sc ? 1° = 10g
Then, 180° = 200g 9
Sc
180° = Sc ? 1° = 180

Worked out Examples

Example 1. Find the interior angle of a regular pentagon in grade and radian.
Solution:
In a regular pentagon, number of sides (n) = 5
60
Interior angle (x) = n – 2 × 180° = 5 – 2 × 180° = 108°
we have, n 5

1° = 10g
9
10g
or, 108° = 9 × 108 = 120g

again, 1° = Sc
180
Sc Sc
or, 108° = 180 × 108° = 5

? Each angle of a regular pentagon is 120° and 5Sc.

vedanta Excel in Additional Mathematics - Book 8

Measurement of Angles

Example 2. Find the sum of interior angles of a hexagon in grade and radian.

Solution: In a hexagon, number of sides (n) = 6

Sum of angles of a hexagon = (n – 2) × 180° = (6 – 2) × 180°

= 4 × 180° = 720°

we have,

1° = 10g
9
10g
? 720° = 9 × 720 = 800g

Also, 1° = Sc
180
Sc
? 720° = 180 × 720° = 4Sc.

Example 3. The ratio of angles of a quadrilateral is 1 : 2 : 3 : 4. Find the angles of
the quadrilateral is degrees and grades.

Solution: Let the angles of the quadrilateral be k, 2k, 3k and 4k. Then sum of
angles of a quadrilateral = 360°

k + 2k + 3k + 4k = 360°

or, 10k = 360°

? k = 36°

Now, the angles of quadrilateral The angles in grades are
in degrees are:
36° = 36 × 10g = 40g
k = 36° 9
10g
2k = 2 × 36° = 72° 72° = 72 × 9 = 80g

3k = 3 × 36° = 108° 108° = 108 × 10g = 120g
9
4k = 4 × 36° = 144° 10g
9
144° = 144 × = 160g

Example 4. The ratio of an interior and exterior angle of a regular polygon is
3:1. Find the number of sides and express each interior and exterior
angle of the polygon in grade and radian.

Solution: Let 3k and k be the interior and exterior angles of given regular polygon

Then, interior angle + exterior angle = 180°

or, 3k + k = 180°

or, 4k = 180°

? k = 45°

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Measurement of Angles

? Interior angle = 3k = 3 × 45° = 135°

Exterior angle = k = 45°

Let n be the number of sides.

Exterior angle of regular polygon = 360°
n
360°
or, 45° = n

? n=6

we have:

1 ° = 10g
9
10g
? 135° = 9 × 135° = 150g

Also, 1° = Sc
180
Sc 3Sc
? 135° = 180 × 135° = 4

? Each interior angle is 135° or 150g or 4Sc.

Again, expressing external angle in grade and radian, we have

45° = 45 × 10g = 50g
9
Sc Sc
45° = 45 × 180 = 4

Exercise 4.2

Short Questions :

1. Find the size of interior angle of each angle of the following regular polygons in
degrees and grades:

(a) Pentagon (b) Hexagon (c) Heptagon (d) Octagon

(e) Nonagon (f) Decagon (g) Dodecagon

2. Find the size of exterior angles of the following regular polygons in degrees and
grades:

(a) Pentagon (b) Hexagon (c) Heptagon (d) Octagon

3. Find the sum of interior angles of the following polygons in degrees:

(a) Quadrilateral (b) Pentagon

(c) Hexagon (d) Octagon

62 vedanta Excel in Additional Mathematics - Book 8

Measurement of Angles

4. From the given exterior angles of regular polygons find the number of sides:

(a) 60° (b) 120° (c) 72° (d) 36°

5. Find the (i) interior and (ii) exterior angle of the given regular polygons in
degrees and radian:

(a) Octagon (b) Hextagon (c) Pentagon

Long Questions :

6. (a) The angles of a pentagon are in the ratio 2 : 3 : 4 : 5 : 6. Find the angles in
(i) degrees and (ii) grades.

(b) The angles of a hexagon are in the ratio of 1 : 2 : 3 : 4 : 5 : 5. Find the angles
in

(i) degrees (ii) grade

(c) The angles of a octagon are in the ratio of 1 : 2 : 3 : 2 : 2 : 5 : 6 : 6. Find the
angles in (i) degrees and (ii) radian.

7. (a) The ratio of interior and exterior angles of a regular polygon is 2 : 1. Find
the number sides of the polygon. Also express the angles in degrees and
radian.

(b) The ratio of interior and exterior angles of a regular polygon is 7 : 2. Find
the number of sides. Also express the angles in degrees and radian.

8. Draw a regular octagon of side 3 cm. Find its interior and exterior angles in
degrees, grades and radian.

1. (a) 108°, 120g (b) 120°, 133.33g (c) 128.57°, 148.86g (d) 135°, 150g

(e) 140°, 155.56g (f) 144°, 160g (g) 150°, 166.67g

2. (a) 72°, 80g (b) 60°, 66.67g (c) 51.43°, 57.14 (d) 45°, 50g

3. (a) 360° (b) 540° (c) 720° (d) 1080°

4. (a) 6 (b) 3 (c) 5 (d) 10

5. (a) (i) 135°, 3Sc (ii) 45°, Sc (b) (i) 120°, 2Sc (ii) 60°, Sc
4 4 3 3
6. (a) (i) 54°, 81°, 108°, 135°, 162° (ii) 60g, 90g, 120g, 150g, 180g

(b) (i) 36°, 72°, 108°, 144°, 180°, 180° (ii) 40g, 80g, 120g, 160g, 200g, 200g

(c) (i) 40°, 80°, 120°, 80°, 80°, 200°, 240°, 240°

(ii) 29Sc, 49Sc, 23Sc, 29Sc, 29Sc, 109Sc, 43Sc, 4Sc
3
2Sc Sc 7Sc 2Sc
7. (a) n = 6, 120°, 60°, 3 , 3 (b) n = 9, 140°, 9 , 40°, 9

vedanta Excel in Additional Mathematics - Book 8 63

Trigonometry

5Trigonometry

5.1 Trigonometry

Trigonometry is the branch of mathematics which literally consists hypotenuse A
of three words. Tri-refering to three, 'gono' meaning angle, and
'metry' which means measurement. Hence, trigonometry deals
with the measurement of triangles. We only deal with right-
angled triangles.

CB

Right-angled Triangle

A triangle having one angle 90° is known as right-angled triangle. The side
opposite to the right-angle is called hypotenuse. It is denoted by 'h'. In figure,

AC is the hypotenuse.

Note :

Hypotenuse is the longest side in a right angled triangle.

5.2 Reference Angle

An angle under consideration before finding out perpendicular and base in a right-
angled triangle is known as reference angle. It is usually denoted by Greek Alphabets
like D T E, etc.

The side in front of the reference angle is called perpendicular and the remaining
side is called base. Perpendicular and base are denoted by p and b respectively. So,
in short, p, b, and h are called the sides of right-angled triangle.

