Optional Math 9

7. i) 1 + SinA = (SecA + TanA)2
1 – SinA

ii) 1 + 1 + 1 = 2Cosec2A
CosA 1 – CosA

iii) 1 – 1 – 1 = 2TanA.SecA
SinA 1 + SinA

iv) 1 C+oSsiAnA + CosA = 2SecA
1 + SinA

v) 1 – CosA + SinA = 2CosecA
SinA 1 – CosA

PRIME more creative questions:

8. Prove that the following identities.
Sin³q – Cos³q
a. i) Sinq – Cosq = 1 + SinqCosq

ii) (1 + Sina + Cosa)2 = 2(1 + Sina)(1 + Cosa)

iii) (1 – SinA – CosA)2 = 2(1 – SinA)(1 – CosA)

iv) 1 T–aCnoAtA + CotA = 1 + SecA.CosecA
1 – TanA

v) SSinin4AA – Cos4A = SinA + CosA
– CosA

b. i) (1 + Cotq – Cosecq) (1 + Tanq + Secq) = 2

ii) Sec²q + Cosec²q = Tanq + Cotq

iii) (SinA + CosecA)2 + (CosA + SecA)2 = Tan2A + Cot2A + 7

iv) SeTcaqn–q 1 + Tanq = 2Cosecq
Secq + 1

v) Sin6A + Cos6A = 1 – 3 Sin2A.Cos2A

Answer

Show to your teacher.

PRIME Opt. Maths Book - IX 197

5.5 Complex identities:

We discuss about the simple trigonometrical identities in previous exercise where as we
introducing some difficult and technical trigonometrical identities in this chapter.

Hence,
We try to explain the procedures to prove them through some worked out examples.

Worked out Examples

1. Prove that: TanA + CotA = 1 + SecA. CosecA.
1 – CotA 1 – TanA

L.H.S. = TanA + CotA
1 – cot A 1 – TanA

SinA CosA

= CosA + SinA

1 – CosA 1 SinA
SinA CosA

SinA CosA

CosA SinA

= SinA – CosA + CosA – SinA
SinA CosA

= SinA × SinA – CosA × CosA
CosA SinA – CosA SinA SinA – CosA

= Sin²A – Cos²A
CosA.SinA(SinA – CosA)

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

= 1 + SinA CosA
CosA SinA

= 1 + SinA CosA
CosA SinA SinA CosA

= SecA. CosecA + 1.
= R.H.S. is proved

2. Sin2A Cos2B – Cos2A Sin2B = Sin2A – Sin2B

L.H.S. = Sin2A Cos2B – Cos2A Sin2B

= Sin2A (1 – Sin2B) – (1 – Sin2A) Sin2B

= Sin2A – Sin2 A Sin2 B – Sin2B + Sin2 A Sin2 B

= Sin2A – Sin2B

= R.H.S. Proved.

198 PRIME Opt. Maths Book - IX

3. SecA – TanA – 1 = 1 – SecA – TanA
SecA – TanA + 1 1 + SecA + TanA

L.H.S. = SecA – TanA – 1
SecA – TanA + 1

= (SecA – TanA)–(Sec²A – Tan²A)
(SecA – TanA)+(Sec²A – Tan²A)

= (SecA – TanA) – {(SecA + TanA)(SecA – TanA)}
(SecA – TanA) + {(SecA + TanA)(SecA – TanA)}

= (SecA – TanA)(1 – SecA – TanA)
(SecA – TanA)(1 + SecA + TanA)

= 1 – SecA – TanA
1 + SecA + TanA

= R.H.S. is proved

4. CosecA + CotA – 1 = 1 + CosA
1 – CosecA + CotA SinA

L.H.S = CosecA + CotA – 1
1 – CosecA + CotA

= (CosecA + CotA) – (Cosec²A – Cot²A)
1 – CosecA + CotA

= (CosecA + CotA) – {(CosecA + CotA) (CosecA – CotA)}
1 – CosecA + CotA

= (CosecA + CotA)(1 – CosecA + CotA)
1 – CosecA + CotA

= 1 + CosA
SinA SinA

= 1 + CosA
SinA

= R.H.S. Proved

5. (3 – 2Sin2q) (2Cot2q – 3) = (1 + 3Cot2q) (2 – 5Sin2q).

