Final-for-print-Illustrated-OPT-Math-9-2076-2-1

Example 9 : Verify that: sec A 1 A – 1 = 1 – sec A 1 A
+ tan cos A cos A – tan
Solution: Here,
RHS = 1 – 1
1 1 cos A sec A + tan A
LHS = + – cos A
sec A tan A
1
1 1 = 1 –
= + – cos cos A 1 sin
1 sin A cos – cos A
cos A cos A A

= 1 = 1 = 1 A – 1 cos A × 1 + sin A
1 + sin cos cos – sin A 1 + sin A
A
cos A
= 1 – cos A(1 + sin A)
cos A 1 cos A 1 – sin2 A
= 1 + sin A – cos
A
= 1 – cos A(1 + sin A)
= cos A × 1 – sin A – 1 cos A cos2 A
1 + sin A 1 – sin A cos A
= 1 – (1 – sin A)
= cos A(1 – sin A) – 1 cos A
1 – sin2 A cos A

= cos A(1 – sin A) – 1 = – tan A.
cos2 A cos A

= 1 – sin A – 1 = – tan A
cos A

Hence, LHS = RHS Proved.

''Alternatively'' TRIGONOMETRY

LHS = sec A 1 tan A – 1
+ cos A

= sec A – tan A – sec A [ Conjugating by sec A – tan A]
sec2 A – tan2 A

= sec A – tan A – sec A = – tan A

RHS = 1 A – 1 [ Conjugating by sec A – tan A]
cos sec A – tan A

= sec A – sec A + tan A
sec2 A – tan2 A

= sec A – sec A – tan A = – tan A

Hence, LHS = RHS. Proved.

''Next method''

Here, 1 – 1 = 1 – 1
sec A + tan A cos A cos A sec A – tan A

or, 1 + 1 = 1 + 1 [ By transposition]
sec A + tan A sec A – tan A cos A cos A

or, sec A +1 tan A + sec A 1 tan A = 2 A
– cos

Now, LHS = 1 + 1
sec A + tan A sec A – tan A

= sec A – tan A + sec A + tan A= 2sec A= 2 A = RHS. Proved.
sec2 A + tan2 A 1 cos

Trigonometric Ratios 201

Example 10 : Prove that: 1 cos A – 1 sin A A = cos A – sin A
+ tan A + cot

Solution: Here, = cos2 A – sin2 A
LHS = cos A – sin A cos A + sin A sin A + cos A

1 + tan A 1 + cot A = cos2 A – sin2 A
sin A + cos A
= 1 +cocssoinAs AA – sin A
= (cos A + sin A) (coa a – sin A)
1 + cos A cos A + sin A
sin A

= cos cos A A – sin A = cos A – sin A = RHS. Proved.
A + sin sin A + cos A

cos A sin A

Example 11 : Prove that: (sec A + tan A – 1) (sec A – tan A + 1) = 2tan A

Solution: Here,

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

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

= sec2 A – (tan A – 1)2

= sec2 A – tan2 A + 2tan A – 1

= 1 + 2tan A – 1

= 2tan A = RHS. Proved.

Example 12 : Show that: tan A – sec A + 1 = cos A
tan A + sec A – 1 1 + sin A

TRIGONOMETRY Solution: Here,

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

= tan A – sec A + 1
(tan A + sec A) – (sec2 A – tan2 A)

= 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)

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

= cos A = RHS. Proved.
1 + sin A

Example 13 : Verify that: cosec6 A – cot6 A = 3cot4 A + 3cot2 A + 1

Solution: Here,

LHS = cosec6 A – cot6 A = (cosec2 A)3 – (cot2 A)3

= (cosec2 A – cot2 A) [(cosec2 A)2 + cosec2 A . cot2 A + (cot2 A)2]

= 1 . [(cosec2 A – cot2 A)2 + 2cosec2 A . cot2 A + cosec2 A . cot2 A)

= (12 + 3cosec2 A . cot2 A) = 1 + 3(1 + cot2 A) cot2 A = 1 + 3cot2 A + 3cot4 A

= 3cot4 A + 3 cot2 A + 1 = RHS. Proved.

202 Illustrated Optional Mathematics-9

Example 14 : Prove that: sin cos A B + sin cos B A = sin cos A B + cos B
A+ cos B– cos A– cos sin B + cos A

Solution: Here, sin cos A A + sin cos B A = sin cos A A + sin cos B A
A+ cos B– cos A– cos B+ cos

or, sin cos A A – sin cos A B = sin cos B A – sin cos B A
A+ cos A– cos B+ cos B– cos

or, cos(sAin(sAin+Ac–oscoBs)B(s–insAin–Ac–oscoBs) B) = cos B(sin B – cos A – sin B – cos A)
(sin B + cos A) (sin B – cos A)

or, cossinA2 A. (–– 2cos B) = cos A . (– 2cos B)
cos2 B sin2 B– cos2 A

or, –si2nc2oAsA– .ccooss2 B) = – 2cosA . cos B)
B sin2 B – 1 + sin2 A

or, –si2nc2oAsA– .ccooss2 B) = – 2cosA . cos B)
B sin2 A – cos2 B

Hence, LHS = RHS. Proved.

EXERCISE - 5.2 (C)

1. Prove the following trigonometric identities.

(a) tan A . cos A = sin A (b) sec A . sin A = tan A

(c) sin θ . sec θ . cot θ = 1 (d) 1 + tan2 A . cos A = 1 TRIGONOMETRY
(e) sin2 A – cos2 B = sin2 B – cos2 A (f) tan2 θ – sin2 θ = sin2 θ . tan2 θ

(g) cos α . cosec α sec2 α – 1 = 1 (h) sin2 β + cos 4β = cos2 β + sin4 β

(i) tan2 A – cot2 A = sec2 A – cosec2 A (j) tan A + cot A = sec A . cosec A

(k) sec2 A + cosec2 A = sec2 A . cosec2 A (l) tan2 A + cot2 A = sec2 A + cosec2 A – 2

2. Show the relations of the following trigonometric identities:
(a) (cos A – sin A)2 = 1 – 2sin A . cos A (b) tan2 A + cot2 A + 2 = sec2 A . cosec2 A

(c) sec θ – tan θ . sin θ = cos θ (d) cosec4 A + cot4 A = 1 + 2cot2 A . cosec2 A

(e) (1 – tan A)2 + (1 – cot A)2 = (sec A – cosec A)2

(f) (1 + sin A + cos A)2 = 2(1 + sin A) (1 + cos A)

(g) (1 + sin α – cos α)2 + (1 – sin α + cos α)2 = 4 (1 – sin α . cos α)

(h) (1 + sin α – cos α)2 – (1 – sin α + cot α)2 = 4 cos α (tan α – 1)

(i) cos2 A – sin2 A = cosec A + sec A

sin A . cos2 A – cos A . sin2 A

(j) (sin A + cosec A)2 + (cos A + sec A)2 = tan2 A+ cot2 A + 7

(k) (sec θ – cos θ) (cosec θ – sin θ) (tan θ + cot θ) = 1

(l) (cosec2 A + cot2 A)2 = 1 + 4cosec2 A . cot2 A

3. Prove that: (b) sec A + tan A = 1 + sin A = 1 cos A
(a) cos A . cot A = 1 + cosec A cos A – sin A

1 – sin A

Trigonometric Ratios 203

(c) sec 1 = 1 – sin A (d) 1 = cosec α – cot α
+ cos A cosec α + cot α
A tan A

(e) cot A = cosec A – 1 (f) 1 = sin A . cos A
cosec A + 1 cot A tan A – cot A

(g) 1 – tan α = cot α – 1 (h) 1 – tan2 A = cos2 A – sin2 A = 1 – 2sin2 A
1 + tan α cot α + 1 1 + tan2 A­

(i) tan θ + sin θ = sec θ + 1 (j) cos x + sin x = sin x + cos x
tan θ – sin θ sec θ – 1 1 – tan x 1 – cot x

(k) 2cos2 α –­1 = cot α – tan α (l) 1 – 1 θ – 1 + 1 θ = 2sec θ . sec2 θ
sin α . cos α sin sin

(m) 1 – c1os α – 1 + 1 α = 2cot α . cosec α (n) 1 cos A + 1 cos A = 2 sec A
cos – sin A + sin A

(o) tan x – 1 sin x x = 2 cot x (p) 1 + sin2 θ = 2sec2 – 1
sec x – 1 + cos 1 – sin2 θ

(q) cos x + 1 + sin x = 2 sec x (r) sin x + 1 + cos x = 2 cosec x
1 + sin cos x A– cos sin x
x sec x

(s) cosec α 1 + cosec α 1 = 2 sec2 α (t) sec 1 – 1 + sec 1 + 1 = 2 cot A . cosec A
cosec α – cosec α + A A

4. Show that: 1+ sin θθ (b) 1 – cos A = cosec A – cot A
(a) sec θ + tan θ = 1– sin 1 + cos A

TRIGONOMETRY (c) 11 +– sin A = (sec A – tan A)2 (d) 1+ sin A – 1– sin A = 2 tan A
sin A 1– sin A 1+ sin A

(e) 1+ cos A + 1– cos A = 2 cosec A (f) 1+ sin A – sec A = sec A – 1 – sin A
1– cos A 1+ cos A 1– sin A 1+ sin A

5. Prove the following trigonometric identities:

(a) (sec A – tan A – 1) (sec A + tan A + 1) = – 2 tan A

(b) (1 + sin A + cos A) (1 – sin A – cos A) = – 2sin A . cos A

(c) (tan θ + cot θ +sec θ) (tan θ + cot θ – sec θ) = cosec2 θ

(d) ttaann A + sec A – 1 = 1+ sin A (e) cot α + cosec α – 1 = cosec α + cot α
A – sec A + 1 cos A cot α – cosec α + 1

(f) 1 – sec A + tan A = sec A + tan – 1 (g) sin A + cos A + 1 = sec A + tan A
1 + sec A – tan A sec A + tan A + 1 sin A + cos A – 1

(h) 11 ++ sec θ – tan θ + 1 + sec θ + tan θ = 2sec θ
sec θ + tan θ 1 + sec θ – tan θ

(i) 11 + sin A– cos A 2 1 – cos A = (cosecA– cot A)2
+ sin A+ cos A 1 + cos A
=

(j) ssiinn θ + cos θ + 1 – 1 + sin θ – cos θ = 2cosec θ.
θ – cos θ + 1 1 + sin θ + cos

6. Prove that:

(a) sin4 A + cos4 A = 1 – 2sin2 A . cos2 A (b) tan4 θ + sec4 θ = 1 + 2tan2 θ . sec2 θ

(c) cosec4 A – cot4 A = 1 + 2cot2 A (d) 1 – cos4 A = 1 + 2cot2 A
sin4 A­

204 Illustrated Optional Mathematics-9

(e) sec4 A – 1 = 2tan2 α + tan4 α (f) sin3 α + cos3 θ = 1 – sin θ . cos θ
(g) sin3 α + cos3 α + sin3 α – cos3 α = 2 sin θ + cos θ

sin α + cos α sin α – cos α (h) cos6 θ + sin6 θ = 1 – 3sin2 θ . cos2 θ
(i) sec6 α – tan6 α = 1 + 3tan2 α . sec2 α
(k) sin4 A + cos4 A = tan2 A + cot2 A (j) cosec6 A – cot6A = 3 cosec2 A . cot2 A + 1

sin4 A­ – cos4 A tan2 A – cot2 A

(l) sin6 A – cos6 A = (sin A + cos A) (sin A – cos A) (1 + sin A . cos A) (1 – sin A . cos A)
(m) cos8 A – sin8 A = (cos2 A – sin2 A) (1 – 2sin2 A . cos2 A)
(n) sin8 A + cos8 A = 1 – 4sin2 A cos2 A + 2sin4 A . cos4 A

7. Show that:

(a) 1 + tan2 A = tan2 A (b) 1 – tan2 θ = cos2 θ – sin2 θ
1 + cot2 A 1 + tan2 θ

(c) sec A – tan A = 1 – 2sec A . tan A + 2tan2 A (d) tan α + cot α = 1 + tan α + cot α
sec A + tan A 1 – cot α 1 – tan α

(e) 1 + 1 = 0 (f) sec θ – tan θ = cos2 θ

cosec A(1 – tan A) sec A(1 – cot A) sec θ + tan θ (1 + sin θ)2

(g) tan α – sin α = 2cot α (h) sec A – cot A = sin2 A
sec α – 1 1 + cos α cosec A – cot A (1 – cos A)2

(i) 1+ tan2 θ = 1 – tan θ 2 (j) tan2 α – cot2 α – sin2 α – cos2 α
1 + cot2 θ 1 – cot θ 1 + cot2 α cos2 α

(k) cot θ . cos θ = cot θ – cos θ TRIGONOMETRY
cot θ + cos θ cot θ . cos θ

(l) cot(1 1 α) + cot(1 1 α) = 1 + sec α . cosec α
– cot – tan

(m) 1 + cosec 1 cot A) = sin A + cos A
sec A(1 – tan A) A(1 –

(n) cosseeccAA++tacnotAA –cosseeccAA––tacnotAA = 2 (sec A – cosec A)

8. Prove the following trigonometric identities:

(a) sin α + sin β + cos α – cos β = 0 (b) cot A + tan B = cot A . tan B
cos α + cot β sin α – sin β tan A + cot B

(c) cos A + cos B = cos A + cos B
sin A – cos B sin B + cos A sin A + cos B sin B – cos A

(d) (sin α . cos β – cos α . sin β)2 + (cos α . cos β + sin α . sin β)2 = 1

(e) (sin θ + sec θ)2 + (cos θ + cosec θ)2 = (sec θ + cosec θ)2

(f) 1 + (cosec α . tan β)2 = 1 + (cot α . sin ββ))22 (g) sin2A.sec2 B + tan2 B.cos2A = sin2A + tan2 B
1+ (cosec α . tan β)2 1 + (cot α . sin

Project Work

(i) Draw a right-angled triangle on an A4 sheet. Label it and name its sides on the base on the
reference angle. Also, list the trigonometric ratios from the triangle.

(ii) List the trigonometric relations in a full page.
(iii) Prepare the converting chart of one trigonometric ratio into other trigonometric ratios.

Trigonometric Ratios 205

51.31 TRIGONOMETRIC RATIOS OF
STANDARD ANGLES

5.3 (A) Values of Trigonometric Ratios of Standard Angles

Learning Objectives

At the end of this topic, the students will be able to:

• find the values of the trigonometric ratios of standard angles.

• use the values of the trigonometric ratios of standard angles in the problems.

In this topic, we derive the values of the trigonometric ratios of some standard angles like as 0o, 30o, 45o, 60o

and 90o by geometrical interpretation. After then we compute the values of trigonometric ratios greater than

90o up to 360o like as 120o, 135o, 150o, 180o, 210o, 225o, 240o, 270o, 300o, 315o, 330o, 360o., ... on the basis of

the values of the trigonometric ratios of the standard angles in next topic.