In figure, AB = Perpendicular (p) A
BC = Base (b)

AC = Hypotenuse (h)

B TC

64 vedanta Excel in Additional Mathematics - Book 8

Trigonometry

Worked out Examples

Example 1. Find out p, b, and h from the following right-angled triangles with

the given angle of reference A
(i) C = T be the reference angle.

Solution: (ii) A = D be the reference angle. D
Here, In right-angled 'ABC, ‘ABC = 90°

(i) ‘C = T is the angle of reference

Then, AC = hypotenuse (h)

AB = perpendicular (p) T

CB = base (b). CB
Again,

(ii) ‘A = D be another reference angle

Then, AC = hypotenuse (h)

CB = perpendicular (p)

AB = base (b)

5.3 Pythagoras Theorem

In any right-angled triangle, the squares made on the hypotenuse is always equal to
the sum of the squares made on the remaining two sides.

Mathematically, h2 = p2 + b2

Example 2. Find the missing sides from the right-angled triangles given below:

(a) P (b) A

6 cm 10 cm 7 cm 6 cm

Q R B D 8 cm C

(c) 12 cm S
P

24 cm

Solution: Q 7 cm R To find PR and SR
(a) Here, in right-angled 'PQR

vedanta Excel in Additional Mathematics - Book 8 65

Trigonometry ‘Q = 90°, PQ = 6 cm, PR = 10 cm, QR = ?
We know,
By using pythagoras theorem,

(b) h2 = p2 + b2
or, PR2 = PQ2 = QR2
or, 102 = 62 + QR2
(c) or, 100 = 36 + QR2
or, 100 – 36 = QR2
66 or, 64 = QR2
or, QR = 64 = 8
? QR = 8 cm
Hence, QR is 8 cm.
Here, two right-angled 'ADC and 'ABD are given.
So, the sides AC and BD are to be calculated
In ' ABD, using pythegoras theorem,

AB2 = AD2 + BD2
or, 72 = 62 + BD2
or, 49 – 36 = BD2
or, BD2 = 13
or, BD = 13
Similarly, in 'ADC,

AC2 = AD2 + DC2
or, AC2 = 62 + 82
or, AC2 = 36 + 64
or, AC2 = 100
? AC = 10 cm
Hence, AC = 10 and BD = 13
So, In right-angled ' PQR, using pythegoras theorem

PR2 = PQ2 + QR2
or, PR2 = 242 + 72
or, PR2 = 576 + 49

vedanta Excel in Additional Mathematics - Book 8

Trigonometry

5.4 or, PR2 = 625
or, PR2 = 252
? PR = 25 cm
Now, again
In right-angled 'PSR,

SR2 = PS2 + PR2
or, SR2 = 122 + 252
or, SR2 = 144 + 625
or, SR2 = 769
or, SR = 769 = 27.73 cm
Hence, PR = 25 cm and SC = 27.73 cm.

Converse of Pythagoras Theorem

Converse of Pythagoras theorem states that if sum of squares two sides of a triangle
is equal to the square of third side then the triangle must be a right-angled triangle.

Example 3. Examine whether the following triangles are right-angled?

(a) D (b) In 'PQR, PQ = 3 cm, QR = 3 cm
and PR = 5 cm, check whether
'PQR is a right-angled .

13 cm 12 cm

F 5 cm E

Solution: (a) Here, in 'DEF, P

DF = 13 cm, DE = 12 cm and FE = 5 cm. 5 cm 3 cm
Q
We have to see whether the triangle is right-
angled.

So, for ' DEF to be a right-angled triangle, R 3 cm
Using pythagoras theorem, we get

DF2 = DE2 + EF2

or, 132 = 122 + 52

or, 169 = 144 + 25

vedanta Excel in Additional Mathematics - Book 8 67

Trigonometry

or, 169 = 169 (True)
Since, Pythagoras theorem holds true in 'DEF it is a
right-angled triangle.
(b) Here, in 'PQR

PQ = 3 cm, QR = 3 cm, and PR = 5 cm
As PR is the longest which is 5 cm, it is considered as the hypotenuse
So, for 'POT to be a right-angled triangle, h2 = p2 + b2 must
hold true.
or, PR2 = PQ2 + QR2
or, 52 = 32 + 32
or, 25 = 9 + 9
or, 25 = 18 (False)
25 can never be equal to 13.
So, the statement is false.
Since, Pythagoras theorem does not hold true in 'PQR it is not a
right-angled triangle.

Exercise 5.1

Short Questions :

1. Fill in the blanks from the following right-angled triangles considering the
reference angle as black marked once.

(a) P (b) P R

perpendicular = ........

base = ........ perpendicular = .....

hypotenuse = ........ Q base = .....

R Q hypotenuse = ......
(c) B C
(d) P PQ = .........
A PR = .........
68 AB = ............. QR = .........
BC = .............
AC = ............. Q R

vedanta Excel in Additional Mathematics - Book 8

Trigonometry

2. In the adjoining figure find out perpendicular, base, and P
hypotenuse from the given right-angled triangle with D D
and T as the reference angles. Also give your reason.
Q
3. Find the length of missing side in each of the following X T
S
right-angled triangles. R
(a) A (b)

3 13

B 4 C Z 12 Y
(c) D F
(d) 41 O
2 M

40

E 4 N
(e) P (f) X

? 10 cm 12

Q 6 cm R 6 cm S U
3
(g) A V 4W
(h) A
12 15
D
13 5
C
B D 13
3 C B

vedanta Excel in Additional Mathematics - Book 8 69

Trigonometry

(i) Q (j) Y

8 6 12
P R 25 O

8 S

X 20 Z

4. (a) In ABC, AC = 12 cm, AB = 13 cm and BC = 5 cm. Is ∆ABC a right-angled
triangle? Give reasons for your answers.

(b) PQR is a triangle with sides PQ = 7 cm, PR = 5 cm and QR = 6 cm. Check
whether the triangle is right-angled. Give reasons for your answers.

1. and 2. Show to your teacher.

3. (a) 5 units (b) 5 units (c) 2 3 unit (d) 9 units

(e) PR = 8 cm, PQ = 10 cm (f) UW = 5 cm, WX = 13 cm

(g) BD = 5 cm, CD = 4 cm (h) BC = 12 cm, AB = 9 cm

(i) PR = 10 units, RS = 6 units (j) YZ = 15 units, OZ = 9 units

4. (a) Right angled triangle (b) Not a right angled triangle

5.5 Trigonometric Ratios

A right-angled triangle consists of three sides: the perpendicular, the base, and the
hypotenuse. As we talk about ratios, 6 ratios can be obtained from these three sides
which are as follows:

(i) p (ii) b (iii) p (iv) b
h h b p
h h
(v) b (vi) p

These six ratios are called trigonometric ratios and they are given certain names.