L.H.S. = (3 – 2Sin2q) (2Cot2q – 3)

= (3 – 2sin2q) c 2Cos2 i – 3m
Sin2 i

= (3 – 2Sin2q)( 2Cos²q – 3Sin²q )
Sin²q

= ( 3 – 2Sin²q ){2(1 – Sin2q) – 3Sin2q}
Sin²q Sin²q

= {3 (1 + Cot2q) – 2} {2 – 2Sin2q – 3Sin2q}

= (3 + 3Cot2q – 2) (2 – 5Sin2q)

= (1 + 3Cot2q) (2 – 5 Sin2q)

= R.H.S. Proved.

PRIME Opt. Maths Book - IX 199

Exercise 5.5

1. i) (Cosq + Sinq) (Secq + Cosecq) = 2 + Secq.Cosecq

ii) (1 + Sinq – Cosq)2 + (1 – Sinq + Cosq)2 = 4(1 – Sinq.Cosq)

iii) 1 – Cot2a + Cot4a = Sin2a(1 + Cot6a)

iv) (CosecA – SinA)(SecA – CosA)(TanA + CotA) = 1

v) CSiont22BA – Cosec2A = Cot2A – Cot2B
Tan2B

2. Prove that the followings:

i. Sin2aCos2b – Cos2aSin2b = Cos2b – Cos2a.

ii. TCaons22 A – Tan2 B = Tan2A – Tan2B.
B Cos2 A

iii. Cot2qCosec2b – Cosec2q . Cot2b = Cosec2q – Cosec2b

iv. (xCosa + ySina)2 + (xsina – yCosa)2 = x2 + y2.
1 + 3Cosq – 4Cos3q
v. (1 + 2Cosq)2 = 1 – Cosq

3. Prove that the followings:

i. 1 CotA + TanA = SecA.CosecA + 1
– TanA 1 – CotA

ii. TTaanna2–a1 + Cot2a = Seca.Coseca + 1.
Cota – 1

iii. 1S–inC2oata + Cos2 a = Sina.Cosa + 1.
1 – Tana

iv. (3 – Tan2q) (2 – 3Sin2q) = (3 – 4Sin2q) (2 – Tan2q)
v. (3 – 2Cot2q) (2 – Cos2q) = (3 – 5Cos2q) (2 + Cot2q)

4. Prove that the followings:

i. 11 + Secq – Tanq = 1 – Sinq
+ Secq + Tanq Cosq

ii. S1e–cqSe+cqTa+nTqa–n1q = 1 + Sini
Cosi

iii. 1Co–sCecoqse+cqCo+tqCo–t1q = 1 – Cosi
Sini

iv. C1o–sCecoqse+cqCo+tqCo–t1q = Sini
1 – Cosi

v. S1e–cqSe–cTqa–nTqa–n1q = Cosi
1 – Sini

5. Prove that the followings:

i. CCoosseeccqq – Cotq + 1 = 1 + Cosecq + Cotq
– Cotq – 1 1 – Cosecq – Cotq

ii. 11 – Secq – Tanq = Secq – Tanq – 1
+ Secq + Tanq Secq – Tanq + 1

200 PRIME Opt. Maths Book - IX

iii. CosecA1– CotA – 1 = 1 – 1 CotA
SinA SinA CosecA +

iv. Co1si – 1 = 1 – 1
Seci + Tani Seci–Tani Cosi

v. 1C–oSsiAnA + CosA = 2TanA
1 + SinA

6. PRIME more creative questions:
a. i. Sin6A – cos6A = (1 – 2Cos2A)(1 – Cos2A + Cos4A)

ii. Sin8q – Cos8q = (2Sin2q – 1)(1 – 2Sin2q + 2Sin4q)