Trigonometric Ratios of 0o

In the adjoining right-angled triangle POM, ∠PMO = 90o and ∠POM = θ. When the angle θ becomes

smaller and smaller and the line segment PM also becomes smaller in length. In figure, when θ becomes

0o, the point P coincides with M, then PM = 0 and OM = OP. P

TRIGONOMETRY Now, we have

(i) sin 0o = PM = 0 = 0 [ sin θ = p ]
OP OP h

(ii) cos 0o = PM = OM = 1 [ cos θ = b ] θ
OP OM h
(iii) tan 0o = PM = 0 = 0 p O M
OM OM [ b ]
tan θ =

[ (iv) cot 0o = OM = OM = ∞ (undefined) cot θ = b ]
PM 0 p
sec 0o = OP = h
(v) OM OP = 1 [ sec θ = b ]
OP

OP = OP = ∞ (undefined) h
PM 0 p
[ (vi) ]cosec
cosec 0o = θ =

Here, (i) sin 0o = tan 0o = 0 ii) cos 0o = sec 0o = 1 iii) cot 0o = cosec 0o = ∞

Trigonometric Ratios of 30o and 60o A

In the adjoining figure, ∆ABC is an equilateral triangle and AB = BC = CA = 2a 2a 30° 3a 2a
(suppose). We have, ∠ABC = ∠ACB = ∠BAC = 60o. Now, draw AD ⊥ BC, then
BD = CD = a and ∠BAD = 30o. From the right-angled triangle ABD,

We have, AD = AB2 – BD2 [ Pythagoras theorem]

= (2a)2 – a2 60° 60°
= 4a2 – a2 = 3a2 = 3.a B aD a C

206 Illustrated Optional Mathematics-9

Now, we have

For Ratios of 30o:

(i) sin 30° = BD = a = 1 (iv) cot 30° = AD = 3a = 3
AB 2a 2 BD a

(ii) cos 30° = AD = 23aa = 3 (v) sec 30° = AB = 2a = 2
AB 2 AD 3a 3

(iii) tan 30° = BD = a = 1 (vi) cosec 30° = AB = 2a = 2
AD 3a 3 BD a

For Ratios of 60o:

(i) sin 60° = AD = 3a 3 (iv) cot 60° = BD = BD = a = 1
AB 2a = 2 AD AB 3a 3

(ii) cos 60° = BD = a = 1 (v) sec 60° = AB = a = 2
AB 2a 2 BD 2a

(iii) tan 60° = AD = 3a 3 (iv) cosec 60° = AB = 2a = 2
BD a= AD 3a 3

Here, (i) sin 30° = cos 60° = 1 (iv) cot 30° = tan 60° = 3
2
2 2
(ii) cos 30° = sin 60° = 3 (v) sec 30° = cosec 60° = 3

(iii) tan 30° = cot 60° = 1 (vi) cosec 30° = sec 60° = 2 TRIGONOMETRY
3

Trigonometric Ratios of 45°

In the adjoining figure, ∆ABC is an isosceles right-angled triangle in 2a A
which AB = BC = a (suppose), ∠ABC = 90° and ∠ACB = ∠BAC = 45°.
a
So, AC = AB2 – BC2 [ Pythagoras theorem] B

a2 – a2 = 2a 2 = 2 . a 45o
Now, we have Ca

(i) sin 45° = AB = a = 1 (iv) cot 45° = BC = a = 1
AC 2a 2 AB a

(ii) cos 45° = BC = a = 1 (v) sec 45° = AC = 2a 2
AC 2a 2 BC a=

(iii) tan 45° = AB = a = 1 (vi) cosec 45° =AC = 2a 2
BC a BC a=

Here, (i) sin 45o = cos 45o = 1
2

(ii) tan 45o = cot 45o = 1

(iii) sec 45o = cosec 45o = 2

Trigonometric Ratios of Standard Angles 207

Trigonometric Ratios of 90o A

In the adjoining right-angled triangle ABC, ∠ACB = 90o and ∠ACB = θ, when the
angle θ increases greater and greater, then the line segment BC becomes smaller in
length. Finally, when θ is 90o, the point C coincides with the point B. i.e. BC = 0 and
AC = AB.

Now, we have, θ B
C

(i) sin 90o = AB = AB = 1 (iv) cot90o = BC = 0 = 0
AC AB AC AB

(ii) cos 90o = BC = 0 = 0 (v) sec 90o = AC = AC = ∞ (undefined)
AC AC BC 0

(iii) tan 90o = AB = AB = ∞ (undefined) (vi) cosec 90o = AC = AC = 1
CC 0 AB AC

Here, (i) sin 90o = cosec 90o = 1 (ii) cos 90o = cot 90o = 0 (iii) tan 90o = sec 90o = ∞

As the trigonometric ratios of the standard angles 0o, 30o, 45o, 60o and 90o are very often used, they should
be remembered very carefully. So, we discuss the method of computing values of the trigonometric ratios
of 0o, 30o, 45o, 60o and 90o in the form of the table as follows:

Angle 0o 30o 45o 60o 90o

Rule 0 1 2 = 1 3 4
Ratio 4 4 4 2 4 4

TRIGONOMETRY sin 0 11 31
cos 1 22 2
tan 0
cot ∞ 31 1 0
sec 1 22 2
cosec ∞
1 1 3∞
3

31 1 0
2 3
32
2∞

22 2 1
3

In this table, at first we suppose 0, 1, 2, 3 and 4 for the columns of the angles 0o, 30o, 45o, 60o and 90o
respectively and divide every number by 4 and then take square root on them. Simplify them and the
computed number is the values of the sine of the standard angles respectively. Reverse the value of sine
of the standard angles, we obtain the value of cosine of the standard angles. For computing the values of
tangent of the standard angles, divide the values of the sine of the standard angle by that of cosine of the
same angle. Similarly, for computing the values of cotangent of the standard angles, reverse the values of
the tangent. For the values of secant, take the reciprocal of the values of the cosine and then reverse them,
we obtain the values of cosecant of the standard angles.

208 Illustrated Optional Mathematics-9

Example 1 : Find the value of (sin 30o + cos²30o – tan 0o) sin 90o. Use of scientific calculator

Solution: Here, (sin 30o + cos² 30o – tan 0o) sin 90o (i) For sin 30o 0 = 0.5 = 1
Press: sin 3 2
= 1 + 32
2 2 –0 1 (ii) For in cos 30o 0 = 0.8661 =
3
= 1 + 3 =2+3 = 5 Press: cos 3 2
2 4 4 4

Example 2 : If πc = 180o, find the value of cot2 πc – 2 cos2 πc – 3 sin2 πc – 4sin2 πc .
6 3 4 4 6
Solution: Here, cot2 πc – 2cos2 πc– 3 sin2 πc – 4sin2 πc
6 34 4 6

= cot2 180° – 2 cos2 180° – 3 sin2 180° – 4sin2 180°
6 34 4 6
= cot2 30o – 2cos2 60o – 3 sin2 45o – 4sin2 30o
4

2 1 2–3 1 2 1 2
2 2
= 3 –2. 24 – 4 .

= 3 – 2 × 1 – 3 × 1 – 4 × 1 = 3 – 1 – 3 – 1 = 2 – 1 – 3 = 16 –4 – 3 = 9 = 118 .
4 4 2 4 2 8 2 8 8 8
Example 3 : If θ = 30o and α = 60o prove that: tan(α – θ) = tan α – tan θ
1 + tan θ . tan α
Solution: Here, θ = 30o , α = 60o

Now, LHS = tan(α – θ) = tan(60o – 30o) = tan 30o = 1 TRIGONOMETRY
3
1
3 – 3 –1
tan α – tan θ tan 60° – tan 30° 3 3 1 3 2 1
RHS = 1 + tan α . tan θ = + tan 60° . tan 30° = = +1 = 3×2 = 3
1 1 + × 1
3

Hence, LHS = RHS. Proved.

Example 4 : Find the value of x if 3 sin 60o + x cos 30o . tan 45o = x cot 30o.

Solution: Here, 3 sin 60o + x cos 30o . tan 45o = x cot 30o

33 3 or, 3 3 = 2 3x – 3x
or, 3 × 2 + x . 2 × 1 = x .
or, 3 3 + 3 . x = 3 . x or, 3 3 = 3x

2 or, x = 3 3
or, 3 3 + 3x = 2 3x 3

∴ x=3

EXERCISE - 5.3 (A)

1. (a) What is the value of sin 60°? (b) Which angle of trigonometric ratio is equal to sin 30°?
(d) Which trigonometric ratios have the value 13?
(c) Write the value of tan 45°?
2. Find the value of: (b) sec 0o (sin 30o + tan 45o)
(a) sin 0o – cos 0o + tan 0o (d) sin2 45o + 3 tan2 30o – cot2 30o

(c) sin2 30o + cos2 60o – tan2 30o 2

Trigonometric Ratios of Standard Angles 209

(e) sin2 60o (2cos2 45o – cosec2 30o – sec2 45o) (f) sin2 45° – tan2­ 30° + cos 60°
sec2­ 30° + sin2 60°
3. Find the value of: (πc = 180o)
(a) sin πc – cos πc + tan πc (b) tan2 πc + cos2 πc – cosec2 πc
634 32 4

(c) sin πc – cos πc × cosec πc – sec πc (d) sin πc + cos πc sin πc – cos πc
66 44 3 36 6

(e) sin2 πc – sin2 πc – sin2 πc + 3cos2 πc (f) 3 sec2 πc + 2 sec2 πc – 4cos2 πc + 2tan2 πc
6 34 6 4 43 3 3 4

4. If A = 90o, B = 60o, C = 45o, D = 30o and E = 0o, find the value of:

(a) sin2 A + cos2 B (b) 2cos A . sin B (c) sin A + cot B + tan C + cot D – sec E

(d) 1 2tan B (e) sin2 C – cos2 D (f) sin2 B – sin2 C
– tan2 B cos2 C – cos2 D

5. Prove the following:

(a) 1 – sin 30° = 1 – tan 3600°° (b) cot2 30° – 1 = cot 60o
1 + sin 60° 1 + tan 2cot 30°

(c) cos2 45o – sin2 45o = cos 90o (d) sin 90o = 3 sin 30o – 4sin3 30o

(e) sin πc + cos πc sin πc – cos πc = 12
66 33

(f) cos πc . cos πc + sin πc . sin πc = sin πc . cos πc + cos πc . sin πc
3 4 34 4 6 46
6. If A = 60o and B = 30o, verify that:

TRIGONOMETRY (a) sin (A + B) = sin A . cos B + cos A . sin B (b) cos (A – B) = cos A . cos B + sin A . sin B

(c) tan (A + B) = tan A + t.atnanBB (d) cot (A – B) = cot B . cot A + 1
– tan A cot B – cot A
1

(e) cos2 B = 1 – 2sin2 B (f) cos3 B = 4cos3 B – 3cos B

(g) sin2 A = 2sin A cos A = 1 2tan A (h) (1 + sin A) (1 – sin A) = cot2 A
+ tan2 A (1 + cos A) (1 – cos A)

7. Find the value of x if:

(a) 2 sin 45o. sec 60o – 2xcos 60o = 1 (b) cos 60o + 2tan2 30o + xsin2 60o = 9 + cos 60o

(c) tan2 45o – sec2 30o = x sin 45o . cos 45o . tan2 30o

(d) 2xsin π – 2cos2 π + tan π = sin π + x cos π
6 642 3
(e) x cos πc . tan πc + 3sin πc = x tan πc (f) x sin πc . cos πc + tan2 πc = x tan2 πc – cos2 πc
64 3 3 44 3 43

ANSWERS

2. (a) – 1 (b) 32 (c) 1 (d) – 2 (e) – 15 (f) 8
(b) 1 6 (d) – 1 4 25
3. (a) 1
4. (a) 5 (c) 1 2 (e) 5 (f) 31
2 (d) – 3 4 6
4 (e) – 1
7. (a) 1 (b) 0 (c) 4 + 3 (d) 3 (f) – 1
4
3
100 1
(b) 9 (c) – 2 (e) 3 (f) 6 2

210 Illustrated Optional Mathematics-9

5.3 (B) Solution of Right-angled Triangle

Learning Objectives

At the end of this topic, the students will be able to:
• solve the right-angled triangle by using Pythagoras theorem and trigonometric ratios.

If ∆ABC is the right-angled triangle in which ∠ABC = 90o suppose, A h
∠ACB = θ, ∠BAC = α, AC = h, AB = p and BC = b. αo
Then we have,
p

θ + α = 90o ……… (i) [ sum of acute angles in right triangle] θo
and h² = p² + b² ……... (ii) [ Pythagoras' theorem ]
Bb C

If we know one side and any one of the remaining four parts of the right-angled triangle ABC, we find the

other parts. The process of calculating the unknown parts of the right-angled triangle is called the solution

of the right-angled triangle. There are two methods for the solution of the right-angled triangle. They are:

(i) Using Pythagoras' Theorem

(ii) Using Trigonometric Ratios.

There are two cases for solution of the right-angled triangle:

(i) When two sides are given TRIGONOMETRY
When two sides of the right-angled triangle are given, then we use the Pythagoras' theorem,

h² = P² + b².

If the sides p and b are given, then we find h by h = p2­+ b2.

Otherwise, we use p = h2­– b2 or b = h2­– p2, and then, apply the trigonometric ratio;

sin θ = p or, cos θ = b or, tan θ = p to find θ.
h h b

At last, α = 90o – θ.

(ii) When one side and one acute angle are given

When one side and one acute angle of the right-angled triangle are given, then we can use the
suitable trigonometric ratio to find the measure of any one side of the remaining two sides of the

right-angled triangle. If one side p and one acute angle θ are given, then we use any one ratio which

has p,

as sin θ = p or tan θ = p ⇒ h = p θ or b = p θ.
h b sin tan

After then, we use

b = h2­– p2 or h = p2­+ b2 after finding h or b.

At last, α = 90o – θ.

Trigonometric Ratios of Standard Angles 211

Example 1 : In the given right-angled triangle ABC, if ∠ABC = 90o, AC = 13 units and BC = 12
units, solve ∆ABC.

Solution: Here, in the given right-angled triangle ABC, ∠ABC = 90o, AC = 13 units, BC = 12 units,

AB = ? ∠BAC = α = ?, ∠ACB = θ = ? A

Now, we have αo 13 units

AB = AC2­– BC2 [ Pythagoras' theorem ]

= 132­– 122 = 169­– 147 = 25 = 5 units θo
B 12 units C
Again,
BC p
sin α = AC sin α = h Use of scientific calculator

(i) For sin 45o
Press: sin 4 5 = 0.7071 = 1
= 12 = sin 67.38o 2
∴ 13 (ii) For a = sin–1 12
13
α = 67.38o and

∴ θ = 90o – α = 90o – 67.38o = 22.62o Press: shift sin ( 12 ÷ 13 ) = 67.38°

Hence, AB = 5 units, ∠BAC (α) = 67.38o and ∠ACB (θ) = 22.62o.