A

Let 'ABC be a right-angled triangle where

‘ABC = 90°,

‘ACB = T, be the reference angle.

Then, T B
Side AB = perpendicular (p) C

70 vedanta Excel in Additional Mathematics - Book 8

Trigonometry

Side AC = hypotenuse (h)

Side BC = base (b)

Now, we can get six ratios here as well.

(i) p AB (ii) b = BC
h= AC h AC

(iii) p = AB (iv) b = BC
b BC p AB

(v) h = AC (vi) h = AC
b BC p AB

Now, let's introduce the names for these ratios with reference angle T.

S.No. Ratio Nomenclature Abbreviation

1. p sine T sin T
h

2. b cosine T cos T
h

3. p tangent T tan T
b

4. b cotangent T cot T
p secant T sec T
cosec T/csc T
5. h cosecant T
b

6h
p

Worked Out Examples

Example 1. In 'PQR given alongside, find all the trigonometric ratios for the
Solution:
reference angle T. P

Here in, 'PQR, ‘Q = 90°

and ‘PRQ = T is the reference angle.

So, PQ = perpendicular (p), QR = base (b),

and PR = hypotenuse (h) T

sinT = p = PQ cosT = b = QR R Q
h PR h PR
?

tanT = p = PQ
b QR

Similarly, cosecT = h = PR secT = h = PR
p PQ b QR
b QR
cotT = p = PQ

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Example 2. Find six trigonometric ratios for the adjoining figure with respect to
Solution: D and E.

Example 3. Here, ' PQR is a right-angled triangle with ‘PQR = 90°
Solution: P
Taking ‘PRQ = E as the reference angle, we get,

Perpendicular (p) = PQ D

Base (b) = QR

Hypotenuse (h) = PR E
R
Required trigonometric ratios are given below: Q

(i) sinE = p = PQ (ii) cosE = b = QR
h PR h PR
p b QR
(iii) tanE = b = PQ (iv) cotE = p = PQ
QR
h PR h PR
(v) secE = b = QR (vi) cosecE = p = PQ

Now,

Taking ‘QPR =D as the reference angle.

Perpendicular (p) = QR

Base (b) = PQ

Hypotenuse (h) = PR

Required trigonometric ratios are given below:

(i) sinD = p = QR (ii) cosD = b = PQ
h PR h PR
p b PQ
(iii) tanD = b = QR (iv) cotD = p = QR
PQ
h PR h PR
(v) secD = b = PQ (vi) cosecD = p = QR

In the right-angled 'ABC,

(i) Find the length of AB

(ii) Find all trigonometric ratios for reference angle T. C

Here, in right-angled ABC,

‘ABC = 90° 25 cm

‘CAB = T, the reference angle 7 cm

So, Perpendicular (p) = BC = 7 cm T B
Hypotenuse (h) = AC = 25 cm A
Base (b) = AB = ?

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We know, in a right-angled triangle, using pythagoras theorem,
AC2 = AB2 + BC2

or, AB2 = AC2 – BC2
or, AB = 255 – 72

= 625 – 49
= 576
? AB = 24 cm
Now, p = BC = 7 cm, b = AB = 24 cm and h = AC = 25 cm

sinT = p = 7 cosT = b = 24
h 25 h 25

tanT = p = 7 cotT = b = 24
b 24 p 7

secT = h = 25 cosecT = h = 25
b 24 p 7

Example 4. In the figure 'PQR, ‘PRQ is a right-angled triangle. Find sinT and
Solution: cos D.

Here, In right-angled 'PQR, 12 cm DQ
‘PRQ = 90° R

So, h2 = p2 + b2 9 cm

or, PQ2 = PR2 + QR2 T
or, PQ2 = 92 + 122

or, PQ2 = 81 + 144 P

or, PQ2 = 225

or, PQ = 225

? PQ = 15 cm

Now, taking 'T' as the reference angle,

sinT = p = RQ
h PQ

= 12 = 3
15 5

Again, taking 'D' as the reference angle,

cosD =hb = RQ
PQ

= 12 = 3
15 5

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Example 5. In right-angled triangle PQR, if PQ = 3 cm and RP = 4 cm, find sinQ,
Solution:
cosQ, and tanR. Given that P is a right angle.

PQR is a right-angled triangle with P as the right angle. Q

So, from figure we can see QR has to be calculated first.

So, we know, In right-angled triangle PQR, 3 cm

using pythagoras theorem

QR2 = PQ2 + RP2 R 4 cm P
or, QR2 = PQ2 + RP2 = 32 + 42

= 9 + 16 = 25 = 5 cm

? QR = 5 cm

Now, sinQ = PR
QR

or, sinQ = 4 sinT = p
5 h

cosQ = PQ cosT = b
RP h

? cosQ = 3
5

Example 6. Similarly, tanR = PQ tanT = p
Solution: PR b

? tanR = 3
4

From the adjoining figure, prove tanT = 12 P
5 Q

In right-angled triangle PQS,

PS2 = QS2 + PQ2 1513

or, 152 = 92 + PQ2 T
R
or, PQ2 = 225 – 81 S 9
or, PQ2 = 144

or, PQ = 12

Again, in right-angled 'PQR,

PR2 = PQ2 + QR2

or, 132 = 122 + QR2

or, 169 = 144 + QR2

74 vedanta Excel in Additional Mathematics - Book 8

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or, 169 – 144 = QR2
or, QR2 = 25
or, QR = 5

Now, tanT = PQ
QR

? tanT = 12 Proved.
Example 7. 5

Solution: In a right-angled triangle PQR, right angle is at R. If PQ = 32 cm and
5
sinQ = 8 , find QR

Representing the information in diagram, we get,

PQ = 32 cm P
R
sinQ = 5
8
QR = ?

So, sinQ = PR 32 cm
PQ
PR
or, sinQ = 32

But, sinQ = 5 Q
8
5 PR
So, 8 = 32

or, PR = 5 × 32 = 20 cm
8

Now, in right-angled 'PQR, using Pythagoras theorem

PQ2 = PR2 + QR2

or, QR2 = PQ2 – PR2

or, QR = PQ2 – PR2

= 322 – 202

= 1024 – 400

= 624

= 4 39

= 24.98

? QR = 24.98 cm

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Exercise 5.2

Short Questoins : A

1. From the given right angled triangle ABC, write down the
name of six basic trigonometric ratios.

2. From the figure given below, calculate the mentioned T

trigonometric ratios. C B

sinT = sinE = P
tanT = tanE =

secT = secE = E

cosecT = cosecE =

cot T = cotE = RT Q
cos T = cosE =

3. Find all six trigonometric ratios with respect to the reference angles given in
the triangles.

(a) A (b) M N

B TC T
P

4. Find all six trigonometric ratios with respect to the reference angles given in
the Greek alphabet.

(a) A DB (b) P

T

E

C RJ Q

5. Find all the trigonometric ratios with reference angles D and E.

(a) P (b) P
D E

Q ER S
S
D
R Q

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6. Calculate all trigonometric ratios from the right-angled triangles given below by
first calculating the unknown sides.