iii. (1 + Cotq – Cosecq) (1 + Tanq + Secq) = 2

iv. SeTcaan–a1 – Sina = 2Cota
1 + Cosa

v. 1 + SinA – 1 – SinA = 2TanA
1 – SinA 1 + SinA

vi. TTananA2–A1 + CotA = 1 + SecA.CosecA
1 – TanA

b. i. Sec6 i + Tan6 i = 1+Sec2q.Tan2q
Sec2 i + Tan2 i

ii. 2(Sin6A+Coc6A) – 3(Sin4A+Cos4A) + 1 = 0

iii. Tan3a+Cot3a = Sec3aCosec3a-3Seca.Coseca
iv. SSiinnAA–+CCoosAsA+–11 = SecA + TanA

v. CCoossAA–+SSininAA+–11 = 1 + CosA
SinA

c. i. (3 – 4Sin2a) (1 – 3Tan2a) = (3 – Tan2a)(1 – 4Sin2a)

ii. SinxCo+sCxosy + Cosy = Cosx + Cosy
Siny – Cosx Sinx – Cosy Siny + Cosx

iii. 11 ++ Sec2A.Cot2C = 1 + Tan2A.Cos2C
Sec2B.Cot2C 1 + Tan2B.Cos2C

iv. 11 ++ SinA – CosA = 1 + SinA + CosA = 2 CosecA
SinA + CosA 1 + SinA – CosA

v. CSoinsii + Cosb + Sini + Sinb =0
– Sinb Cosi – Cosb

Answer

Show to your subject teacher.

PRIME Opt. Maths Book - IX 201

5.6 Use of given trigonometrical ratio to find the rest ratios:

In this chapter, we discuss the process of finding all the trigonometrical ratios, if one of
them is given either directly or in the form of algebric expression.

Worked out Examples

1. Find the ratio of Sinq and Cotq from the given right angled triangle.

Solution :

In right angled triangle ABC

\ B = 90° A

BC = 4cm = b

AC = 5cm = h q 5cm
4cm
AB = p = h2 – b2

= 52 – 42

= 9 B C

= 3cm

Then,

Sinq = p = BC = 4cm = 4
h AC 5cm 5

Cotq = b = AB = 3cm = 3
p BC 4cm
4

2. If Sin A= 3 , find the vale of Sec2A – Tan2A.
5
Solution :

In right angled DABC

\ B = 90°
\ A = reference angle

SinA = 3 or p = 3
Here, 5 h 5

p = 3cm

h = 5 cm

Then,

b = h2 – p2 = 52 – 32 = 16 = 4 cm

Again, = a h 2 – ` p j2
Sec2A – Tan2A b b
k

= a 5 2 ` 3 2
4 4
k– j

= 25 – 9
16

= 16
16

=1

202 PRIME Opt. Maths Book - IX

3. If a Tanq = b find the value of Sin2 i – Cos2 i
Solution: Sin2 i + Cos2 i

or, aTanq = b

or, Tanq = b
a

or, p = b
b a

p=b

b=a

h = p2 + b2 = a2 + b2

Then, p
h
sin2 i – cos2 i ` j2 – a b 2
sin2 i + cos2 i h
k

= ` p j2 b 2
h h
+ a k

c b b² 2 – a a b² 2
a² + a² +
= m k
c
b 2 + a 2
a² + a² +
b² m a b² k

= b2 – a2 × a2 + b2
a2 + b2 b2 + a2

= b2 – a2
b2 + a2

4. Express all the trigonometrical ratios in terms of SinA.

Solution:

Let, Sin A = K
p
h = k
1
p=k

h=1

b = h2 – p2 = 1 – k2

Then,

Sin A = Sin A

Cos A = b = 1 - k2 = 1 – Sin2 A
h 1

Tan A = p = k = SinA
b 1 – k2 1 – Sin2 A

Cosec A = h = 1 = 1
p k SinA

Sec A = h = 1 = 1
b 1 – k2 1 – sin2 A

Cot A = b = 1 – k2 = 1 – sin2 A
p k sin A

PRIME Opt. Maths Book - IX 203

5. If SinA – CosA = 0, find the value of SecA.
Solution :

SinA – CosA = 0
or, SinA = CosA

or, SinA =1
CosA

or, TanA = 1
p

or, b = 1
\ p = 1

b=1

h = p2 + b2

=2

Then,

SecA = h
b

= 2
1

= 2

SecA = 2

6. If 2cos2q – Sinq – 1 = 0, find the value of sinq.
Solution :
2cos2q – Sinq – 1 = 0
or, 2(1 – Sin2q) – Sinq – 1 = 0
or, 2 – 2Sin2q – Sinq – 1 = 0
or, 2Sin2q + Sinq – 1 = 0
or, 2Sin2q + (2 – 1)Sinq – 1 = 0
or, 2Sin2q + 2sinq – Sinq – 1 = 0
or, 2Sinq(Sinq + 1) – 1(Sinq + 1) = 0
or, (Sinq + 1) (2Sinq – 1) = 0

Either OR 2Sinq – 1 = 0
Sinq + 1 = 0

\ Sinq = – 1 Sinq = 1
2
1
\ Sinq = – 1 or 2

7. If 5cosq + 12 Sinq = 13, find the value of Tanq.
Solution :
5Cosq + 12Sinq = 13
or, 5Cosq = 13 – 12 Sinq
Squaring on both sides,
or, 25Cos2q = 132 – 312 Sinq + 144Sin2q
or, 25 – 25Sin2q = 169 – 312 Sinq + 144Sin2q

204 PRIME Opt. Maths Book - IX

or, 169Sin2q – 312Sinq + 144 = 0

or, (13Sinq – 12)2 = 0

or, 13Sinq – 12 = 0

or, Sinq = 12
13

\ Sinq = 12
13

or, p = 12
h 13

\ p = 12

h = 13

\ b = h2 – p2 = 169 – 144 = 5
p
\ Tanq = b = 12
5

Alternative method

5cosq + 12sinq = 13

Dividing both sides by cosq

or, 5Cosi + 12Sini = 13
Cosi Cosi Cosi

or, 5 + 12Tanq = 13Secq

Squaring on both sides,

or, (5 + 12Tanq)2 = (13Secq)2

or, 25 + 120Tanq + 144Tan2q = 169 + 169Tan2q

or, 25Tan2q – 120Tanq + 144 = 0

or, (5Tanq – 12)2 = 0

or, 5Tanq – 12 = 0

\ Tanq = 12
5

8. If Sinq + Cosq = 2 Sinq, prove that Sinq – Cosq = ± 2 Cosq.
Solution :

Sinq + Cosq = 2 Sinq
Squaring on both sides,
or, (Sinq + Cosq)2 = ( 2 Sinq)2
or, Sin2q + 2SinqCosq + Cos2q = 2Sin2q
or, 1 + 2SinqCosq = 2Sin2q
or, 2SinqCosq = 2Sin2q – 1
or, – 2SinqCosq = 1 – 2Sin2q
or, 1 – 2SinqCosq = 2 – 2Sin2q
or, Sin2q + Cos2q – 2SinqCosq = 2(1 – Sin2q)
or, (Sinq – Cosq)2 = 2Cos2q

\ Sinq – Cosq = ± 2 Cosq is proved

PRIME Opt. Maths Book - IX 205

Exercise 5.6

1. i) Is a triangle having sides 6cm, 8cm and 10cm a right angled?
ii) If Sinq – 1 = 0, find Cosq.
iii) If Sinq – Cosq = 0, find Sinq.
iv) If Tana = k, what will be the value of Cosa?
v) If SecA = CosecA, find the value of CotA.

2. Find the trigonometric ratios of Tana, Cosa Cosecq and Cotq from the following

diagrams. 8cm
q
i) 6cm L ii) P Q
K
a

10cm q 17cm
a

iii) MR A
P iv) 4cm

3cm7 2cma
7cm BC
12cm
S a 3cm q Q 13cm q
v) 17cm D
A
a

13cm D

Cq B

3. i) If Sin A = 3 , find the value of TanA and Sec A.
5
ii)
iii) If Cos A = 5 , find the value of Cosec2A – Cot2A.
iv) 13
v)
If 4Tan A = 3, find the value of 5(Sin A + Cos A)
206 If 17 Sin A = 15, find Sec2A – Tan2A.