Example 2 : In the adjoining right-angled triangle PQR, ∠PQR = 90o, ∠QPR = 30o and QR = 7 cm,
find the remaining parts of ∆PQR.

Solution: Here, in the given right-angled triangle PQR, ∠PQR = 90o, ∠QPR = 30o, QR = 7 cm,

TRIGONOMETRY PQ = ?, PR = ? and ∠PRQ = ? P
30o
Noro, w, s21in=3Q07Ro = QPRR [ sin θ = hp]

or, PR = 14 cm

or, PQ = PR2­– QR2 = 142­– 72 = 169 ­– 49 = 147 = 12.12 cm Q 7 units R
∠PRQ = 90o – ∠QPR = 90o – 30o = 60o.

Hence, the remaining parts of ∆PQR are PR = 14 cm, PQ = 12.12 cm and ∠PRQ = 60o.

Example 3 : In the adjoining figure, ∆ABC and ∆ADC are the two right-angled triangles and BC =
5 cm. Find the lengths of CD.

Solution: Here, in the given right-angled triangles ABC and ADC, ∠ABC = 90o, ∠ADC = 90o,

BC = 5 cm and ∠ACB = 30o, ∠DAC = 60o, CD = ? D

Now, in right-angled ∆ABC,

cos 30o = BC A 60o ?
AC
or, 23 = A5C 10
or, AC = 3 cm 30o
B 5 cm
Again, in right angle ∆ACD, C

sin 60o = CD or, 3 = 3 CD or, CD = 5cm
AC 2

Hence, the length of CD is 5cm. 10

212 Illustrated Optional Mathematics-9

Example 4 : Find the value of cos 2θ and sin 3θ, if 3tan θ = 3 and θ is an acute angle

Solution: Here, 3tan θ = 3 ⇒ tan θ = 1
3
1
We have, 3 = tan 30o ∴ tan θ = tan 30o ⇒ θ = 30°

Now, cos 2θ = cos 2 × 30o = cos 60o = 21, sin 3θ = sin 3 × 30o = sin 90o = 1.

EXERCISE - 5.3 (B)

1. (a) Write the relation between hypotenuse (h), perpendicular (p) and base (b) of a right-angled triangle.

(b) What is the sum of acute angles of a right triangle?

2. Find the unknown sides of the following right-angled triangles. X

(a) A (b) P (c) L N (d)

6 58 97

17

B8 C Q 4R M Y 72 Z

3. Find the unknown angles of the following right triangles:

(a) E (b) A (c) Q R (d) E

12 6
G
30° BC 58° F
FG P A

4. (a) In the given right-angled triangle ABC, ∠ABC = 90o, AB = 12 P
cm and BC = 5 cm, find the side AC, ∠ACB and ∠BAC.
12 cm 17 cm TRIGONOMETRY

(b) Find the length of PQ, the measures of ∠QPR and ∠PRQ from

the given ∆PQR. C 5 cm B Q 8 cm R
Q.N. 4 (b)
(c) Find the remaining parts of the adjoining right-angled triangle Q.N. 4 (a)
X
LMN. L

(d) In the adjoining right-angled triangle XYZ, ∠Y = 90o, 8 cm 30°
∠YXZ = 30o, YZ = 12 cm, solve ∆XYZ.
60°
(e) Solve the triangle ABC when ∠A = 90o, a = 20 cm, c = 16 cm.
M N Y 12 cm Z
(f) Solve the triangle ABC when ∠B = 90o, ∠C = 60o, a = 9cm.
Q.N. 4 (c) Q.N. 4 (d)
5. (a) Find the length of AB from the adjoining figure when ∆ABC
and ∆ACD are right-angled triangles. 3 cm D A
A 60°
(b) In the given figure, ∆ABC and ∆BCD are two right-angled CB
triangles with right angles at B and C respectively. If CD = 24
cm, find the length of AB. 30° 24 cm 30°

ANSWERS 45° D
BC Q.N. 5 (b)

Q.N. 5 (a)

2. (a) 10 (b) 3 (c) 15 (d) 65 3. (a) 60o (b) 45o, 45o (d) 32o (e) 30o, 60o

4. (a) 13 cm, 67.38o, 22.62o (b) 15 cm, 28.07o, 61.93o (c) 30o, 16 cm, 8 cm (d) 60o, 24cm, 12 3 cm
33

(e) 12 cm, 36.87o, 53.13o (f) 30o, 18 cm, 9 3 cm 5. (a) 3 2 cm (b) 8 cm

Trigonometric Ratios of Standard Angles 213

51.41 TRIGONOMETRIC RATIOS OF
ANY ANGLE

Learning Objectives

At the end of this topic, the students will be able to:

• find the trigonometric ratios of any angle.

In the adjoining figure, two straight lines XX' and YY' Y
perpendicularly intersect at O. These two lines divide the
entire plane into four equal parts, each part is called quadrant. P'
The plane included by XOY is called first quadrant, the
plane included by YOX' is called second quadrant, the plane X' θ +ve angle
included by X'OY' is called third quadrant and the plane 90o X
included by XOY' is called fourth quadrant. O θ -ve P

angle

P''

Take a revolving line OP which starts from the line OX in the Y'
anti-clockwise direction. It makes positive angle, otherwise
negative angle. In one complete rotation of the revolving line 2nd 1st
OP, it turns 360o which coincides with OX.
Quadrant Quadrant

180o 0o, 360o

Thus, the range of the angle in the first quadrant is 0o to 90o. 3rd 4th
Similarly, in the second quadrant, 90o to 180o, in the third
quadrant, 180o to 270o and in the fourth quadrant, 270o to 360o Quadrant Quadrant
as shown in the figure.
TRIGONOMETRY 270o

Trigonometric Ratios of Negative Acute Angle (– θ)

Let the revolving line OP makes a positive acute angle θ with x-axis Y
in the position of P1 and makes an negative acute angle – θ with x-axis
in the position of P1(x, y). Draw P1M ⊥ OX and P2M ⊥ OX. So, the P1(x, y)
right-angled triangles OP1M, and OP2M are congruent.
X' O θ P X
–θ M

Then OP1 = OP2, P1M = – P2M. Y' P2(x, – y)

Now, in right-angled ∆OP1M, P1M
p PO1PM1 b OM p OM
(i) sin θ = h = (ii) cos θ = h = OP1 (iii) tan θ = b =

(iv) cot θ = b = OM (v) sec θ = h = OP1 (vi) Cosec θ = h = OP1
p P1M b OM p P1M

Again, in right-angled ∆OP2M,

(i) sin(– θ) = p = P2M = – P1M = – sin θ (ii) cos(– θ) = b = OM = OM = cos θ
h OP2 OP1 h OP2 OP1

(iii) tan(– θ) = p = O2M = O1M – tan θ (iv) cot(– θ) = b = OM = – OM = – cot θ
b OM OM p O2M O1M

(v) sec(– θ) = h = OP2 = OP1 = sec θ (vi) cosec(– θ) = h = OP2 = OP1 = – cosec θ
b OM OM p O2M – P1M

Hence, the cosine and secant of negative angle θ i.e., cos (– θ) and sec (– θ) are positive in sign as the same

ratios and the others are negative in sign as the same ratios.

214 Illustrated Optional Mathematics-9

Trigonometric Ratios of Complementary Angle (90o – θ)

Trigonometric Ratios of (90o – θ) Y
Let the revolving line OP make an angle θ with x-axis in the anti P1(x, y)
clockwise direction at the position of P1 and draw P1M ⊥ OX.
Then ∠OP1M = 90 – θ. 90 – θ

X' O θ MP X

Now, in right-angled ∆OP1M for angle θ. Y'

(i) sin θ = p = PO1PM1 (ii) cos θ = b = OM
h h OP1

(iii) tan θ = p = P1M (iv) cot θ = b = OM
b OM p P1M

(v) sec θ = h = OP1 (vi) cosec θ = h = OP1
b OM p P1M

Now, in right-angled ∆OP1M for angle (90o – θ).

(i) sin(90o – θ) = OM = cos θ (ii) cos(90o – θ) = P1M = sin θ
OP1 OP1

(iii) tan(90o – θ) = OM = cot θ (iv) cot(90o – θ) = P1M = tan θ
P1M OM

(v) sec(90o – θ) = OP1 = cosec θ (vi) cosec(90o – θ) = OP1 = sec θ
P1M OM

Since the angle (90o – θ) lies in the first quadrant where θ is an acute angle. So, all the trigonometric ratios TRIGONOMETRY

are positive in sign and the ratios are changed in form below: These are only in
sin → cos, cos → sin tan → cot, CAST rule.
cot → tan sec → cosec, cosec → sec

Trigonometric Ratios of Any Angles

Trigonometric Ratios of (90o + θ)

Let the revolving line OP make an angle θ with x-axis in anti-clockwise direction at the position of P1. The

revolving line OP further revolves in the same direction and makes 90o with the position of P1. Now draw

P1M ⊥ OX and P2M1 ⊥ OX'. Y

Then ∠P1OM = θ, ∠POP2 = 90o + θ, P2 P1

∴ ∠P2OM' = 90o – θ and ∠OP2M1' = θ. 90° + θ
So, ∆OP1M ≅ ∆OP2M' by ASA congruent condition.
Then P1M = – OM' and OM' = P2M' and OP1 = OP2. θ
θ
X' M' O MP X

Now, in right-angled ∆P1OM Y'

(i) sin θ = p = PO1PM1 (ii) cos θ = b = OM
h h OP1

(iii) tan θ = p = P1M (iv) cot θ = b = OM
b OM p P1M

(v) sec θ = h = OP1 (vi) cosec θ = h = OP1
b OM p P1M

Trigonometric Ratios of Any Angle 215

In right-angled ∆P2OM'

(i) sin(90° + θ) = P2M' = P1M = cos θ
OP2 OP1

(ii) cos(90° + θ) = OM' = – P1M = – sin θ
OP2 OP1

(iii) tan(90° + θ) = P2M' = OM = – cot θ
OM' – P1M

(iv) cot(90° + θ) = OM' = – P1M = – tan θ
O2M' OM

(v) sec(90° + θ) = OP2 = OP1 = – cosec θ
OM' – OM

(vi) cosec(90° + θ) = OP2 = OP1 = sec θ
O2M' OM

Hence, the angle (90o + θ) lies in the second quadrant, so the ratios and their signs are changed as CAST

rule, i.e. all the trigonometric ratios are changed as sin → cos, tan → cot, sec → cosec and vice versa and

sin and cosec are only positive in sign, but others are negative.

Trigonometric Ratio of (180o – θ)

Let the revolving line OP make an angle θ with OX in the position of P1 and further make is the same

angle θ with OX'. Y

TRIGONOMETRY So, ∠POP2 = 180o – θ and draw P1M ⊥ OX and P2M' ⊥ OX'. P2 P1
Then, OP1 = OP2 and ∆P1OM ≅ P2OM', by ASA congruent condition.
So, P1M = P2M', and OM = – OM'. 180o – θ
θ
X' θ M X
M' O

Now, in right-angled ∆P1OM Y'

(i) sin θ = p = PO1PM1 (ii) cos θ = b = OM
h h OP1

(iii) tan θ = p = P1M (iv) cot θ = b = OM
b OM p P1M

(v) sec θ = h = OP1 (vi) cosec θ = h = OP1
b OM p P1M

In right-angled ∆P2OM'

(i) sin(180° – θ) = P2M' = P1M = sin θ (ii) cos(180° – θ) = OM' = – OM = – cos θ
OP2 OP1 OP2 OP1

(iii) tan(180° – θ) = P2M' = P1M = – tan θ (iv) cot(180° – θ) = OM' = – OM = – cot θ
OM' – OM O2M' P1M

(v) sec(180° – θ) = OP2 = OP1 = – sec θ (vi) cosec(180° – θ) = OP2 = OP1 = cosec θ
O2M' – P1M O2M' P1M

Hence, the angle (180o – θ) lies in the second quadrant and all the trigonometric ratios are unchanged, but
their signs follow the CAST rule.

216 Illustrated Optional Mathematics-9

Trigonometric Ratios of (180o + θ) TRIGONOMETRY
(i) sin(180o + θ) = sin{90o + (90o + θ)} = cos(90o + θ) = – sin θ
(ii) cos(180o + θ) = cos{90o + (90o + θ)} = – sin(90o + θ) = – cos θ
(iii) tan(180o + θ) = tan{90o + (90o + θ)} = – cot(90o + θ) = tan θ
(iv) cot(180o + θ) = cot{90o + (90o + θ)} = – tan(90o + θ) = cot θ
(v) sec(180o + θ) = sec{90o + (90o + θ)} = – cosec(90o + θ) = – sec θ
(vi) cosec(180o + θ) = cosec{90o + (90o + θ)} = sec(90o + θ) = – cosec θ

Hence, the angle (180o + θ) lies in the third quadrant and all the trigonometric ratios are unchanged and
their signs follow the CAST rule.
Trigonometric Ratios of (270o – θ):

(i) sin(270o – θ) = sin{180o + (90o – θ)} = – sin(90o – θ) = – cos θ
(ii) cos(270o – θ) = cos{180o + (90o – θ)} = – cos(90o – θ) = – sin θ
(iii) tan(270o – θ) = tan{180o + (90o – θ)} = tan(90o – θ) = cot θ
(iv) cot(270o – θ) = cot{180o + (90o – θ)} = cot(90o – θ) = tan θ
(v) sec(270o – θ) = sec{180o + (90o – θ)} = – sec(90o – θ) = – cosec θ
(vi) cosec(270o – θ) = cosec{180o + (90o – θ)} = – cosec(90o – θ) = – sec θ

Hence, the angle (270o – θ) lies in the third quadrant and all the trigonometric ratios are changed but their
signs follow the CAST rule.
Trigonometric Ratios of (270o + θ):

(i) sin(270o + θ) = sin{180o + (90o + θ)} = – sin(90o + θ) = – cos θ
(ii) cos(270o + θ) = cos{180o + (90o + θ)} = – cos(90o + θ) = sin θ
(iii) tan(270o + θ) = tan{180o + (90o + θ)} = tan(90o + θ) = – cot θ
(iv) cot(270o + θ) = cot{180o + (90o + θ)} = cot(90o + θ) = – tan θ
(v) sec(270o + θ) = sec{180o + (90o + θ)} = – sec(90o + θ) = cosec θ
(vi) cosec(270o + θ) = cosec{180o + (90o + θ)} = – cosec(90o + θ) = – sec θ

Here, the angle (270o + θ) lies in the fourth quadrant and all the trigonometric ratios are changed and their
signs follow the CAST rule.