(a) P (b) A 41 cm C

T

13 cm 5 cm 40 cm

QT R B

7. From the adjoining figure, calculate:

(a) sinD (b) tanD (c) cosecD

(d) cosT (e) cotT (f) secT

8. In a right-angled PQR, right-angled at P

(a) If PQ = 8 cm, PR = 6 cm find sin R, cos R and tan Q.

(b) If QR = 5 2 cm, PR = 7 cm find sec Q, cosec Q, tan Q and cot R.

(c) If QR = 41 cm, RP = 40 cm find sin Q, cos Q and tan R.

9. (a) From the figure given below, prove that sin T = 3 .
5

(i) P (ii) D

12 cm 13 cm A 13 cm 15 cm

QT R 5 cm
3 cm
S BT C
A
(b) From the figures given below, prove that: cos D = 5 .
13

(i) S (ii)

12 cm cm

P 15
D cm

3 cm 13

C D9 D B

Q 4 cm R

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(c) From the figures given below, prove that tan T = 3 .
4

(i) D (ii) P
T
8 cm 6 cm 25 cm12 cm
C Q
AT
8 cm S

R 20 cm

B

10. (a) In a right-angled triangle PQR right-angled at Q, if sin R = 3 and PR = 28.
7

Find (i) PQ, (ii) QR and (iii) tan R

(b) In a right-angled triangle ABC, right-angled at C if tan B = 4 and BC = 15 cm,
find AB and sin A. 3

1. - 5. Show to your teacher.

6. (a) QR = 12 cm, sinT = 5 , cosT = 12 , tanT = 5 ,
13 13 12
13 13 12
cosecT = 5 , secT = 12 , cotT = 5

(b) AC = 9 cm, sinT = 40 , cosT = 9 , tanT = 40
41 41 9
41 41 9
cosecT = 40 , secT = 9 , cotT = 40

7. AB = 7 cm (a) 24 (b) 24 (c) 25
25 7 24
24 24 25
(d) 25 (e) 7 (f) 24

8. (a) 4 , 3 , 3 (b) 5 2, 52 , 7, 7 (c) 40 , 9 , 9
5 5 4 7 41 41 40

10. (a) (i) 12 cm (ii) 8 10 cm (ii) 3 cm (b) AB = 20 cm, sinA = 4
2 10 5

5.6 Operations on Trigonometric Ratios

Trigonometric ratios can be operated as in algebra. Addition, subtraction, multiplication
and division are basic operations in algebra. These operations can be operated in
trigonometric ratios. Let us compare these operations in algebra and trigonometry.

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S.N. Operations In Algebra In Trigonometry
1. Addition x + 2x = 3x sin T + 2sin T = 3sin T

2. Subtraction x2 + 3x2 = 4x2 sin2 T + 3sin2 T = 4sin2 T
3. Multiplication
4. Division 4x3 + x3 = 5x3 4sin3 T + sin3 T = 5sin3 T
2x – x = x 2sin T – sin T = sin T

2x – 5x = – 3x 2sin T – 5sin T = – 3sin T
x . x2 = x3 sin T . sin2 T = sin3 T

2x . 4x2 = 8x3 2sin T . 4sin2 T = 8sin3 T

x3 = x2 sin3 T = sin2 T
x sin T
x2 – y2 (x – y) (x + y) sin2 T – sin2 D
x – y = x–y sin T – sin D

=x+y = (sin T + sin D) (sin T – sin D)
sin T – sin D

= sin T + sin D

S.No. Formula Expanded form Factorised form
1. (a + b)2 a2 + 2ab + b2 (a + b) (a + b)
2. (a – b)2 a2 – 2ab + b2 (a – b) (a – b)
3. a2 – b2 — (a + b) (a – b)
4. a2 + b2 (a +b)2 – 2ab —

5. (a + b)3 (a – b)2 + 2ab (a + b) (a + b) (a + b)
a3 + 3a2b + 3ab2 +b3
6. (a – b)3 (a – b) (a – b) (a – b)
a3 + b3 + 3ab (a + b)
7. a3 + b3 a3 – 3a2b + 3ab2 – b3 (a + b) (a2 – ab + b2)
8. a3 – b3 (a – b) (a2 + ab + b2)
a3 – b3 – 3ab(a – b)
(a + b)3 – 3ab (a + b)
(a – b)3 + 3ab (a – b)

Worked out Examples

Example 1. Simplify the following:
(a) sinA + 5 cosA + 7 sinA – 10 cosA
(b) (cosD + sinD) (cosD – sinD)

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(c) (sinT + cosT) (sin2T – sinT. cosT + cos2T)

(d) 1 1 + 1
– sinD 1+ sinD

Solution: (a) sinA + 5 cosA + 7 sinA – 10 cosA

= sinA + 7 sinA + 5 cosA – 10 cosA

= 8 sinA – 5 cosA.

(b) (cosD + sinD) (cosD – sinD)

= cosD (cosD – sinD) + sinD (cosD – sinD)

= cos2D – cosD.sinD + sinD.cosD – sin2D

= cos2D – sin2D

(c) (sinT + cosT) (sin2T – sinT . cosT + cos2T)

= sinT(sin2T – sinT.cosT+cos2T)+cosT(sin2T – sinT.cosT + cos2T)

= sin3T – sin2T.cosT+sinT.cos2T+cosT.sin2T – sinT.cos2T + cos3T

= sin3T + cos3T

(d) 1 + 1
1 – sinT 1+ sinT

= (1 + sinT) + (1 – sinT)
(1 –sinT) (1 + sinT)

= 1 + sinT + 1 – sinT [ (a + b) (a – b) = a2 – b2]
1 –sin2T
2
= 1– sin2T

Example 2. Expand the following:
Solution:
(a) (cosT + 1)2 (b) (cosT – sinT)3

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

= cos2T + 2cosT + 1

(b) (cosT – sinT)3 = cos3T – 3cos2T . sinT + 3cosT.sin2T – sin3T

Example 3. Factorise the following:
Solution:
(a) sin2T + sinT (b) sec2T – cot2T

(c) sinD – sin4D (d) tan2T + 5 tanT + 6

(e) sin2T – 3sinT + 2

(a) sin2T + sinT = sinT(sinT + 1)

(b) sec2T – cot2T = (secT + cotT) (secT – cotT)

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(c) sinD – sin4D = sinD(1 – sin3D)
= sinD {(1)3 – (sinD)3}
= sinD{(1 – sinD) (1 + sinD + sin2D)}

(d) tan2T + 5 tanT + 6 = tan2T + (3 + 2) tanT + 6
= tan2T + 3tanT + 2tanT + 6
= tanT (tanT + 3) + 2(tanT + 3)
= (tanT + 3) (tanT + 2)