If 3 TanA = 4, frind the value of 3CosA – SinA
CosA + 2SinA

PRIME Opt. Maths Book - IX

4. i) If n Tan A = m, find the value of Sin2A + Cos2A.
ii)
If p CosA = q, prove that p2 – q2 Cosec A = p.
iii)
iv) If 1 – Cos q = 1 , find the value of Cotq – Tanq.
v) 2
If Sinq – Cosq = 0, find Cosecq.
If Secq – Cosecq = 0, find the value of Secq.

5. i) Express all the trigonometrical ratios in terms of Cos A.
ii) Express all the trigonometrical ratios in terms of Tan A.
iii) Express all the trigonometrical ratio in terms of Sin A.
iv) Express all the trigonometrical ration in terms of Cosec A.
v) If 3CosA = 4, find all other trigonometrical ratios.

6. i) If Sinq = a –1 , find cot2q – Cosec2q.
a +1

ii) If 41CosA = 40, find TanA
1 – Tan2 A

iii) If CosA = a , find CotA + TanA
a2 + b2

iv) If (m2 + n2)SinA = m2 – n2, prove that (m2 – n2)CotA = 2mn.

v) If yTanq = x, find the value of xSini – yCosi
xSini + yCosi

7. i) If 5TanA = 4, prove that 5SinA – 3CosA = 5
SinA + 2CosA 14

ii) If a Cotq = b, prove that aSinq + bCosq = a2 + b2

iii) If SinA = 3 and SinB = 12 , find the value of SinACosB + CosASinB.
5 13

iv) If ( 3 + 1 ) Tanq = 3 – 1 , Find the value of Sinq + Cosq.
v) If x2 + y2 Cosq = x, find the value of xSinq – yCosq.

8. PRIME more creative questions:
a. i) If 2Sin2q = 2 + Cosq, find the value of Cosq.
ii) If 2Sin2q – Cosq – 1 = 0, find the value of Cosq where ‘q’ lies in first quadrant.
iii) If Secq.Tanq = 2 , find the value of Sinq.
iv) If 2 3 Cos2q = Sinq find the value of Sinq where q lies in first quardrent.
v) If Tan2q + ( 3 – 1 ) Tanq = 1, find the value of Tanq where ‘q’ lies in first
3
quadrent.

b. i) If 3Cosq + 4Sinq = 5, find the value of Sinq.
ii) If 4Cosq + 3Sinq = 5, find the value of Cotq.
iii) If 8Sinq + 15Cosq = 17, find the value of Tanq.
iv) If Cota + Coseca = m, find the value of Cosa where ‘a’ lies in first quadrent.
v) If Cos4A + Cos2A = 1, prove that Tan4A + Tan2A = 1

PRIME Opt. Maths Book - IX 207

c. i) If CosA + SinA = 2 CosA, prove that CosA – SinA = ± 2 SinA.
ii) If Cotq + Cosq = a and Cotq – Cosq = b, prove that a2 – b2 = 4 ab
iii) If Secq + Tanq = a, prove that (a2 + 1) sinq = a2 – 1. Where ‘q’ lies in first quadent.
iv) If Tanq + Cotq = 3, find the value of Sec2q + Cosec2q.
v) If TanA = 1 in a right angled DABC, find the value of SinACosA.