Trigonometric Ratios of (360o – θ):
(i) sin(360o – θ) = sin{180o + (180o – θ)} = – sin(180o – θ) = – sin θ

(ii) cos(360o – θ) = cos{180o + (180o – θ)} = – cos(180o – θ) = cos θ

(iii) tan(360o – θ) = tan{180o + (180o – θ)} = tan(180o – θ) = – tan θ

(iv) cot(360o – θ) = cot{180o + (180o – θ)} = cot(180o – θ) = – cot θ

(v) sec(360o – θ) = sec{180o + (180o – θ)} = – sec(180o – θ) = sec θ

(vi) cosec(360o – θ) = cosec{180o + (180o – θ)} = – cosec(180o – θ) = – cosec θ

Here, the angle (360o – θ) lies in the fourth quadrant and all the trigonometric ratios are unchanged and
their signs follow the CAST rule.

Trigonometric Ratios of Any Angle 217

Trigonometric Ratios of (360o + θ):
(i) sin(360o + θ) = sin{180o + (180o + θ)} = – sin(180o + θ) = sin θ
(ii) cos(360o + θ) = cos{180o + (180o + θ)} = – cos(180o + θ) = cos θ
(iii) tan(360o + θ) = tan{180o + (180o + θ)} = tan(180o + θ) = tan θ
(iv) cot(360o + θ) = cot{180o + (180o + θ)} = cot(180o + θ) = – cot θ
(v) sec(360o + θ) = sec{180o + (180o + θ)} = – sec(180o + θ) = sec θ
(vi) cosec(360o + θ) = cosec{180o + (180o + θ)} = – cosec(180o + θ) = cosec θ

Here, the angle (360o +θ ) lies in the first quadrant and all the trigonometric ratios are unchanged and their
signs follow the CAST rule.

General Rule for Finding Trigonometric Ratios of Any Angle with reference to 90o
The trigonometric ratios of any angle can be first generalized with referred to 90o only for sine of the given
angle in the following ways.

TRIGONOMETRY Ratio of Angle Result General Form Quadrant
sin θ sin θ sin(0 × 90o + θ) First
sin(90o – θ) cos θ sin(1 × 90o – θ)
sin(90o + θ) cos θ sin(1 × 90o + θ) Second
sin(180o – θ) sin θ sin(2 × 90o – θ) Third
sin(180o + θ) – sin θ sin(2 × 90o + θ)
sin(270o – θ) – cos θ sin(3 × 90o – θ) Fourth
sin(270o + θ) – cos θ sin(3 × 90o + θ) First
sin(360o – θ) – sin θ sin(4 × 90o – θ) and so on...
sin(– θ) – sin θ sin(0 × 90o – θ)
sin(360o + θ) sin θ sin(4 × 90o + θ)

From the above table it is clear that as we see in the column 'General form' when the number or multiple of
90o is odd (i.e., 1, 3,…), the ratio is changed with the rule as sin → cos, tan → cot, sec → cosec and vice
versa. Similarly, when the number or multiple of 90o (i.e. 2, 4,….) is even, the ratio is not changed. But the
sign of the ratio is arranged on the CAST rule. Hence, the general form of converting the trigonometric
ratios of any angle into the standard angle is (n × 90o ± θ), when n is a whole number and the sign of the
ratios follows the above mentioned rule.

Note: 1) If n is an odd number in (n × 90o ± θ), then the trigonometric ratio will be changed.

2) If n is an even number or zero in (n × 90o ± θ), then the trigonometric ratio will not be changed.

218 Illustrated Optional Mathematics-9

2nd Quadrant 90o
coSinec (+ve)
1st Quadrant
All ratios (+ve)

90o + θ S A θ, 90o – θ
180o – θ 360o + θ
+ve +ve

180o +ve +ve 0o, 360o

θ
–ve +ve

180o + θ O –ve – θ, 270o + θ
270o – θ –ve 360o – θ
+ve +ve

3rd Quadrant TC 4th Quadrant

coTan (+ve) seCos (+ve)

270o

Changing Ratio as follows: sin → cos, tan → cot, sec → cosec and vice versa.

We can easily derive the trigonometric ratios of 120o, 135o, 150o, 180o, 210o, 225o, 240o, 270o, 300o, 315o,
330o, 360o, 390o and so on by using the above mentioned general rule. Now, we derive the trigonometric
ratios of some angles as follows:

Now, we make the table which shows the trigonometric ratios of the standard angles up to 360o by using TRIGONOMETRY
the method of reversing the previous values and the CAST rule as follows:

Y

3 120o
(i) sin 120o = sin(1 × 90o + 30o) = cos 30o = 2
X' O X

"Alternatively"

sin 120o = sin(2 × 90o – 60o) = sin 60o = 3 Y'
2 Y

(ii) cos 150o = cos(1 × 90o + 60o) = – sin 60o = – 3 240o X
2 X' O

Y'

(iii) cot 240o = cot(3 × 90o – 30o) = tan 30o = 1 Y
3 600o
X' O
(vii) sin 600o = sin(6 × 90o + 60o) = –t sin 60o = – 3 X
2 Y'

Trigonometric Ratios of Any Angle 219

1st Quadrant 2nd Quadrant 3rd Quadrant 4th Quadrant
30o 45o 60o
Angle 0o 90o 120o 135o 150o 180o 210o 225o 240o 270o 300o 315o 330o 360o
Ratio

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

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

tan 0 1 1 3 ∞ –3 –1 – 1 0 11 3 ∞ –3 -1 – 1 0
3 3 3
3

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

sec 1 2 22 ∞ –2 – 2 – 2 –1 – 2 – 2 -2 ∞ 2 221
3 3 3 3

cosec ∞ 2Method of Reversing223 1 2 22 ∞ –2 – 2 – 2 –1 – 2 –2 –2 ∞
220 Illustrated Optional Mathematics-9333

TRIGONOMETRY

Example 1 : Find the value of cos 855o. cos(180o – 45o ) 90o

Solution: Here, cos 855o 180o – 45o 45o 0o
= cos(9 × 90o + 45o) O 360o

= – sin 45o

= – 12. 270o


Example 2 : Simplify: sin(180° + A) × sec(270° + A) × ccoosse(c3(6900°°+–AA)).
tan(90° + A) cot(180° + A)

Solution: Here, sin(180° + A) × sec(270° + A) × cos(360° + A)
tan(90° + A) cot(180° + A) cosec(90° – A)

1

= – sin A × cosec A × cos A = sin A × sin A × cos A
– cot A cot A sec A cos A cos A 1

sin A sin A cos A

= sin A × sin A × 1 × sin A × cos A × cos A = sin2 A
cos A sin cos A 1
A

Example 3 : Find the value of tan 21o tan 69o – cos 112o . cosec 158o.

Solution: Here, tan 21o . tan 69o – cos 112o cosec 158o TRIGONOMETRY

= tan 21o. tan(90o – 21o) – cos(90 + 22o) cosec(180o – 22o)

= tan 21o. cot 21o – (– sin 22o) cosec 22o

= tan 21o . cot 21o + sin 22o. cosec 22o = 1 + 1 = 2

Example 4 : Prove that: sin 35o + cos 65o = cos 55o + sin 25o.
Solution: Here, LHS = sin 35o + cos 65o
= sin(90o – 55o) + cos(90o – 25o)
= cos 55o + sin 25o = RHS. Proved.

Example 5 : If sin 4A = cos 2A, then find the value of A.

Solution: Here, sin 4A = cos 2A "Alternatively"

or, sin 4A = sin(90o – 2A) or sin(90o + 2A) sin 4A = cos2A

∴ 4A = (90o – 2A) or (90O + 2A) or, cos(90o – 4A) = cos2A

Either, 4A = 90o – 2A ∴ 90o – 4A = 2A or, 90o = 6A

or, 6A = 90o ∴ A = 15o
∴ A = 15o. Again, cos(270o + 4A) = cos 2A
OR, 4A = 90o + 2A ∴ 270o + 4A = 2A or, 270o = – 2A
or, 2A = 90o ∴ A = – 135o
∴ A = 45o ∴ A = (180o – 135o) or, (360o – 135o)
or, A = 45o or 225o

Hence, A = 15o or 45o. ∴ A = 15o, 45o, or 225o

Trigonometric Ratios of Any Angle 221

Example 6 : Find the value of x if 2 cot 120o – x sin 120o . cos 180o = tan 150o

Solution: Here, 2 cot 120o – xsin 120o cos 180o = tan 150o

or, 2cot(90o + 30o) – xsin(90o + 30o) . cos(90o + 90o) = tan(90o + 60o)

or, 2(– tan 30o) – xcos 30o (– sin 90o) = – cot 60o

or, –2× 1 +x. 3 1
3 2 .1=– 3

or, – 2 + 3x =– 1
3 2 3

or, 23x = 23 – 1
3

or, 23x = 13

or, x= 1 × 2
3 3

or, x = 23.

Example 7 : If A, B and C are the angles of a triangle then show that sin(A + B) – cos C = cos(A + B) + sin c.

Solution: Here, Since A, B and C are the angles of a triangle, then A + B + C = 180o

or, A + B = 180o – C

Now,

TRIGONOMETRY LHS = sin (A + B) – cos C RHS = cos(A + B) + sinC
= cos(180o – C) + sin C
= sin(180o – C) – cos C = – cos C + sin C
= sin C – cos C
= sin C – cos C

= sin C – cos C

Hence, LHS = RHS. Proved.

EXERCISE - 5.4

1. (a) Which trigonometric ratio is equal to sine of angle – A?
(b) Which trigonometric ratio is equal to tangent of (90° + A)?
(c) Write the trigonometric ratio that is equal to cosecant of (270° – θ).

2. Determine the values of the following trigonometric ratios:

(a) sin 135o (b) cos 150o (c) cot 225o (d) sec 114πc
(h) cot(– 1950o)
(e) cosec 660o (f) tan 1035o (g) sin 1680o

3. Simplify:

(a) sin (180o – A) . tan (90o + A) – tan (– A) . sec (270o + A)

(b) cos(270° – A) × tan(360° – A) × cosec(90° + A)
sin(180° + A) sec(90° + A) cot(180° – A)

222 Illustrated Optional Mathematics-9

(c) sin(– A) × cos(180° + A) × cot(540° + A)
sin(90° – A) cosec(450° – A) sec(630° – A)

(d) sin 480o. cos 330o – sec 420o + cos 630o

(e) 3ssiinn22112500°°–+ccoos2t21­ 12305°° (f) sin3(π + θ) × tan(2π – θ)
4. Find the values of the following: π sec2( π – θ
cos2 2 + θ

(a) sin 780o . cos 390o – sin 330o . cos 300

(b) sin 420o . cos 390o + cos(– 300o) sin(– 330o)

(c) tan2 45o + 4sin2 60o + 2cos2 45o + sec 180o + cosec 135o

(d) sin 250° + tan 290°)
cot 200° + cos 340

(e) 2 cos2 3πc + sin 5πc + 1 cos πc + tan2 5πc
4 6 2 3
πc 3πc 5πc 7πc
(f) sin2 4 + sin2 4 + sin2 4 + sin2 4

5. Prove that:

(a) cot (– 1358o) = tan 8o (b) tan 9o. tan 27o = cot 63o . cot 81o

(c) cos 20o . cos 40o = sin 70o . sin 50o (d) sin 140o . cos 165o = – sin 40o . cos 15o

(e) cos 9o + cosec 36o + cot 72o = sin 81o + sec 54o + tan 18o

6. Prove that: TRIGONOMETRY

(a) sin 55o . cos 35o + sin 35o . cos 55o = 1

(b) cos 65o . cos 25o – sin 65o . sin 25o = 0

(c) cos 20o + cos 40o + cos 140o + cos 160o = 0

(d) cos(90° + θ) sec(– θ) tan(180° – θ) = – 1
sec(360° + θ) sin(180° + θ) cot(90°­ – θ)

(e) sin(270° + A) cos(90° – A) = 1
sin(180° – A) cos(180° – A)

(f) cot (90o – A) . tan (90o – A) . cot (90o + A) . tan (270o – A) . cosec (90o + A) . sin(270o + A) = 1

7. Show that:

(a) sin 40o + cos 35o + cos 130o + sin 235o = 0

(b) cos πc + cos 3πc + cos 5πc + cos 7πc =0
8 8 8 8

(c) cot πc . cot 3πc . cot 5πc . cot 7πc . cot 9πc = 1
20 20 20 20 20

(d) sin2 πc + sin2 3πc + sin2 5πc + sin2 7πc =2
8 8 8 8

(e) cos2 πc + cos2 3πc + cos2 5πc + ocs2 7πc = 2
16 16 16 16

(f) tan πc + tan 3πc + tan 5πc + tan 7πc = 0
8 8 8 8

Trigonometric Ratios of Any Angle 223

8. Determine the angle A when:

(a) sin 3A = sin 2A (b) cos A = sin 4A (c) tan 2A = cot 4A

(d) sec 6A = cosec 3A (e) cot 4A – tan 2A = cos 90o (f) cosec 3A = sec 6A

9. Find the value of x if:

(a) tan 225o – cosec2 30o = x . sin 315o . cos 135o . tan2 60o

(b) 2 cot 120o – x . sin 120o . cos 180o = tan 150o

(c) tan2 135o – sec2 60o = x sin 135o . cos 45o . tan2 60o

(d) cosec(90o + θ) + x cos θ . cot(90o + θ) = sin(90o + θ)

(e) x cos(90o + α) . cos(90o – α) + tant(180o – α) . cot(90o + α) = sin α . sin(180o – α)

(f) x sec(90o + A) . cosec A + tan (90o + A) cot(180o + A) = x cot A . tan(90o + A)

10. If A, B and C are the angles of ∆ABC, then prove that:

(a) cos(A + B) = – cos C

(b) tan(A + C) = – tan B

(c) sin(A + B) + sin(B + C) + sin(C + A) = sin A + sin B + sin C

(d) cos(A + B) + cos(B + C) + cos(C + A) + cos A + cos B + cos C = 0

(e) cosec A + B = sec C
2 2 2

(f) tan(2A + 2C) = – tan 2B

11. If A, B, C and D are the angles of a quadrilateral ABCD, then show that:

TRIGONOMETRY (a) cos(A +C) – cos(B + D) = 0 (b) sin(A + B) + sin(C + D) = 0

(c) cos(A + B) – cos(C + D) = 0 (d) cos C +A + cos B + D =0
2 2

ANSWERS

2. (a) 1 (b) – 3 (c) 1 (d) – 2 (e) – 2 (f) –1
2 2 3
(f) sin2 θ . cos θ
(g) – 3 (h) 3 (f) 2
2
2 (f) – cot2 A
3. (a) – sin A . tan A (b) tan3 A (c) – sin A . cos2 A (d) – 5 (e) 7
(c) 4 + 2 4
4. (a) 1 (b) 1 (e) 4
(d) – 1

8. (a) 0o, 36o or 360o (b) 18o or 30o (c) 15o or 45o

(d) 10o or 90o (e) 15o or 45o (f) 9o, 67.50°, 157.50o
9. (a) – 2
(b) 2 (c) – 2 (d) tan θ (e) tan2 α
3

Project Work

Prepare the chart of the values of trigonometric ratios of the angles from 0° to 360°.