(e) sin2T – 3sinT + 2 = sin2T – 2sinT – sinT + 2
= sinT(sinT – 2) – 1(sinT – 2)
= (sinT – 2) (sinT – 1)

Exercise 5.3

Short Questions :

1. Simplify the following :

(a) 9 sinA + 10 sinA

(b) 3 cos2T + 5cos2T

(c) (cosecT – sinT) (cosecT + sinT)

(d) (2sinx + cosx) (2sinx – cosx)

(e) (sinT + cosT) (sin2T – sinT.cosT + cos2T)

(f) (1 – tanE) (1 + tanE) (1 + tan2E)

(g) (tanD – cotD) (tan2D + tanD . cotD + cot2D)

(h) (1 + sinT) (1 – sinT) (1 + sin2T)

(i) (tanT + cotT)2 – cot2T

(j) (sinx + cosx)2 – (sinx + 2sinx.cosx)

(k) 1 + 1
1 – cosT 1 + cosT

(l) sinE 1 – sinE 1 cosE
– cosE +

(m) tan2T – cot2T
tanT – cotT

(n) sin3T + cos3T
sinT + cosT

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(d) (tanT – cotT)2
2. Expand the following : (f) (tanT + cotT)3
(a) (1 + tanT)2
(c) (sinT + cosT)2 (b) sinT.cosT + 2sin2T.cos2T
(e) (cosecT – 1)2 (d) sinT.cos2T + 3cosT.sin2T
(g) (sinT + cosT)3 (f) 4 sin2B – 25 cos2B
(h) sin4T – 9cos4T
3. Factorise the following : (j) sec3E – cosec3T
(a) 10 cosA – 10 (l) cos2T + 3 cosT + 2
(c) sin2D + 3sinD.cosD (n) sin2T – sinT – 6
(e) 4cos2T – 9
(g) cosec2T – 4 cot2T
(i) tan4T – 25cot4T
(k) tan3T – cot3T
(m) tan2T – 3 tanT + 2
(o) 3sin2T – 8sinT + 4

1. (a) 19sinA (b) 8cos2T (c) cosec2T – sin2T (d) 4sin2x – cos2x

(e) sin3T + cos3T (f) 1 – tan4E (g) tan3D – cot3D (h) 1 – sin4T

(i) tan2T + 2tanT.cotT (j) sin2x + cos2x – sinx (k) 2
1 – cos2T
2 cosE
(l) sin2E – cos2E (m) tanT+ cotT (n) sin2T – sinT.cosT + cos2T

2. (a) 1 + 2tanT + tan2T (b) sin2T – 2sinT.cosT + cos2T

(c) sin2T+2sinT.cosT+cos2T (d) tanT – 2 + cot2T

(e) cosec2T – 2cosecT + 1 (f) tan3T + 3tan2T.cotT 3tanT.cot2T + cot3T

(g) sin3T + 3sin2T.cosT + 3sinT.cos2T + cos3T

3. (a) 10(cosA – 1) (b) sinT.cosT(1 + 2sinT.cosT)

(c) sinD(sinD + 3cosD) (d) sinT.cosT(cosT + 3sinT)

(e) (2cosT + 3) (2cosT – 3) (f) (2sinB + 5cosB) (2sinB – 5cosB)

(g) (cosecT + 2cotT) (cosecT – 2cotT) (h) (sin2T + 3cos2T) (sin2T – 3cos2T)

(i) (tan2T – 5cot2T) (tan2T + 5cot2T)

(j) (secE – cosec E) (sec2E + secE.cosecE + cosec2E)

(k) (tanT – cotT) (tan2T + tanT.cotT + cot2T) (l) (cosT + 2) (cosT + 1)

(m) (tanT – 2) (tanT – 1) (n) (sinT – 2) (sinT + 2) (o) (sinT – 2) (3sin – 2)

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5.7 Trigonometric Identities

The three sides of a right-angled triangle are: A

Perpendicular (p), Hypotenuse (h), and Base (b). Hypotenuse (h) Perpendicular (p)

If we show them in a right-angled triangle, they appear as
in the figure aside.

Based on these three sides of the triangle, we have six T

trigonometrical ratios which are as follows: C Base (b) B

(a) sinT = p (b) cosT = b (c) tanT = p
h h b
h h
(d) cotT = b (e) secT = b (f) cosecT = p
p
Now, if we observe deep patterns within them, we have the following relations :

(I) Reciprocal Relations

(a) We know that

sinT = p = 1 = 1
h h cosecT
1p
? sinT = cosecT ............................. (i)

cosecT × sinT = 1 .........................(ii)

cosecT = 1 = ........................ (iiii)
sinT

Equation (i), (ii) and (iii) are the formula for the reciprocal relation between

sinT and cosecT.

(b) Again, we have

cosT = b = 1 = 1
h h secT

1b
So, cosT = secT ........................ (iv)

cosT × secT = 1 ................... (v)

secT = 1 ....................... (vi)
cosT

Equation (iv), (v), and (vi) are the formula for the reciprocal relation between
cosT and secT.

(c) Similarly, we have

tanT = p = 1 = 1
b b cot T
p
tanT = 1 T ................... (vii)
cot

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tanT × cotT = 1 ..................... (viii)

cotT = 1 = ...................... (ix)
tan T

Equations (vii), (viii) and (ix) are the formula for the reciprocal relation between

tanT and cotT.

The following table summarizes the relations this:

sinT × cosecT = 1 cosT × secT = 1 tanT × cotT = 1

sinT = 1 cosT = 1 tanT = 1
cosecT secT cotT

cosecT = 1 secT = 1 cotT = 1
sinT cosT tanT

(II) Quotient Relations

We have divide sinT by cosT

sinT = p and cosT = b
h h

p p p
h h× b
sinT = b = h = = tanT
cosT b
h

? tanT= sinT ................. (x)
cosT

Similarly, dividing cosT by sinT, we get

b h b
cosT h p p p
sinT = p = h × = = cotT
h
cosT
? cotT = sinT ................. (xi)

Equation (x) and (xi) are called the quotient relations of tanT and cotT
respectively.