Answer

1. Show to your teacher.

2. i) Tana = 4 , Cosa = 3 , Cosecq = 5 , Cotq = 4
3 5 3 3

ii) Tana = 8 , Cosa = 15 , Cosecq = 15 , Cotq = 8
15 17 17 15

iii) Tana = 7 , Cosa = 24 , Cosecq = 1 , Cotq = 1
24 25 2

iv) Tana = 4 , Cosa = 3 , Cosecq = 5 , Cotq = 12
3 5 13 5

v) Tana = 5 , Cosa = 12 , Cosecq = 3 , Cotq = 4
12 13 5 3

3. i) TanA = 3 , SecA = 5
4 4

ii) 1 iii) 7 iv) 1 v) 5
11

4. i) 1 iii) 2 iv) 2 v) 2
3

5. i) Show to your subject teachers.

6. i) –1 ii) 360 iii) a2a+bb2 x2 – y2
1519 v. x2 + y2

7. iii) 6635 iv) 32 v) o

8.a. i) 0, 1 ii) 12 iii) 12 ,– 2 iv) 23 v) 1
2 3

b. i) 4 ii) 34 iii) 185 iv) 0, 2m
5 m2 + 1

c. iv) 9 v) 1
2

208 PRIME Opt. Maths Book - IX

5.7 Trigonometrical Ratios of some standard angles

The trigonometrical ratios of some standard angles 0°, 30°, 45°, 60° & 90° are discussed
and geometrical interpretations are given in this chapter.

i) Trigonometrical ratios of 0°. A

Cq B

In the right angled DABC,
\ B = 90°
\ C = q (reference angle)

If A approaches to B, the value of q decreases and at q = 0 ‘A’ coincides with ‘B’ such

that

AB = 0 and AC = BC

Sinq = p AB ( Sin0° = 0 =0
h = AC ( BC
(
Cosq = b = BC Cos0° = BC =1
h AC BC

Tanq = p = AB Tan0° = 0 =0
b BC BC
Also,

Cosec0° = ∞,

Sec0° = 1,

Cot0° = ∞

ii) Trigonometrical ratios of 30° & 60°. A 2a
Let, ABC is an equilateral triangle 30° aC
In an equilateral DABC, 2a
AD⊥BC 209
AB = BC = AC = 2a (say) 60° D
BD = DC = a Ba
\ A = \ B = \ C = 60°
AD = AB2 – BD2

= ^2ah2 – ^ah2

= a 3

Then, In rt. \ ed DADB,
\ BAD = 90° – 60° = 30°
AB = 2a = h
BD = a = p

AD = a 3 = b

PRIME Opt. Maths Book - IX

Sin30° = p = BD = a = 1
h AB 2a 2

Cos30° = b = AD = a3 = 3
h AB 2a 2

Tan30° = p = BD = a = 1
b AD a3 3

Also,

Cosec30° = 2
Sec30° = 2

3

Cot30° = 3

Again, In rt. \ ed DABD,

\ ABD = 60°

AD = a 3 = p

BD = a = b

AB = 2a = h

Sin60° = p = AD = a3 = 3
h AB 2a 2

Cos60° = b = BD = a = 1
h AB 2a 2

Tan60° = p = AD = a3 = 3
b BD a

Also,
Cosec60° = 2

3

Sec60° = 2

Cot60° = 1
3

iii) Trigonometrical ratios of 45°
A

BC PRIME Opt. Maths Book - IX
Let DABC is an isoceles right angled triangle where
\ B = 90°
AB = BC = a (say)
\ \ A = \ C = 45°
Then
\ AC = AB2 + BC2
= a2 + a2
= a 2

210

Also, Taking reference angle A. A
p BC
Sin45° = h = BC = a = 1
AC a2 2

Cos45° = b = AB = a = 1
h AC a2 2

Tan45° = p = BC = a =1
b AB a

Cosec45° = h = AC = a2 = 2
p BC a

Sec45° = h = AC = a2 = 2
b AB a

Cot45° = b = AB = a =1
p BC a

iv) Trigonometrical ratios of 90°.

Let, ABC is a right angled triangle.

Where, \ B = 90°

\ C = Reference angle = q

When C approaches to B, then reference angle increases and at q = 90°, C coincides

with B.