224 Illustrated Optional Mathematics-9

51.51 COMPOUND ANGLES

Learning Objectives

At the end of this topic, the students will be able to:
”” compute trigonometric ratios of compound angles (A ± B).
”” solve the problems related to compound angles.

Z

The sum or difference of two or more than two angles is called a B A–B Y
compounded angle. Let A and B be two angles then A + B or A – B is called A A+B X
the compound angle. O –B W

Trigonometric Ratio for cosine of angles (A + B) and (A – B):

Let OP be a revolving line with a unit distance, starting from Y
OX, describes ∠POQ = A, ∠QOR = B and ∠POS = – B. Then R [cos (A + B), sin (A + B)]
∠POR = A + B and ∠QOS = A – B. So, the coordinates of P, Q, R
and S are (1, 0), (cos A, sin A), (cos(A + B), sin(A + B)) and (cos B, 11A+B Q (cos A, sin A)
– sin B) respectively. Since PR and QS are the opposite sides of the X'
same measurement of angles (A + B) and [A+ (– B)] respectively, B P (1, 0) TRIGONOMETRY
PR = QS. A S (cos B, – sin B)

Now, we have O –B
1

Y'

d = (x2 – x1)2 + (y2 – y1)2
or, PR = (cos (A + B) – 1)2 + (sin (A + B) – 0)2

= (cos2 (A + B) – 2cos (A + B) + 1 + sin2 (A + B)

= 1 – 2cos (A + B) + 1 = 2 – 2cos (A + B)

QS = cos B – cos A)2 + (– sin B – sin A)2

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

= 1 + 1 – 2cos A . cos B + 2sin A . sin B Q
= 2 – 2 (cos A . cos B – sin A . sin B)
Since PR = QS so, 1 MP X
2 – 2 cos (A + B) = 2 – 2 (cos A . cos B – sin A . sin B) A
or, 2 – 2cos (A + B) = 2 – 2(cos A . cos B – sin A . sin B). O
∴ cos(A + B) = cos A.cos B – sin A.sin B.
Taking B = – B, we get cos A = b = OM
cos[A + (– B)] = cos A . cos (– B) – sin A . sin(– B) h 1
∴ cos(A – B) = cos A . cos B + sin A . sin B.
or, OM = cos A ∴ x = cos A
p QM
sin A = h = 1

or, QM = cos A ∴ y = sin A

∴ (x, y) = (cos A, sin A).

Compound Angles 225

Trigonometric Ratios for Sine Angles of (A + B) and (A – B)
We have,
sin (A + B) = cos[90o – (A + B)] = cos[(90o – A) – B]

= cos(90o – A).cos B + sin(90o – A).sin B
= sin A . cos B + cos A . sin B.
∴ sin (A + B) = sin A . cos B + cos A . sin B.
Again, taking B = – B, we get,
sin[A + (– B)] = sin A . cos (– B) + cos A . sin (– B)
∴ sin (A – B) = sin A . cos B – cos A . sin B.

Trigonometric Ratios for Tangent Angles of (A + B) and (A – B)

We have,

tan (A + B) = sin (A + B) = sin A . cos B + cos A . sin B
cos (A + B) cos A . cos B – sin A . sin B

= sin A . cos B + cos A . sin B
cos A . cos B – cos A . cos B
sin A . sin B
cos A . cos B cos A . cos B
cos A . cos B

= tan A +A.tatnanBB. [Diving by cos A.cos B on numerator & denominator.]
1 – tan

∴ tan(A + B) = tan A +A.tatnanBB.
1 – tan
TRIGONOMETRY
Again, taking B = – B, we get

∴ tan(A – B) = tan A – tan B .
1 + tan A. tan B

Trigonometric Ratios for Cotangent Angles of (A + B) and (A – B)

We have, cot (A + B) = tan [90o – (A + B)] = tan [(90o – A) – B]

cot A – 1
+ cot A
= tan (90° – A) – t.atnanBB. = cot B
+ tan (90° – A) 1
1 1 . cot B

cot A . cot B – 1

= cot B = cot A . cot B – 1
cot B + cot A cot B + cot A

cot B

∴ cot(A + B) = cot A . cot B – 1
cot B + cot A

Again, taking B = – B, we get

cot [A + (– B)] = cot A . cot (– B) – 1
cot (– B) + cot A

or, cot (A – B) = – cot A . cot B – 1
– cot B + cot A

∴ cot(A – B) = cot A . cot B + 1
cot B – cot A

226 Illustrated Optional Mathematics-9

Example 1 : Without using calculator or table, find the values of :

(a) sin 15o (b) sin 75o + cos 105o

Solution : (a) We have,

sin 15o = sin (45o – 30o) = sin 45o.cos 30o – cos 45o. sin 30o

= 1 . 3 – 1 . 1 = 3–1
2 2 2 2 22

∴ sin 15o = 3–1
22

(b) We have,

sin 75o + cos 105o = sin (45o + 30o) + cos(60o + 45o)

= sin 45o.cos 30o + cos 45o.sin 30o + cos 60o.cos 45o – sin 60o.sin 45o

= 1 . 3 + 1 . 1 + 1 . 1 – 3 . 1
2 2 2 2 2 2 2 2

= 1+1 = 222 = 1
22 2

∴ sin 75o + cos 105o = 1
2

Example 2 : Prove that: sin 22o – cos 22o = – 2 sin 23o.

Solution : Here, LHS = sin 22o – cos 22o = sin(45o – 23o) – cos(45o – 23o) TRIGONOMETRY

= sin 45o.cos 23o – cos 45o.sin 23o – (cos 45o.cos 23o + sin 45o.sin 23o)

= 1 cos 23o – 1 sin 23o– 1 cos 23o – 1 sin 23o
2 2 2 2

= – 1 +1 sin 23o = – 2 sin 23o = – 2sin 23o = RHS.
2 2

Example 3 : Prove that: sin 35° + cos 35° = – tan 80o.
sin 35° – cos 35°
Solution: Here,

LHS = sin 35° + cos 35° = sin (45° – 10°) + cos (45° – 10°)
sin 35° – cos 35° sin (45° – 10°) – cos(45° – 10°)

= sin 45°.cos 10° – cos 45°.sin 10° + cos 45°.cos 10° + sin 45°.sin 10°
sin 45°.cos 10° – cos 45°.sin 10° – cos 45°.cos 10° – sin 45°.sin 10°

1 cos 10° – 1 sin 10° + 1 cos 10° + 1 sin 10°
2 2 2 2
=
1 1 1 1
2 cos 10° – 2 sin 10° – 2 cos 10° – 2 sin 10°

1 (cos 10° + cos 10°) 2cos 10°
2 2sin 10°
= = – = – cot 10o
1
– 2 (sin 10° + sin 10°)

= – cot(90o – 80o) = – tan 80o = RHS.

Compound Angles 227

Example 4 : Show that: tan 17o + tan 28o + tan 17o.tan 28o = 1.
Solution: We have,
17o + 28o = 45o
Taking tan on both sides, we have
tan(17o + 28o) = tan 45o

or, 1ta–nta1n7°17+°.tatnan2288°° = 1
or, tan 17o + tan 28o = 1 – tan 17°.tan 28o

or, tan 17o + tan 28o + tan 17o.tan 28o = 1.

Example 5 : Find the value of cos 68o.cos 23o + sin 68o.sin 23o + sin πc .
Solution: 4
πc
Here, cos 68o.cos 23o + sin 68o.sin 23o + sin 4
=
cos(68o– 23o) + sin 180°
4
1 1 1 +1 2
= cos 45o + sin 45o = 2 + 2 = 2 = 2 = 2.

Example 6 : Prove that: cot πc – A = cos A + sin A
4 cos A – sin A
πc
πc πc cot 4 . cot A +1
4 4
Solution: LHS = cot – A = cot – A = πc
4
cos A cos A + sin A cot A – cot
sin A
1 . + 1 sin A cos A + sin A
1 cos A – sin A cos A – sin A
TRIGONOMETRY = cos A = = = RHS.
sin A
– sin A

Example 7 : Prove that: tan 32o + tan 51o + tan 97o = tan 32o.tan 51o.tan 97o.

Solution: We have,

32o + 51o + 97o = 180o

or, 32o + 51o = 180o – 97o

Taking tan on both sides, we have

tan(32o + 51o) = tan(180o – 97o)

or, 1ta–nta3n2°3+0°t.atann5511°° = – tan 97o
or, tan 32o + tan 51o = – tan 970 + tan 32o.tan 51o.tan 97o
or, tan 32o + tan 51o + tan 97o = tan 32o.tan 51o.tan 97o.

Example 8 : Prove that: cos(60o – A).cos(30o – B) – sin(60o – A).sin(30o – B) = sin(A + B).
Solution: Here,

LHS = cos(60o – A).cos(30o – B) – sin(60o – A).sin(30o – B)
= cos(60o – A + 30o – B) [ cos A cos B – sin A sin B = cos(A + B)]
= cos[90o – (A + B)]

= sin(A + B) = RHS.

228 Illustrated Optional Mathematics-9

Example 9 : GIfisviennA, s=in54Aa=nd45caonsdBc=os15B3 , find the values of : (a) cos(A + B) (b) tan(A – B).
Solution:
= 5
13
∴ cos A = 1 – sin2 A and sin B = 1 – cos2 B

or, cos A = 1– 4 2 1– 5 2
5 13
and sin B =

or, cos A = 25 – 16 and sin B = 169 – 25
25 169

or, cos A = 9 and sin B = 114649 or, cos A = 3 and sin B = 12
25 5 13

4 12

∴ tan A = sin A = 5 = 4 and tan B = sin B = 13 = 152.
cos A 3 3 cos B 5

5 13

“Alternatively”

sin A = 4 or, p1 = 4 =K cos B = 5 or, b2 = 5 = L
5 h1 5 13 h2 13
∴ p1 = 4K , h1 = 5K ∴ b2 = 5L, h2 = 13L.

We have, We have,

b1 = h12 – p12 p2 = h22 – b22
= 25 k2 – 16 k2 = 169 L2 + 25 L2

= 9 k2 = 3K = 144 L2 = 12L

∴ cos A = b1 = 3k = 53, sin B = p2 = 12L = 1132, tan A = p1 = 34, tan B = p2 = 12 TRIGONOMETRY
h1 5k h2 13L b1 b2 5

Now, (a) cos(A + B) = c1ota+snAtaA.nc–oAst.atBann–BBsin=A1.s34+in–43B1.521=5235=. 5 – 4 . 12 = 15 – 48 =– 33
(b) tan(A – B) = 13 5 13 65 65
16
20 – 36 × 15 = – 63
15 15 + 48

Example 10 : Simplify: sin(A + B) + sin(A – B).

Solution: Here, sin(A + B) + sin(A – B)

= sin A.cos B + cos A.sin B + sin A.cos B – cos A.sin B = 2 sin A.cos B.

Example 11 : Prove that: tan(A + B) + tan(A – B) = sin 2A
cos2 A – sin2 B

Solution: LHS = tan(A + B) + tan(A – B) = sin (A + B) + sin (A – B)
cos (A + B) cos (A – B)

= sin (A + B). cos(A – B) + cos (A + B). sin (A – B)
cos(A + B). cos(A – B)

= sin (A + B + A – B)
(cos A.cos B – sin A.sin B) (cos A.cos B + sin A. sin B)

= cos2 sin 2A = cos2 A(1 – sin 2A cos2 A) sin2 B
A.cos2 B – sin2 A.sin2 B sin2 B) – (1 –

= cos2 A – cos2 sin 2A B + cos2 A.sin2 B = sin 2A = RHS. Proved.
A.sin2 B – sin2 cos2 A – sin2 B

Compound Angles 229

Example 12 : Prove that: sec(A + B) = 1 sec A.sec B .
– tan A.tan B
Solution: Here,

1 . 1
cos
RHS = sec A.sec B = A cos B
– tan A.tan
1 B 1 – sin A . sin B
cos A cos B
1

= cos cos A cos B B = 1 + B) = sec(A + B) = LHS.
A cos B – sinAsin cos(A

cos A.cos B

Example 13 : Prove that: sin (α – b) + sin (b – γ) + sin (γ – α) = 0.
sin α. sin b sin b. sin γ sin γ. sin α

Solution: Here, LHS = sin (α – b) + sin (b – γ) + sin (γ – α)
sin α. sin b sin b. sin γ sin γ. sin α

= sin α.cos b – cos α. sin b + sin b.cos γ – cos b. sin γ + sin γ.cos α – cos γ. sin α
sin α. sin b sin b. sin γ sin γ. sin α

= sin α.cos b – cos α. sin b + sin b.cos γ – cos b. sin γ + sin γ.cos α – cos γ. sin α
sin α. sin b sin α. sin b sin b. sin γ sin b. sin γ sin γ. sin α sin γ. sin α

= cot β – cot α + cot γ – cot β + cot α – cot γ = 0 = RHS.

Example 14 : Prove that: tan 50o = tan 40o + 2tan 10o.

Solution: We have, 50o – 40o = 10o

∴ tan(50o – 40o) = tan 10o [ Taking tan on both sides]

TRIGONOMETRY or, tan 50° – tan 40° = tan 10o or, 1 + tan 50° – tan 40° 50°) = tan 10o
1 + tan 50°.tan 40° tan 50°.tan (90° –

or, 1ta+nta5n0°5–0°t.acnot4500°° = tan 10o or, tan 501° – tan 40° = tan 10o
+ 1

or, tan 50o – tan 40o = 2tan 10o or, tan 50o = tan 40o + 2tan 10o.

Example 15 : Prove that: sin(x + 1)x.cos(x – 1)x – cos(x + 1)x.sin(x – 1)x = sin 2x.

Solution: Here,

LHS = sin(x + 1)x. cos(x – 1)x – cos(x + 1)x . sin(x – 1)x

= sin[(x + 1)x – (x – 1)x] = sin(x2 + x – x2 + x) = sin 2x = RHS.

Example 16 : If A + B = 45o, then prove that: (1 + tan A) (1 + tan B) = 2.

Solution: Here, A + B = 45o

Taking tan on both sides, we have

tan(A + B) = tan 45o

or, tan A + tan B = 1
1 – tanA.tan B

or, tan A + tan B = 1 – tan A.tan B

or, tan A + tan B + tan A.tan B = 1

or, 1 + tan A + tan B + tan A.tan B = 1 + 1 [ Adding 1 on both side.]

or, 1(1 + tan A) + tan B(1 + tan A) = 2

or, (1 + tan A) (1 + tan B) = 2.