In bird's eye view

tanT = sinT
cosT

cotT = cosT
sinT

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(iii) Trigonometric Relations from Pythagoras Theorem.
(a) We have,

In a right-angled triangle,
h2 = p2 + b2

Dividing by h2 on both the sides, we get,

h2 = p2 + b2
h2 h2 h2

or, 1= p 2+ b2
h h

or, p 2+ b2 =1
h h
p b
or, (sinT)2 + (cosT)2 = 1 ? sinT = h and cosT = h

or, sin2T + cos2T = 1 ................ (xii)

Also, we have

sin2T = 1 – cos2T [On transposition]
cos2T = 1 – sin2T
sinT = 1 – cos2T
cosT = 1 – sin2T
(b) In the same right-angled triangle, we have
h2 = p2 + b2

Dividing by b2 on both the sides,

We get,

h2 = p2 + b2
b2 b2 b2

or, h 2= p 2 +1
b b

or, h 2– p 2=1
b b

or, (secT)2 – (tanT)2 = 1
or, sec2T – tan2T = 1 ..................... (m)
Also, we have,

sec2T = 1 + tan2T [On transposition]
tan2T = sec2T – 1

secT = 1+tan2T
tanT = sec2T – 1

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(c) In the same way,

In the same right-angled triangle,

h2 = p2 + b2

Dividing by p2 on both the sides, we get,

h2 = p2 + b2
p2 p2 p2

or, h2 = 1 + b2
p2 p2

or, h 2=1+ b2
p p

or, h 2– b 2=1
p p

or, (cosecT)2 – (cotT)2 = 1

or, cosec2T – cot2T = 1

Note :
cosec2T = 1 + cot2T
cosecT = 1 + cot2T
cotT = cosec2T – 1

5.8 Equation and Identity

It has always been observed that students have a confusion regarding equations
and identities. So, if we see the examples given below, it clears the doubt to a great
extent.
Let's see the example given below:

(x + 3)2 = x2 + 6x + 9
Put x = 0, we get,

(0 + 3)2 = 02 + 6 . 0 + 9
? 9 = 9, (true)
Again, put x = 2, we get,

(2 + 3)2 = 22 + 6 . 2 + 32 = 25 (True)
Hence, if we observe the pattern here, all the random values of x satisfies the
statement.
Such a mathematical statement is known as an identity.

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Let's observe another example:
x+3=7

Put x = 0, we get
0+3=7

or, 3 = 7 (False)
Again, put, x = 4, we get,

4+3=7
or, 7 = 7 (True)
Also, put x = 2, we get

2+3=7
or, 5 = 7 (False)
So, if we observe this pattern, only certain value of x satisfies the statement. Such
mathematical statement which is true for certain values of the variables is known
as an equation.

5.9 Trigonometric Identity

Identities involving trigonometric ratios are known as trigonometric identities.

Proving trigonometric identities

In trigonometry, we prove the identities using certain tips and tricks. A trigonometric
identity consists of basically two sides, left hand side (L.H.S). and right
hand side (R.H.S). We prove either of the sides, using various techniques.
Example: cotT . sinT = cosT

Here, cotT . sinT is the right hand side of the identity

LHS = cotT . sinT

= cosT . sinT
sinT

RHS = cosT

We use certain technique to prove LHS = RHS.

5.10 Methods of Proving a Trigonometric Identity

We can prove trigonometrical identities by various methods. As this is the preliminary
stage, we expect the students to follow the given steps to prove trigometrical
identities:

1. Take the identity on the left hand side (L.H.S.) and show it equal to R.H.S.

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2. Take the identity on the R.H.S and show it equal to L.H.S.
Students get better idea and remember the formula as well if they write the formula
on the side whenever they use it.

Worked Out Examples

Example 1. Prove the following :
Solution:
(a) cosecT × sinT = 1 (b) tanT × cosT = sinT

(c) cosec2T – cot2T = 1

(d) cosec2T × tan2T – sec2T × sin2T = 1

(a) LHS = cosecT × sinT

= 1 . sinT = 1
sinT

(b) LHS = tanT × cosT

= csoinsTT. cosT = sinT
(c) LHS = cosec2T – cot2T

= 1 – cos2T = 1 (1 – cos2T)
sin2T sin2T sin2T

= 1 . sin2T = 1 = RHS. Proved.
sin2T

(d) LHS = cosec2T × tan2T – sec2T × sin2T

= 1 . sin2T – 1 . sin2T
sin2T cos2T cos2T

= 1 – sin2T = 1 (1 – sin2T)
cos2T cos2T cos2T

= 1 . cos2T
cos2T

= 1 = RHS. Proved.

Example 2. Prove the following trigonometrical identities:

Solution: (a) sinT . secT . cotT = 1 (b) sin2A – cos2A = 1 – 2cos2A

(c) sec4T – sec2T = tan2T + tan4T (d) sin2D – sin2E = cos2E – cos2D

(e) 1 – 2sinT . cosT = cosT – sinT

(a) LHS = sinT . secT . cotT

= sinT . 1 . cosT = 1 = RHS
cosT sinT

? LHS = RHS proved.

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(b) LHS = sin2A – cos2A
= 1 – cos2A – cos2A = 1 – 2 cos2A = RHS.

? LHS = RHS, proved.
(c) LHS = sec4T – sec2T = sec2T (sec2T – 1)

= (1 + tan2T) tan2T
= tan2T + tan4T = RHS proved.
(d) LHS = sin2D – sin2E
= (1 – cos2D) – (1 – cos2E)
= 1 – cos2D – 1 + cos2E
= cos2E – cos2D = RHS
? LHS = RHS proved.
(e) LHS = 1 – 2sinT . cosT
= (cos2T + sin2T) – 2sinT . cosT
= cos2T – 2sinT . cosT + sin2T
= (cosT – sinT)2
= cosT – sinT = RHS
LHS = RHS, proved.

Example 3. Prove the following identities:
Solution:
(a) tanA 1 cotA = cosA . sinA
+

(b) sin3T – cos3T = 1 + sinT . cosT
sinT – cosT

(c) 1 1 + 1 – 1 = 2cosec2T
+ cosT cosT

(d) secT 1 tanT = secT + tanT
–

(e) 1 – sin4A = 1 + 2tan2A
cos4A

(f) 1 + tan2T = 1 + tanT 2
1 + cot2T 1 + cotT

(a) LHS = tanA 1 cotA = sinA 1 cosA
+ cosA + sinA

= sin2A 1 cos2A [ sin2T + cos2T = 1]
+
cosA . sinA

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Trigonometry

= 1 × cosA + sinA
1 sin2A + cos2A

= cosA . sinA = cosA . sinA = RHS
1

? LHS = RHS proved.

(b) LHS = sin3T – cos3T
sinT – cosT

= (sinT – cosT) (sin2T + sinT . cosT + cos2T)
sinT – cosT

= (sin2T + cos2T) + sinT . cosT

= 1 + sinT . cosT = RHS

? LHS = RHS proved.

(c) LHS = 1 1 + 1 – 1
+ cosT cosT
1 – cosT + 1 + cosT
=
= 2 1 – cos2T = RHS
sin2T = 2cosec2T

? LHS = RHS proved.

(d) LHS = secT 1 tanT = sec2T – tan2T
– secT – tanT

= (secT + tanT) (secT – tanT)
(secT – tanT)
= secT + tanT = RHS

? LHS = RHS proved.

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

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

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

= sec2A + tan2A = 1 + tan2A + tan2A

= 1 + 2tan2A = RHS

? LHS = RHS proved.

(f) LHS = 1 + tan2T = 1 + tan2T
1 + cot2T 1 + 1

1 + tan2T tan2T
tan2T + 1
=

= tan2T × 1 tan2T = tan2T
(1 + tan2T) + tan2T

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Trigonometry

RHS = 1 + tanT 2
=
1 + cotT 2
1 + tanT

1+ 1
tanT

= (1 + tanT) × tanT 2
tanT +
1

= tan2T = RHS

? LHS = RHS proved.

Exercise 5.4

Short Questions :

1. Prove the following :

(a) tanD . cotD = 1 (b) 1 = secE . cosE
(d) 1 + cot2T = cosec2T
(c) secD = tanD
cosecD
2. Prove the following identities:

(a) cosA . cosecA . tanA = 1 (b) secT = tanT
cosecT
sin$
(c) tan$ = cosA (d) tan2T . cos2T = sin2T

(e) sinT . cosecT = cosT (f) tanT . cotT = sinT . cosT
secT secT . cosecT

(g) sinT . secT = tan2T (h) cosD . cosecD . tanD =1
cosT . cosecT sinD . secD . cotD

(i) cot2$ . sec$ = secA . cosec4A
cos2$ . sin2$

3. Prove the following identities:

(a) sinA + 2cosA = cosecA
sin2A + 2sinA . cosA

(b) (sinT ÷ secT) ÷ (cosT ÷ sinT) =sin2T

(c) sinT . cos2T + sin3T = sinT

(d) sinT . cotT + sinT . cosecT = 1 + cosT

(e) (1 – sin2D) . cosec2D = cot2D

(f) sec2A – cosec2A = tan2A – cot2A

(g) sin2T + sin2T . tan2T = tan2T

(h) cosec4D – cosec2D = cot4D + cot2D

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(i) tan2A – sin2A = tan2A . sin2A

(j) cosA . 1 + cot2$ = cosec2A – 1
(k) cosA. 1 + tan2$ + sinA . 1 + cot2$ = 2
(l) sec2A + cosec2A = secA . cosecA
(m) sinA 1 – sin2A + cosA . 1 – cos2A = 2 sinA . cosA
4. Prove the following identities:

(a) (a cosA + b sinA)2 + (a sinA – b cosA)2 = a2 + b2

(b) (1 + tanA)2 + (1 – tanA)2 = 2 sec2A

(c) 1 = cosecT + cotT
cosecT – cotT

(d) (1 + cosT) (1 – cosT) = tan2T
(1 + sinT) (1 – sinT)
secT – 1 1 – cosT
(e) secT + 1 = 1 + cosT

(f) sin4T – cos4T = sinT + cosT
sinT – cosT

(g) 1 + tanA = cotA + 1
1 – tanA cotA – 1

(h) (1 – sinA) (1 + sinA) = cot2A
(1 + cosA) (1 – cosA)

5. Prove the following :

(a) 1 + 1 = 2 cosec2A
1 + cosA 1 – cosA

(b) 1 – 1 = 2 cotD . cosecD
1 – cosD 1 + cosD

(c) sinA + sinA = 2 cosecA
1 + cosA 1 – cosA

(d) cosA + cosA = 2 secA
1 + sinA 1 – sinA

(e) cosT – cosT = 2 tanT
1 – sinT 1 + sinT

(f) 1 + cosT + sinT = 2 cosecT
sinT 1 + cosT

(g) cosE + 1 + sinE = 2 secE
1 + sinE cosE

(h) 1 – sin4D = 1 + 2 tan2D
cos4D

(i) 1 – cos4D = 1 + 2 cot2D
sin4D

(j) 1 – sin$ = secA – tanA
1 + sinA

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Trigonometry

(k) 1 – cosT = cosecT – cotT
1 + cosT

l) secT + 1 = cosecT + cotT
secT – 1

(m) 1 cotT + 1 cotT = 2cosecT
cosecT + cosecT –

(n) secA 1 tanA – 1 = 1 – secA 1 tanA
– cosA cosA +

5.11 Conversion of trigonometric ratios

Conversion simply means process of changing from one form to another. So,
changing the given trigonometrical ratios in terms of other ratios for the same angle
is possible. We can work it out in the following ways :
(i) Using Trigonomtric relations
(ii) Using Pythagoras theorem

Worked out Examples

Example 1. Express all trigonometric ratios of angle T in terms of sinT.
Solution:
Using trigonometric relations

We know,

sinT = sinT

cosecT = 1
sinT

cosT = 1 – sin2T

secT = 1 = 1
cosT 1 – sin2T

tanT = sinT = sinT
cosT 1 – sin2T
1 – sin2T
cotT = 1 =
tanT sinT

Alternative method

Using Pythagoras theorem

Let ABC be a right-angled triangle right-angled at B.

Let ‘ACB = T be the reference angle.

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Trigonometry

Then, sinT = k = p A
1 h
So, p = k, h = 1, b = ?

We know, k
h2 = p2 + b2

or, b2 = h2 – p2 T
or, b2 = 1 – k2 BC

? b = 1 – k2 = 1 – sin2T

Now, cosT = b = 1 – sin2T = 1 – sin2T
h 1

tanT = p = k sinT
b = 1 – sin2T

1 – k2
1
cosecT = h= sinT
p
1
secT = h = = 1
b 1 – sin2T
1 – k2

cotT = b = 1 – k2 = 1 – sin2T
p k sinT

Example 2. 1
Solution: Find the values of remaining trigonometric ratios if cosT = 2
1 b
Here, cosT = 2 = h A

So, b = 1, h = 2, p = ?

We know in a right-angled triangle ABC, 2

h2 = p2 + b2 T C
or, p2 = h2 – b2 B1

= 22 – 12 = 4 – 1 = 3

? p =3

p3
Now, sinT = =
cosT = h = 2
b 1
h 2

tanT = p 3 = 3
b= 1
h
cosecT = p = 2
3

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secT = h = 2 =2 Trigonometry
b 1
b 95
cotT = p = 1
3
Example 3. If 2 sinT = 3 , find cosT, tanT and cotT.

Solution: We have,

2 sinT = 3

3p
or, sinT = 2 = h
? p= 3,h=2,b=?

In a right-angled triangle

h2 = p2 + b2

or, b2 = (2)2 – ( 3 )2 = 4 – 3 = 1

? b =1

Now, cosT = b = 1
h 2

p3 3
tanT = b = 1 =
b
cotT = p = 1
3

Example 4. If 3 cotT = 4, find the value of 5sinT – 3cosT
Solution: sinT + 2cosT
We have,

3 cotT = 4

or, cotT = 4 = b
3 p
b = 4, p = 3, h = ?