230 Illustrated Optional Mathematics-9

Example 17 : If cot A = 1 and cos B = x 1, then verify that A and B are complementary angles,
x x2 +
πc
Solution: i.e., A + B = 2 . , cos B = x 1 = b
Given, cot A x2 + h
= 1
x

∴ p = h2 – b2 = x2 + 1 – x2 = 1.

∴ cot B = b = x = x.
Now, p 1
∴ 1
cot A cot B – 1 x . x–1 1–1 0 πc
cot A + cot B +x + 2
cot(A + B) = = 1 = 1 + x = 1 x = 0 = cot
x x
A + B = πc . x
2
Hence, A and B are complementary angles.

Example 18 : Prove that: tan 8A – tan 5A – tan 3A = tan 8A.tan 5A.tan 3A.

Solution: Here, 8A = 5A + 3A

∴ tan 8A = tan(5A + 3A)

or, tan 8A = tan 5A + tan 3A
1 – tan 5A tan 3A

or, tan 8A – tan 8A.tan 5A.tan 3A = tan 5A + tan 3A

or, tan 8A – tan 5A – tan 3A = tan 8A.tan 5A.tan 3A.

Example 19 : Prove that : tan 1 tan A – cot 1 cot A = cot 2A.
3A – 3A –

Solution: Here, LHS = 1 – cot 1 cot A TRIGONOMETRY
tan 3A – tan A 3A –

= 1 1 1 – 1
cot 3A cot A cot 3A – cot A
–

= 1 – 1 = cot 3A. cot A – 1
cot A – cot 3A cot3 A – cot A cot A – cot 3A – (cot A – cot 3A)

cot 3A.cot A

= cot 3A.cot A + 1 = cot(3A – A) = cot 2A = RHS.
cot A – cot 3A

Example 20 : If tan α = k.tan b, verify that: (k + 1).sin(α – b) = (k – 1).sin(α + b).

Solution: Given, tan α = k. tan β.
To Prove: (k + 1).sin(α – β) = (k – 1).sin(α + β)

sin (α – b) k – 1
or, sin (α + b) = k + 1

sin (α – b) sin α.cos b – cos α. sin b
Now, LHS = sin (α + b) = sin α. cos b + cos α. sin b

sin α . cos b cos α . sin b
cos α . cos b – cos α . cos b tan α – tan b
= sin α . cos b cos α . sin b = tan α + tan b

cos α . cos b + cos α . cos b

k.tan b – tan b tan b (k – 1) k – 1 = RHS.
= k.tan b + tan b = tan b (k + 1)= k + 1

Compound Angles 231

“Next Method”
tan α k

Given, tan α = k. tan b or, tan b = 1

Using componendo and dividendo rules, we have

sin α sin b
tan α – tan b cos α –
or, tan α + tan b = k – 1 or, sin α + cos b = k – 1
k + 1 cos α sin b k + 1
cos b

sin α.cos b – cos α.sin b

cos α. cos b b = k – 1 or, sin (α – b) = k – 1
or, sin α.cos b + cos α.sin k + 1 sin (α + b) k + 1

cos α.cos b

or, (k + 1) sin(α – β) = (k – 1) sin(α + β). 1
2
Example 21 : If cos θ + sin φ = m and sin θ + cos φ = n then prove that: sin(θ + φ) = (m2 + n2 – 2).

Solution: Here, cos θ + sin φ = m and sin θ + cosφ = n

Now, RHS = 1 (m2 + n2 – 2) = 1 [(cos θ + sin φ)2 + (sin θ + cos φ)2 – 2]
2 2
1
= 2 [cos2 θ + 2cos θ.sin φ + sin2 φ + sin2 θ + 2sin θ.cos φ + cos2 φ – 2]

RHS. = 12 [1 + 1 + 2(sin θ.cos φ + cos θ.sin φ) – 2] = 1 [2 + 2sin(θ + φ) – 2] = sin(θ + φ) =
2

“Alternatively”

TRIGONOMETRY Here, cos θ + sin φ = m and sin θ + cos φ = n

Squaring adding both conditions, we get

cos2 θ + 2cos θ.sin φ + sin2 φ + sin2 θ + 2sin θ.cos φ + cos2 φ = m2 + n2

or, 1 + 1 + 2(sin θ.cos φ + cos θ.sin φ) = m2 + n2

or, 2 + 2sin(θ + φ) = m2 + n2

or, sin(θ + φ) = 1 (m2 + n2 – 2).
2

EXERCISE - 5.5

1. (a) What is compound angle?

(b) Which trigonometric expression is equal to cos (A + B)?

(c) What is the trigonometric ratio of tangent of (A – B)?

2. Find the value of trigonometric ratios without using table and calculator. 7πc
12
(a) cos 15o (b) sin 75o (c) cos 75o (d) sin (e) tan 15o
(f) tan 75o (h) tan 105o (j) sin 255o
(g) cot 5πc (i) cos 195o
12

3. Without using trigonometric table and calculator, calculate the value of the following trigonometric

expressions.

(a) sin 15o + cos 105o (b) sin 75o + sin 105o (c) cos 15o – sin 75o

(d) sin 15o.cos 105o (e) cos 195o ÷ sin 15o (f) cot 165o – tan 105o

(g) sin2 15o – sin2 45o (h) cos2105o + tan2 45o (i) tan2 15o . tan2 75o

232 Illustrated Optional Mathematics-9

4. Prove that: (b) cos 25o – sin 25o = 2sin 20o
(a) sin 28o – cos 28o = – 2sin 17o

(c) sin 65o – sin 25o = 2sin 20o (d) 3cos 20o + sin 20o = 2sin 80o

(e) ccooss 10° – sin 10° = tan 35o (f) ssiinn 2200°° + cos 20° = cot 155o
10° + sin 10° – cos 20°

5. Evaluate the values of the following expressions without using calculator or table:

(a) sin 19o40′. cos 25o20′ + cos 19o40′.sin 25o20′ (b) cos 44o.cos16o – sin 44o.sin 16o – sin 30o

(c) sin(180o – A).cos(90o – A) – cos(180o – A).sin(90o – A)

(d) tan 88° + tan 47° (e) cot 20°.cot 65° + 1 – 1
1 – tan 88°.tan 47° cot 20° – cot 65°

6. Prove the following:

(a) sin 75° – cos 75° = 1 (b) sin (45° + A) = 1 (c) cot (45° – A) = 1
cos 15° – cos 75° cos (45° – A) tan (45° + A)

7. Verify that:
(a) sin(60o – x).cos(30o + x) + cos(60o – x).sin(30o + x) = 1

(b) cos(60o – A).cos(30o – B) – sin(60o – A).sin(30o – B) = sin(A + B)

(c) sin(45o + A).cos(45o – B) + cos(45o + A).sin(45o – B) = cos(A – B)

(d) sin(2n + 1)x.cos(2n – 1)x – cos(2n + 1)x.sin(2n – 1)x = sin 2x

(e) cos(120o + A) + cos A + cos(120o – A) = 0

8. Show that:

(a) tan 32o + tan 13o + tan 32o.tan 13o = 1 (b) 1 + cot 22o + cot 23o = cot 22o.cot 23o

(c) cot 29o – cot 74o – cot 29o.cot 74o = 1 (d) tan 35o + tan 65o + tan 80o = tan 35o.tan 65o.tan 80o TRIGONOMETRY

(e) cot 110o + cot 85o + cot 75o = cot 110o.cot 85o.cot 75o

(f) tan 15o.tan 25o + tan 25o.tan 50o + tan 50o.tan 15o = 1

(g) tan 7A – tan 5A – tan 2A = tan 7A.tan 5A.tan 2A

(h) cot 8A.cot 4A – cot 12A.cot 4A – cot 12A.cot 8A = 1

9. Prove that:

(a) 1ta+nta5nA5–At.atann33AA = tan 2A (b) tan nx + tan x = tan(n + 1)x
1 – tan n x.tanx

(c) cot2 A.cot2 B – 1 = cot (A + B) (d) tan(A + B).tan(A – B) = tan2 A – tan2B
cot2 B – cot2 A tan(A – B) 1 – tan2 A.tan2 B

(e) cot (A – B).cot B – 1 = cot A (f) tan (A + B + C) – tan (A – B + C) = tan 2B
cot (A – B) + cot B 1+ tan(A + B + C).tan(A – B + C)

10. Prove that:

(a) tan(45o + A) = 1 + tan A = cos A + sin AA (b) cot(45o – A) = cot A + 1 = cos A + sin A
1 – tan A cos A – sin cot A – 1 cos A – sin A

11. Prove that:

(a) sin(α + b) + sin(α – b) = 2sin α cos b

(b) cos(A + B).cos(A – B) = cos2 B – sin2 A = cos2 A – sin2 B

(c) tan(A + B) + tan(A – B) = sin 2A A (d) tan(A + B).tan(A – B) = sin2 A – sin2­ B
cos2 B – sin2 cos2 A – sin2 B

Compound Angles 233

12. Prove that:

(a) 2cos(A – 45o) = cos A + sin A (b) 2cos(45o + A).cos(45o – A) = 2cos2 A – 1 = 1 – 2sin2 A

(c) scions22AA..ccoossAA+– sin 2A.sin A = cot 3A (d) sin2(30o + θ) – sin2(30o – θ) = 3 sin 2θ
cos 2A.sin A 2

13. Prove that:

(a) ssiinn (A – B) = cot B – cot A (b) sin (A + B).sin (A – B) = tan2 A – tan2 B
A .sin B cos2 A.cos2 B

(c) sin (α + b).sec α.sec b = tan α + tan b (d) 1 + tan 2A.tan A = sec 2A

sec α + sec b (f) tan A + cot 2A = cosec 2A
(e) sec(α + b ) = 1 – tan α.tan b

(g) csoins((AA + B) + sin(A – B) = tan A (h) tan(A + B) = sin sin2 A – sin2 B B
+ B) + cos(A – B) A.cos A – sin B.cos

(i) ccoosesc58AA – sin 5A = sin 3A (h) sin (α + β) sec α.sec β = tan α + tan β
sec 8A

14. Prove that:

(a) csoisn(BB.c–osCC) + sin(C – A) + sin(A – B) =0
cos C.cos A cos A.cos B

(b) sin A.sin(B – C) + sin B.sin(C – A) + sin C.sin(A – B) = 0

15. Show that:

(a) tan 65o – tan 25o = 2tan 40o (b) tan 80o = 2tan 70o + tan 10o

(c) cot 40o.cot 25o – cot 65o.cot 40o = 2 (d) tan 215o = 2tan 160o + tan 55o

TRIGONOMETRY 16. If sin A = 4 and sin B = 5 , find the values of:
5 13

(a) sin(A + B) (b) cos(A – B) (c) tan(A + B)
(c) cot(α – b )
17. If sin α = 1 and cos b = 25, find the values of:
10

(a) sin(α – b ) (b) cos(α + b )

18. If tan α = 3 and tan b = 5 , prove that:
4 12
(a) tan(α + b ) = 12333 (b) sin(α – b ) = 1665 (c) cot(α – b ) = 31165
4 8
19. If cos A = – 5 and cot B = 12 , where A lies in second quadrant and B in third quadrant, find values of:

(a) sin(A + B) (b) cos(A – B) (c) tan(A + B)

20. (a) If tan α = 2 and tan b = 1 , show that: α + b = πc .
3 5 4
(b) If tan α = x + 1 and tan b = x – 1, then verify that: 2cot(α – b ) = x2

(c) If tan α = m and tan b = 1 , show that: α + β = 90o.
m

21. (a) If tan(x + y) = 23 and tan y = 7, what is the value of tan x?

(b) If cos (A – B) = 84 and sec A = 17 , find the value of sin B.
(c) Evaluate 85 of tan A 8 = 2m +1 and A + B = π4c.

the value if cot B

22. If A + B = 45o then prove that:

(a) 1 – tan A.tan B = tan A + tan B (b) (1 + tan A) (1 + tan B) = 2

(c) (cot A – 1) (cot B – 1) = 2 (d) (cos A – sin A) (cos B – sin B) = 2sin A . sin B

234 Illustrated Optional Mathematics-9

23. If α – β = πc then prove that:
4

(a) (1 + tan α) (1 – tan b) = 2 (b) (1 + cot α) (1 + cot b) = 2cot b

24. If A + B = 225o, Prove that:

(a) (1 + tan A) (1 + tan B) = 2 (b) (1 – cot A) (1 – cot B) = 2

25. If A + B + C = πc then prove that the following relations:

(a) tan A + tan B + tan C = tan A.tan B.tan C

(b) tan A2 .tan B + tan B2 .tan C + tan C2 .tan A = 1
2 2 2

(c) cot 2A.cot 2B + cot 2B.cot 2C + cot 2C.cot 2A = 1

26. (a) If α + b + γ = πc and cos α = cos b .cos γ then show that : tan α = tan b + tan γ.

(b) If A + B + C = πc and sin A = sin B.sin C then show that : cot B + cot C = 1.

(c) If A + B + C = πc , prove that : tan A + cos A.sec B.sec C = tan B + cos B.sec C.sec A
2

= tan C + cos C.sec A.sec B.

27. (a) If tan A : tan B = k :1, prove that : sin(A + B) : sin(A – B) = (k + 1) : (k – 1).

(b) If an angle θ is divided into two parts α and b such that tan α : tan β :: x : y, verify that :
sin θ : sin(α – b) = (x + y) : (x – y).

28. Prove that:

(a) sin(A + B + C) = cos A.cos B.cos C(tan A + tan B + tan C – tan A.tan B.tan C)

(b) cos(A + B + C) = cos A.cos B.cos C(1 – tan B.tan C – tan C.tan A – tan A.tan B)

(c) tan(A + B + C) = tan A + tan B + tan C – tan A.tan B.tan C TRIGONOMETRY
1 – tan B.tan C – tan C.tan A – tan A.tan B

(d) cot(A + B + C) = cot A.cot B.cot C – cot A – cot B – cot –C1.
cot B.cot C + cot C.cot A + cot A.cot B

ANSWERS

2. (a) 3 + 1 (b) 3 + 1 (c) 3 – 1 (d) 3 + 1 (e) 2 – 3
22 22 22 22

(f) 2 + 3 (g) 2 – 3 (h) – (2 + 3) (i) – 1 + 3 (j) – 1 + 3
22 22
3. (a) 0 (b) 1 + 3 (c) 0 (d) 14 ( 3 – 2) (e) – ( 3 + 2)
22

(f) 0 (g) – 3 (h) 6– 3 (i) 1
1 4 4 (d) – 1 (e) 0

5. (a) 2 (b) 0 (c) 1

16. (a) 6635 (b) 6556 (c) 6136 17. (a) – 5 1 2 (b) 1 (c) – 7
2

19. (a) 36 (b) – 1835 (c) 7376 21. (a) 881 (b) 4 , 11346454 (c) m 1
85 5 m+

Project Work

List out the formulae of the trigonometric ratios of compound angles.