We know in a right-angled triangle

h2 = p2 + b2

= (3)2 + (4)2 = 9 + 16 = 25

? h=5

Now, we know,

sinT = p = 3
h 5
b 4
cosT = h = 5

vedanta Excel in Additional Mathematics - Book 8

Trigonometry

So, 5 sinT – 3 cosT = 5 × 3 – 3 × 4
sinT + 2 cosT 5 5

3 + 2 × 4
5 5

= 15 – 12 × 5 = 3
5 + 11
3 8

Example 5. m2 – n2
Solution: If (m2 + n2) sinT = 2mn, prove that cosT = m2 + n2

We have,
Example 6.
Solution: (m2 + n2) . sinT = 2 mn

or, sinT = 2mn = p
m2 + n2 h
p = 2mn , h = m2 + n2, b = ?

We know,

h2 = p2 + b2

or, b2 = (m2 + n2)2 – (2mn)2

= m4 + 2m2n2 + n4 – 4m2n2

= m4 – 2m2n2 + n4

= (m2)2 – 2m2n2 + (n2)2

= (m2 – n2)2

? b = m2 – n2

Now, cosT = b = m2 – n2 proved.
h m2 + n2

If sinT – cosT = 0, prove that secT = 2

Given that,

sinT – cosT = 0

or, sinT = cosT

or, sinT =1
cosT
1 p
or, tanT = 1 = b

p = 1, b = 1, h = ?

In a right-angled triangle,

h2 = p2 + b2

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

? h =2

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Now, secT = h = 2
b 1 =2
Proved.
? secT = 2

Example 7. If 1 – cosA = 1 , show that tan2A + cot2A = 3 1 .
Solution: 2 3

Given that,

Example 8. 1– cosA = 1
Solution: 2
1
or, 1– 2 = cosA

? cosA = 1 = b
2 h
b = 1, h = 2, p = ?

We know in a right-angled ',

h2 = p2 + b2

or, p2 = (2)2 – (1)2 = 4 – 1 = 3

? p = 3

Now, tanA = p = 3 3
b 1=
1
cotA = 3

So, LHS = tan2A + cot2A

=( 3 )2 + 12
3
1 9 + 1 10 1
= 3 + 3 = 3 = 3 = 3 3 = RHS

If tanT = pq, prove that p cosT +q sinT = p2 + q2
p cosT –q sinT p2 – q2

Here, tanT = q , then cotT = p
p q

Now, LHS = p cosT + q sinT
p cosT – q sinT

Dividing namerator and denominator by cosT, we get,

= p +q. tanT = p +q. q
p –q. tanT p –q. p
q
p2 + q2 p p
p –
= × p2 q2

= p2 + q2 = RHS proved.
p2 – q2

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Exercise 5.5

Short Questions :

1. (a) Express all trigonometric ratios of angle T in terms of cosT

(b) Express all trigonometric ratios of angle T in terms of tanT

(c) Express all trigonometric ratios of angle T in terms of cosecT.

(d) Express all trigonometric ratios of T in terms of secT.

2. (a) Find all the values of trigonometric ratios when sinT = 3 .
2
1
(b) If tanA = 2 , find all the values of remaining trigonometric ratios.

3. (a) If secT = 2 find cosecT and cotT.

(b) Find the value of tanT and sinT, whose cosT is 3 .
2
sinD + 2cosD
4. (a) If 5 tanD = 4, find the value of 5sinD – 3cosD .

(b) If 2 tanT = 1, find the value of cos2T – sin2T .
2 cosT sinT
5. (a) If x tanT = y, find the value of x cosT + y sinT.

(b) If x2 + y2 . cosT = y, show that ytanT = x.

6. (a) If cosT – sinT = 0, prove that secT = 2 .

(b) If sinT – x cosT = 0, show that secT . cosecT = x + 1 .
x
7. (a) If 1 – sinT = 1 , then show that 6 sec2T + 9 tan2T = 11.
2
(b) If (a + 1) sinD = a – 1, show that 2 a . tanD = a – 1.

8. (a) If 3 sinT + 4 cosT = 5, find the value of tanT.

(b) If 5 cosT + 12 sinT = 13, find the value of tanT.

1. Show to your teacher.

2. (a) cosT = 1 , tanT = 3, cosecT = 2 , secT = 2, cotT = 1
2 3 3

(b) sinA = 1 , cosA = 2 , cotA = 2, cosecA = 5, secA = 5
5 5 2

3. (a) cosecT = 2, cotT = 1 (b) tanT = 1 , sinT = 1
3 2
14
4. (a) 5 (b) 1 5.(a) x2 + y2
22
3 12
8. (a) tanT = 4 (b) tanT = 5

98 vedanta Excel in Additional Mathematics - Book 8

Trigonometric Ratios of Some Standard Angles

Trigonometric Ratios of 6
Some Standard Angles

Different angles have different value with various trigonometric ratios. We
consider 0°, 30°, 45°, 60°, and 90° as the standard angles and we evaluate value of
these angles in this unit.

6.1 Trigonometric Ratio of 45°

Let ABC be a right-angled isosceles triangle where A
45°
‘B = 90° and ‘A = ‘C = 45°

Also, let BC = AB = a.

We know in right-angled triangle ABC,

h2 = p2 + b2 45° B
C

or, (AC)2 = (AB)2 + (BC)2

or, (AC)2 = a2 + a2

or, (AC)2 = 2a2

? AC = a 2

Taking ‘C as the reference angle, we get,

sin 45° = AB = a a = 1
cos 45° AC 2 2
tan 45° BC
cosec 45° = AC = a = 1
a2 2
a
= AB = a = 1
BC

= AC = a2 = 2
AB a

sec 45° = AC = a2 = 2
cot 45° BC a

= BC = a = 1.
AB a

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Trigonometric Ratios of Some Standard Angles

6.2 Trigonometric Ratio of 30° and 60°

Let ABC be an equilateral triangle where A = B = C = 60° A

and AB = BC = CA = 2a.

Now, let's draw AD perpendicular to BC so that,

BD = DC = a and ‘BAD = ‘DAC = 30°

Now, in right-angled 'ADC, 60° 60°
(AC)2 = (AD)2 + (DC)2 B DC

or, (AD)2 = (AC)2 – (DC)2

= (2a)2 – (a)2

= 4a2 – a2 = 3a2

? AD = a 3

Now, to find out the trigonometric ratio for 60°, let's take ‘ACD = 60° as the reference
angle. From 'ADC, we get,

sin60° = AD = a3 = 3
cos60° AC 2a 2
DC a 1
= AC = 2a = 2

tan60° = AD = a3 = 3
DC a
AC 2a 2
cosec60° = AD = a3 = 3

sec 60° = AC = 2a =2
cot 60° DC a
DC a 1
= AD = a3 = 3

Now, to find out the trigonometric ratio for 30°, let's take ‘DAC = 30° as the reference

angle. From 'ADC, we get,

sin 30° = DC = a = 1
AC 2a 2

cos 30° = AD = a3 = 3
AC 2a 2

tan 30° = DC = a = 1
AD a3 3

cosec 30° = AC = 2a = 2
DC a
AC 2a 2
sec 30° = AD = a3 = 3

cot 30° = AD = a 3 = 3
DC a

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