Compound Angles 235

VECTOR 6UNIT

Estimated Teaching Periods : 12

Competency

”” To demonstrate and use of the concept of vector and operations of vectors.

Learning Outcomes
At the end of this unit, the students will be able to:

”” introduce a vector and differentiate the vector and scalar quantities with examples.
”” find the magnitude and direction of the vector.
”” identify and separate the types of vector.
”” solve the problems by using operations of vectors (addition and subtraction).
”” find the condition of parallelism of two vectors under scalar multiplication of vector.

What do you learn?

Introduction to Vector

� Directed line segment � Displacement and vector representation

� Components of vector � Vector in coordinate plane

� Magnitude and direction of vector � Types of vectors

Operations on Vectors � Subtraction of vectors
� Addition of vectors � Condition for parallel vectors
� Scalar multiplication of vectors

Are you ready?

VECTOR � Line and line segment � Ray and directed line segment

� Initial and terminal points � Plotting ordered pairs in coordinate plane

� Components of coordinate axes � Horizontal distance, vertical distance and oblique distance

� Pythagoras theorem � Tangent ratio of an angle

� Positive and negative angles � Square root of number

� Length of line segment � Angle made by line with axes

� Triangle, parallelogram and polygon

� Add, subtract and multiply the numbers or expressions

� Complementary and supplementary angles

Specification Grid

Unit Chapter Cognitive Knowl- Under- Applica- Higher Total Total
Domain edge standing tion Ability Ques- Marks
tions
Topic 1 mark 2 marks 4 marks 5 marks

6 Vector From All Topics 1 2 - 1 4 10

236 Illustrated Optional Mathematics-9

61.1 INTRODUCTION TO VECTOR

6.1 (A) Basic on Vector

Learning Objectives

At the end of this topic, the students will be able to:

• identify a vector and its components. • draw the vector in coordinate plane.

The word "Vector" originates from the Latin word "Vehere" or "Vectus" , which B
means "to carry". It is physical quantity which has the magnitude (or length) and a
direction. Vector is frequently represented by a line segment with a finite direction
or graphically as an arrow, connecting an initial point A with a terminal point B and A

is denoted by AB as shown in the figure. It is also denoted by a . It is commonly used in mathematics,
physics and engineering fields, etc.

Vector and Scalar D N
C↑
An athlete Mr. Manandar is running around the football ground ABCD in a straight
path. He runs at first from the position A to B, i.e., in the direction from west to east.
After then he turns to North and moves toward it from B to C. Similarly, he runs
from C to D in west and D to A in south.

What did we find from it? We are clear that Mr. Manandhar covered the length of A

to B, B to C, C to D and D to A and also turned related directions. He covered the B
total distance AB + BC + CD + DA. The covering distance by him is called scalar A

quantity. But the covering distance by him is AB or BC or CD or DA in the direction of east, it is called

vector quantity. The scalar quantity has only magnitude, but the vector quantity has both magnitude and

direction.

A quantity which has only magnitude is called scalar quantity. For example; length, area, volume, mass, VECTOR
time, temperature, speed, energy, power, etc. are scalar quantities. A quantity which has both magnitude
and direction is called vector quantity. For example; velocity, acceleration, gravity, force, pressure,
displacement, weight, work, etc. are vector quantities.

Directed Line Segment l
Let us consider a line segment AB taken from a line l with arrow head from A to B.

Such a line segment is called directed line segment. The directed line segment AB is A B

denoted by AB. Here, the initial point A is called initial point and the final point B is A

called terminal point of AB. B

A line segment having initial and terminal points with specific direction is called the directed line segment.

Displacement and Vector Representation P'

A moving point P moves in a certain distance towards specific direction to P
P' as shown in the adjoining figure. Such a movement of the point is called a
displacement.

Introduction to Vector 237

In the square grid, any point P moves 4 units right and 3 units Up (+ve)

up. It moves from O to P, where O is the initial point and P Left P
(-ve) O
is terminal point as shown in the adjoining figure. Then, the Initial point Down (-ve)

point P moves from O to P. This is the displacement of the

point P. It is denoted by OP = 4 in the form of column. Right
3 (+ve)
Each displacement OP is called a vector. For example, when
20 sec
an aeroplane is flying by making the angle of 30o with the 1.5 km

horizontal ground at 540 km/hour in 20 seconds, it travels 3

km and reaches 1.5 km above the ground. Why?

This is also a displacement of the aeroplane. The distance travelled by the
aeroplane in the air is a vector.

A displacement of a point in the specific direction is called a vector. It is 3 km
denoted by AB, CD, YZ , a , b , z , etc. 30o

Components of Vector

Look at the adjoining figure in which the point A moves 4 units right to the B
a
point P and then from the point P moves 3 units up to B. Finally, the point A AP 3 units Up
4 4 units Right
is displaced to B as the vector a . This is symbolized as a = 3 . Here, 4 is the

horizontal movement (displacement) and it is also called x-component of the

vector a and 3 is the vertical movement (displacement) and it is also called

y-component of the vector a .

AB = a = 4 x-component
3 y-component

The horizontal displacement (right or left) of any point along x-asix is called x-component and the

vertical displacement (up or down) of any point along y-axis is called y-component of the vector. Simply,
x
VECTOR the x-component and y-component are denoted by x and y respectively. It is generally denoted by y .

Note: For a Vector x ,
y

i. When x has positive value, it represents forward movement and when x is negative, it represents
backward movement. But when x is zero, it does not move forward and backward.

ii. When y has positive value, it represents upward movement and when y is negative, it represents
downward movement. But when y is zero, it does not move upward and downward.

iii. When x and y are both zeros, it is invariant.

Vector in Coordinate Plane

We discuss the position of a vector in the coordinate plane in the reference of the origin in this section.
There raised two cases if the initial point is at the origin or other than origin. They are as follows:

238 Illustrated Optional Mathematics-9

a. When the initial point of a vector is at origin Y

In the adjoining figure, P(x, y) is any point in the coordinate plane. The P(x, y)
MX
position of the point P(x, y) is OP in which the origin O is as the initial B(x2, y2)

point. Draw PM⊥OX. Then OM = x and PM = y. It is clear from the

figure that the horizontal displacement OM with OM = x and the vertical X' O(0, 0)
Y'
displacement PM with PM = y, represent the displacement OP. So, the
Y
vector OP can be defined by an ordered pair (x, y). Therefore, the vector

OP can also be written as x .
y

b. When the initial point of a vector is at other than origin

)

In the adjoining figure, A(x1, y1) and B(x2, y2) are the initial and y1
terminal points of the vector AB respectively. Draw AL⊥OX, BM⊥OX
,

A(x
1

and AN⊥BM. Then OL = x1, AL = y1, OM = x2, BM = y2 . X' O L N
∴ AN = LM = OM – OL = x2 –x1 and MX

BN = BM – MN = BM – AL = y2 – y1. Y'

It is clear from the figure that the horizontal displacement ANwith AN = x2 – x1 and the vertical

displacement NB with NB = y2 – y1, represent the displacement AB. So, the vector AB can be defined
x2 – x1
by an ordered pair (x2 – x1, y2 – y1). Therefore, the vector AB can be written as y2 – y1 .

Example 1 : Represent the following vectors by directed line segment in a graph.

(a) PQ = –24 (b) ST = –4
0

Solution: (a) Given, PQ = 2 P
–4
2 VECTOR
Here, the horizontal displacement of PQ is 2 units forward and the –4
vertical displacement is 4 units downward.

So, the vector PQ is shown in the adjoining figure. Q

(b) Given, ST = –4
0
Here, the x-component is 4 units backward and the y-component is T – 4 S

0

unchanged.

So, the vector ST is shown in the adjoining figure.

Example 2 : Draw the vector which describes 3 units right and 5 units up. C

Solution: Taking a point A which moves 3 units right to B and 5 units up to C. 3

3 5 5
5 B
Then the point A is finally mapped to C. So, the required vector is AC = .

A3

Introduction to Vector 239

Example 3 : Find the components of the vector represented by the following graph and also its

column vector.

(a) (b) A

B

B
A

Solution: (a) Here, the vector AB moves 6 units right and 2 units up.

So, the required x-component = 6 and y-component = 2.

∴ The column vector of AB, AB = 6 .
2
(b) Here, the vector AB moves 3 units left and 3 units down.

So, the required x-component = – 3 and y-component = – 3

∴ The column vector of AB, AB = – 3 .
– 3

Example 4 : Represent the following vectors by the directed line segments.

(a) A man runs 10 km due east-south.

(b) A bird is flying 550 m/hr due north-west.
(c) A cyclist is cycling 50 km/hr the direction of 75o.

Solution: Here, N E
W O 45o
(a) If a man runs 10 km due east-south then he makes the
angle of 45o with east direction towards south. It is
represented in the directed line segment as shown in the
alongside figure:

(b) If a bird is flying 550 m/hr due north-west, then, it
makes the angle of 45o with the north direction toward
west. It is represented in the directed line segment as
shown in the adjoining figure:
10 km 550 m/hr S E
VECTOR BN

45o
WO

S

(c) If a cyclist is cycling 50 km/hr in a direction of 75o, then Y
its travelling way makes 75o with the horizontal axis.
X' O 75o
EXERCISE - 6.1 (A) X

Y'

1. (a) Define vector and scalar.
(b) What are x-component and y-component of a vector?
(c) What is the vector notation having initial point as origin and terminal point (a, b)?
(d) What is the vector having initial and terminal points as (x1, y1) and (x2, y2) respectively?
2. Represent the following vectors by the directed line segments:

(a) An OX ran 2 km in the north.
(b) An aeroplane is flying 250 km/hr due north-west.
(c) A car is travelling 2 km/min in the direction of 50o.
(d) A girl is walking 4 km in the direction of 30o E.
(e) An eagle is flying 200 m above on the sky being 45o.

240 Illustrated Optional Mathematics-9

3. Draw the vector AB which describes the following components:

(a) 5 units right and 4 units down (b) 3 units left and 2 units up.

(c) 2 units left and 5 units down. (d) x-component = 2 and y-component = – 3

(e) x-component = – 5 and y-component = 4 (f) x-component = – 1 and y-component = 0

4. Find "the components" of the vectors represented by the following graph and also find their column

vectors.

(a) (b) (c)

YC
B

A X (f) D
P
(d) (e)

EN

FM
Q

5. Draw the following vectors on the square grid:

(a) a= 2 (b) b= 6 (c) u= –2
–1 8 –4

(d) v= 0 (e) k= 4 (f) i = –2
–2 0 0

6. Find the vectors represented by the directed line segment joining the given each point with the
origin in column form. Also, represent the vectors in separate square grid.

(a) (0, 3) (b) (– 6, 0) (c) (– 12, 5) (d) (– 15, – 8) (e) (6, – 8) (f) (–7, 4)

7. Find the vectors represented by the directed line segment joining the following points in column
form. Also, show the vector in separate square grid.

(a) A(1, – 2) and B(– 4, 0) (b) C(0, – 2) and D(4, 1) (c) M(0, 0) and N(2, – 1)
(d) P(1, 2) and Q(0, – 1) (e) L(– 2, – 4) and M(2, 4) (f) S(2, – 4) and T(– 4, 5)

8. (a) If the vector AB displaces A(a, – 2) to B(2, 1) and its x-component is 4, find the value of a. VECTOR
(b) If the vector LM displaces L(2, 3) to M(3, – k) and its y-component is – 2, find the value of k.

ANSWERS

1, 2, 3 and 5 show to your teacher

4. (a) 5, 2, 5 (b) 0, 3, 0 (c) 7, – 3, 7 (d) – 5, – 3, – 5
2 3 –3 –3

(e) – 3, 3, – 3 (f) – 2, – 4, – 2 6. (a) 0 (b) – 6
3 –4 3 0

(c) – 12 (d) – 15 (e) 6 (f) – 7
5 –8 –8 –4

7. (a) – 5 (b) 4 (c) 2 (d) –1
2 3 –1 –3

(e) 4 (f) –6 8. (a) – 2 (b) – 1
8 9

Introduction to Vector 241

6.1 (B) Magnitude and Direction of Vector

Learning Objectives
At the end of this topic, the students will be able to:
• find the magnitude and direction of a vector.

The magnitude of a vector is the length of the vector, which is always positive quantity. So, the magnitude
of the vector is also known as the modulus of the vector or the absolute value of the vector. The magnitude

of the vector AB or a is denoted by |AB| or | a |.

The direction of the vector means the angle made by the vector with the x-axis, i.e., the slope of the vector.
The angle made by the vector with x-axis is usually denoted by θ, which is the direction of the vector. In
this section, there are also two cases according as the initial point while defining magnitude and direction
of a vector. They are follows:

a. When the initial point of a vector is at the origin

Let P(x, y) be a point, then OP = x which makes an angle θ.
y

Draw PM ⊥ OX then OM = x and PM = y.

Now, in the right-angled triangle POM, we have Y P(x, y)

OP2 = OM2 + PM2 = x2 + y2 [ Pythagoras theorem] y
MX
∴ OP = x2 + y2 θ

Again, tan θ = y - component = y [ tan θ = bb] X' O x
x - component x Y'
y
∴ θ = tan-1 x

Hence, the magnitude of OP is x2 + y2 i.e., |OP| = x2 + y2 and its direction is tan-1 y ,
x
y
i.e., θ = tan-1 x .

VECTOR b. When the initial point of a vector is other than origin

Let A(x1, y1) and B(x2, y2) be the initial and terminal points of a vector AB respectively,

then AB = x2 – y1 . Y B(x2, y2)
y2 – y1
Draw AL⊥OX, BM⊥OX and AN⊥BM, then
y2 – y1
OL = x1, AL = y1, OM = x2, BM = y2 )
∴ AN = LM = OM – OL = x2 – x1 and
BN = BM – MN = BM – AL = y2 – y1. y
Now, in the right-angled triangle ANB, we have
1

,

A(x θ
1 x2 – x1 N

X' θ MX
CO L

AB2 = AN2 + BN2 [ Pythagoras theorem] Y'

= (x2 – x1)2 + (y2 – y1)2

∴ AB = (x2 – x1)2 + (y2 – y1)2.

242 Illustrated Optional Mathematics-9

Again,

tan θ = y - component = y2 – y1 [∴ tan θ =pb]
x - component x2 – x1

∴ θ = tan-1 y2 – y1 .
x2 – x1

Hence, the magnitude of AB is (x2 – x1)2 + (y2 – y1)2.

i.e., |AB| = (x2 – x1)2 + (y2 – y1)2 and its direction is tan-1 y2 – y1 .
x2 – x1
y2 – y1
i.e., θ = tan-1 x2 – x1 .

Note: Remember the CAST rule for only tangent ratio as;

i. If tan θ = +0ve, then θ = 0.

ii. If tan θ = +0ve, then θ = 90o.

iii. If tan θ = –0ve, then θ = 180o. 2nd Quad. 90o 1st Quad.
0o, 360o
iv. If tan θ = –0ve, then θ = 270o. 180o – θ
-θ θ 90o 4th Quad.
v. If tan θ = + vvee, then 0o < θ < 90o. 180o +θ -θ
+
180o + θ 360o - θ
vi. If tan θ = – vvee, then 90o < θ < 180o.
+ VECTOR270o

vii. If tan θ = – vvee, then 180o < θ < 270o. 3rd Quad.
–

viii. If tan θ = +ve , then 270o < θ < 360o.
-ve

Example 1 : Find the magnitude and direction of the vector whose initial and terminal points are

(2, – 1) and (– 2, 3). Y
Solution: Here, A(2, –1) = (x1, y1) and B(– 2, 3) = (x2, y2) B(–2, 3)

Now, we have

The column vector of AB, AB = x - component X' X
y - component
A(2, – 1)
= x2 – x1 = –2–2 = –4 . Y'
y2 – y1 3 – (– 1) 4

∴ The magnitude of AB, |AB| = x2 + y2

= (– 4)2 + (4)2

= 16 + 16 = 32 = 4 2 units.

Introduction to Vector 243

Again, tan θ = y = 4 = – 1 The value of θ in tangent ratio for 1 is
x –4 45o. i.e., tan 45o = 1. x is -ve and y is +ve.
So, it lies in the 2nd quadrant.
∴ θ = tan-1 (–1) = 180o – 45o = 135o.

Hence, the direction of AB is 135o.

''Alternatively''
Now, we have

∴ The magnitude of AB, |AB| = (x2 – x1)2 + (y2 – y1)2

= (–2 – 2)2 + {3 – (– 1)}2

= (– 4)2 + 42 = 16 + 16

= 32 = 16 × 2 = 4 2 units

The direction of AB, θ = tan– 1 y2 – y1 .
x2 – x1

= tan– 1 3 – (– 1) .
–2–2

= tan– 1 4 = tan– 1 (– 1) = 135o.
–4

Example 2 : If AB displaces A (1, – 2) to B(2, 1) and XY displaces X(3, 1) to Y(4, – 2) then prove

that AB = – XY.

Solution: Here,

AB displaces A(1, – 2) to B(2, 1) and XY displaces X(4, 1) to Y(3, – 2)

Now, AB = xy22 –– yx11 = 2–1 = 1 and
1 – (– 2) 3

XY = x2 – x1 = 3–4 = –1 =– 1
y2 – y1 – 2–1 –3 3

VECTOR Hence, AB = – XY. Proved.

Example 3 : If the magnitude of PQ 3 is 6 units, find the value of y and the direction of PQ .
y
Solution: Here, |PQ| = 6 units Again,

or, x2 + y2 = 6 If PQ = 3 3 then the direction of PQ,
3

or, 32 + y2 = 6 θ = tan– 1 3 3 = tan– 1 ( 3 ) = 60o
3
Squaring on both sides, we get

9 + y2 = 36 If PQ = 3 then the direction of PQ,
–3 3

or, y2 = 36 – 9 –3 3
∴ θ = tan– 1 3
or, y2 = 27

∴ y=±3 3 = tan-1 (– 3 ) = 360o – 60o = 300o.

Hence, the value of y is 3 3 or – 3 3.

244 Illustrated Optional Mathematics-9

Example 4 : The direction of MN is tan-1 4 . If M (m, 1) and N(4, 5) are two points, find the
3

magnitude of MN .

Solution: Here, M(m, 1) and N(4, 5) are two points then the direction of MN,

θ = tan–1 y2 – y1 = tan– 1 5–1 = tan–1 4 4
x2 – x1 4–m –m

By question, the direction of MN, θ = tan–1 4
3
4 4
∴ 4 –m = 3

or, 12 = 16 – 4m

or, 4m = 16 – 12

or, 4m = 4
or, m = 44
or, m = 1.

∴ The magnitude of MN,

|MN| = (x2 – x1)2 + (y2 – y1)2
= (4 – 1)2 + (5 – 1)2

= 32 + 42
= 25 = 5 units.

EXERCISE - 6.1 (B)

1. (a) Write the magnitude and direction of the vector x .
y
VECTOR
(b) What are the magnitude and direction of the vector having initial point (x1, y1) and terminal
vector (x2, y2)?

2. Find the magnitude and direction of the following vectors:

(a) AB = 2 (b) p = –1 (c) k= 2
4 9 0

3. Find the magnitude and direction of the following displacements.

(a) from L(2, 1) to M(– 4, 3) (b) from P(3, 4) to Q(– 2, 0)

(c) from P(0, – 2) to Q(4, 1) (d) from S(4, – 2) to T(3, 0)

4. If the vector XY displaces a point X(2, – 3) to Y(– 2, 4) then,

(a) Express XYin the column vector.

(b) Calculate the length of XY.

(c) Find the direction of XY.

Introduction to Vector 245

5. (a) If a vector PQ displaces a point P(3, – 1) to the point Q(7, 8) and another vector XYdisplaces
a point X(5, – 5) to the point Y(9, 4), then prove that PQ = XY.

(b) If a vector AB displaces a point A(3, – 1) to the point B(7, 8) and another vector MN displaces
a point M(5, – 5) to the point N(9, 4), then prove that AB = – MN.

6. (a) If the magnitude and direction of a vector are 2 3 units and 60o respectively, find the vector.

(b) If the magnitude and direction of a vector a are 2 units and 45o respectively, find the
vector. b

7. (a) If the magnitude of the vector AB = x is 13 units, find the value of x and the direction of
–2

AB.

(b) If the magnitude of the vector PQ having the initial point P(–1, – 2) and the terminal point

(– 3, y) is 2 units, find the value of y and the direction of PQ.

8. (a) The direction of the vector CD = x is 30o. Find the magnitude of CD.
(b) 3
– 10
If the direction of the vector LM having the points L(2,– 3) and M(1, y) is tan-1 3 , find the

value of y and the magnitude of LM.

ANSWERS

2. (a) 2 5 units, 63.43° (b) 82 units, 96.34° (c) 2 units, 0°
(c) 5 units, 36.87°
3. (a) 2 10 units, 161.57° (b) 41 units, 218.66° (c) 119.75° (d) 5 units, 116.57°

4. (a) – 4 (b) 65 units
7

6. (a) 3 (b) 1
3 1

VECTOR 7. (a) ± 3, 213.69°, 326.31° (b) – 2, 0°

8. (a) 6 units (b) 1, 109 units
3

Project Work

Derive the column vector, magnitude and direction of a directed line-segment joining the points (x1, y1)
and (x2, y2).

246 Illustrated Optional Mathematics-9

6.1 (C) Types of Vector

Learning Objectives

At the end of this topic, the students will be able to:

• identify the types of vector. • solve the problems related to the types of vector.

a. Column Vector Y
B(4, 5)

In the adjoining figure, the point A(1, 1) is displaced by AB to B(4, 5).

Draw a right-angled triangle AMB, then AM = x2 – x1 = 4 – 1 = 3 units A(1, 1)
4
3
and BM = y2 – y1 = 5 – 1 = 4 units. Thus, AB = 4 . This type of vector X' O 3M
Y' X

is called column vector.

A vector representing in the form of column only is called column vector. The general form of column vector

is x or x2 – x1 , where (x1, y1) and (x2, y2) are initial and terminal points respectively.
y y2 – y1

Y

b. Position Vector P(5, 4)

In the diagram alongside, the coordinate of the point P is (5, 4). Draw 4 units

PM⊥OX. Then OM = 5 units and PM = 4 units. That means the point

P lies 5 units right and 4 units up. This shows that there is the position X' O X
5 Y' M
of P. So, the vector OP = 4 is called the position vector of P. 5 units

A vector which represents the position of a point in the reference of the origin of coordinate plane is called the

position vector of the point. It is generally defined as x , where (x, y) is the terminal point.
y

c. Row Vector VECTOR

A vector which is written in the horizontal order is called row vector. The general form of the row vector
is (x y) or (x2 – x1 y2 – y1).

d. Null or Zero Vector

A vector having zero magnitude is called null or zero vector. For example, the magnitude of AA which is
0
displaced from the point A to the point A itself, is zero. If a = 0 then | a | = 0. So, a is the null vector.

e. Unit or Identity Vector Y

A vector having the magnitude 1, is called a unit or an identity vector.

For example; OP = 1 , OQ= 0 , OR = –1 and OR = 0 are the unit Q
0 1 0 –1
vectors. X' RO P X
S
If AB = 3 , then | AB | = 5 units.
4 Y'

Introduction to Vector 247

Now, we define the unit vector of AB as AB i.e. 1 × AB
|AB| |AB|
1 3 3/5
= 5 4 = 4/5 , which gives the magnitude 1 unit.

Generally, the unit vector of a vector AB (or a ) is denoted by AB (or a) and is defined by

AB = AB a = | a
|AB | a|
A B
f. Negative Vector A B

A vector having the same magnitude but in opposite direction is B –5
–4
called negative vector. In the adjoining figure, |AB| = |BA|,

but AB ≠ BA.

The negative vector of AB is defined by BA = – AB A5
4
If AB = 5 , then its negative vector BA = – AB =– 5 = –5 .
4 4 –4

Note: The sum of a vector and its negative vector is a null vector. i.e., a + (– a ) = o.

g. Equal and Unequal Vectors B
Two vectors are said to be equal if they have the same magnitude and in the same direction. D

Let AB and CD be the same directed two vectors and AY
C
|AB| = |CD|.
F
Then, AB = CD. If AB = a and CD = c and AB = CD, then, a = c and b = d. X
b d
E

The vectors which are not equal is called unequal vectors. In the adjoining figure, XY and EF are the same

directed two vectors and |XY| ≠ |EF|. Then XY and EF are unequal vectors.

VECTOR Note: The vector and its negative vector are also unequal vectors.

h. Like and Unlike Vectors B
D
A

The vectors are said to be like vectors if these vectors have the same direction C

but their magnitude may be different. In the figure, AB, CD and EF are like E F

vectors. The vectors which do not have the same direction called unlike vectors. Like vectors have are the

same direction but not in unlike vectors.

Note: All equal vectors are like vectors but not vice versa.

i. Parallel and Collinear Vectors M

The vectors are said to be parallel if they have the same or opposite O

direction to each other. In the given figure, LM, NO and PQ are parallel L Q

vectors but not like or unlike vectors. Negative and like vectors are also N

parallel vectors. P

248 Illustrated Optional Mathematics-9

The parallel vectors are said to be collinear if they lie in the same line. The collinear vectors have the same
direction as well as follow whole part axiom.

In the figure, since |AB| + |BC| = |AC| as well as AB, BC and AC have the A B C
same direction. So, AB and BC are collinear vectors. Similarly, in the adjoining P QR S
figure, we can easily show that |PQ| + |QR| = |PR| and |QR| + |RS| = |QS| and
direction of PQ, RS, PR, QS, QR are the same. Hence, the vector satisfies the
whole parts axiom and are also collinear vectors. They are also parallel vectors.

Note: All collinear vectors are parallel vectors but converse of the statement is not true.

j. Perpendicular (or Orthogonal) Vectors A

Two vectors intersect at right angle to each other. They are called perpendicular C BD
vectors. The perpendicular vectors are also called orthogonal vectors. In the figure a

AB and CD are perpendicular to each other. b

k. Localized Vector A A
B
A vector which passes through the given point and parallel to the given vector is C
called a localized vector. In the given figure, a is the given vector and the point A is
the given point. The vector drawn from the point A and parallel to a is the localized

vector. Here, the vector b is called the localized vector.

l. Co-initial Vectors

The vectors having the same initial point are called co-initial vectors. In the adjoining

figure, the vectors OA, OB and OC have the same initial point O. So, they are called

co-initial vectors. O

Example 1 : Find the position vector as the column vector of the point P(2, – 4). VECTOR

Solution: Here,

We have the given point P(2, – 4).

The position vector as the column vector of the point P(2, – 4),

OP = xy - component = 2 .
- component –4

Example 2 : Calculate the unit vector along the direction of EF = –2 .
1

Solution: Here, EF = –2
1
∴ The magnitude of EF, |EF| = x2 + y2

= (– 2)2 + 12 = 4 + 1 = 5 units

Now, we have, –2

The unit vector along the direction of EF, EF = EF = 1 = – 2/ 5 .
|EF| 5 1/ 5

Introduction to Vector 249

Example 3 : Prove that the vectors AB and BC are collinear where AB displaces A(0, 1) to B(3, 2)

and BC displaces B(3, 2) to C(9, 4).
Solution: Here,

For AB: A(0, 1) and B(3, 2) For BC: B(3, 2) and B(9, 4)
|BC| = (9 –3)2 (4 – 2)2
|AB| = (3 – 0)2 (2 –1)2 = 9 + 1 = 10

tan-1 2 – 1 tan– 1 1 = 36 + 4 = 40 = 2 10
3 – 0 3
θ1 = = 4 – 2
9 – 3
θ2 = tan– 1

For AC: A(0, 1) and C(9, 4) = tan– 1 2 = tan– 1 1
|AC| = (9 – 0)2 (4 –1)2 = 81 + 9 = 90= 3 10 6 3

θ3 = tan– 1 4 – 1 = tan– 1 3 = tan– 1 1
9 – 4 9 3

Here, |AB| + |BC| = |AC| and their slopes are the same.

Thus, AB and BC are collinear.
Example 4 : In the given figure, ∆ABC, ∆ABD and ∆ACE are equilateral triangles, and AB = c ,

BC = a and CA = b. Find DB , CE , EA and AD in terms of a , b and c .
Solution: Here, ∆ABC, ∆ABD and ∆ACE are equilateral triangles and D A E
AB = c , BC = a and CA = b.

Now, DB = AC = – CA = – b CE = BA = – AB = – c B C

EA = CB = – BC = – a AD = CB = BC = – a

EXERCISE - 6.1 (C)

1. Define:

VECTOR (a) Column vector (b) Position vector (c) Row vector
(f) Negative vector
(d) Null vector (e) Identity vector (i) Parallel vectors
(l) Co-initial vectors
(g) Equal vectors (h) Like vectors

(j) Orthogonal vector (k) Localized vector

2. (a) What is the magnitude of unit vector?

(b) What is the column vector of a point (3, – 4)?

(c) Write the negative vector of –4 .
5

(d) What is the sum of a vector and its negative vector?

3. Find the position vector as the column vector of the following points:

(a) (2, 0) (b) P(0, – 3) (c) (7, – 3) (d) (4, 2) (e) (– 2, – 5)

4. Find the unit vector along the direction of the following vectors:

(a) AB = 1 (b) c= –2 (c) p= 3 (d) r = – 8 (e) 12
0 1 –4 – 6 –5

250 Illustrated Optional Mathematics-9