Such that

BC = 0

AB = AC

Now, p
h
SinC = = AB & Sin90° = AB =1
AC AB

CosC = b = BC & Cos90° = 0 =0
h AC AC

TanC = p = AB & Tan90° = AB =∞
b BC 0

Also, Sec90° = ∞ Cot90° = 0
Cosec90° = 1

Table for the values w.r.t. angles.

Write down = 0, 1, 2, 3, 4

Dividing by 4 = 0 , 1 , 2 , 3 , 4
4 4 4 4 4

Square root = 0 , 1 , 2 , 3 , 4
4 4 4 4 4

Result = 0, 1 , 1, 3 , 1
2 2 2

PRIME Opt. Maths Book - IX 211

Tabulation of the above values respectively.

Angles 0° 30° 45° 60° 90°
Ratios

Sin 011 31
2 22

Cos 1 3 1 1 0
2 22

Tan 01 1 3∞
Cosec 3
Sec
Cot ∞2 22 1

3

12 22 ∞

3

∞ 31 1 0

3

To remember

Sin0° = Cos90° = Tan0° = Cot90° = 0

Sin90° = Cos0° = Tan45° = Cosec90° = Sec0° = Cot45° = 1

Sin30° = Cos60° = 1
2

Sin60° = Cos30° = 3
2

Tan30° = Cot60° = 1
3

Tan60° = Cot30° = 3

Cosec30° = Sec60° = 2
Cosec60° = Sec30° = 2

3

212 PRIME Opt. Maths Book - IX

Worked out Examples

1. Find the value of sin30°.Cot60°.Cos30°.cosec30°

Solution:

Here, Sin30°.Cot60°.Cos30°.Cosec30°

= 1 × 1× 3
2 3 2 ×2

= 1
2

2. 3Tan230° + Sin260° + Cos245°

Solution:

Here, 3Tan230° + Sin260° + Cos245°

=3× c 1 2 + c 3 2 + c 1 2
2
3 m m 2 m

=3× 1 + 3 + 1
3 4 2

= 4+3+2
4

= 9
4

= 2 1
4

3. Sin2 p + 1 Sec2 p + 2Tan2 p + Cosec2 p
6 2 3 4 2

Solution : p 1 p p p
6 2 3 4 2
Here, Sin2 + Sec2 + 2Tan2 + Cosec2

= Sin21860° + 1 Sec2 180° + 2Tan21840° + Cosec2 180°
2 3 2
1
= Sin230° + 2 Sec260° + 2Tan245° + Cosec290°

= a 1 2 + 1 (2)² + 2(1)² + (1)²
2 2
k

= 1 +2+2+1
4
1 + 20
= 4

= 21
4

= 5 1
4

PRIME Opt. Maths Book - IX 213

4. If A = 30°, B = 60°, C = 45° & D = 90°, find the value of SinA + Cos2B + Tan2C.CosecD.

Solution:

Here, SinA + Cos2B + Tan2C.CosecD

= Sin 30° + Cos260° + Tan245°.Cosec90°

= 1 + a 1 2 + ^1h2 ×1
2 2
k

= 2+1+4
4

= 1 3
4

5. If 2SinA = 1, find the value of A.

Solution:

Here, 2SinA = 1

or, SinA = 1
2

or, SinA = Sin30°

\ A = 30°

Exercise 5.7

1. i) If Sinq = 3 , find Cosq.
2
ii) In an isosceles DABC, ∠B = 90°. What is the value of Tan 45°.

iii) Find the value of Tan260.

iv) Find the value of Sin260° + Cos260°.

v) Find the value of Sec230° – Tan230°

2. Find the value of the followings. ii) Sin90°.Cos60°.Cosec30°
iv) Sec0°.Cosec45°.Cos45°
i) Sin0° + Cos90° + Tan45°

iii) Tan45° + 2Sin60° – 2Cos60°

v) 32 Tan60°.Cot30°.Cos30°

3. Find the value of the followings. ii) 4Cos230° + 3Cot260° + Cosec245°
i) Sin230° + Cos245° + Tan260° p p p
3 6 4
iii) Cos20° + Sin290° + Sec245° iv) Sin .Cos .Tan
p p p
v) 13 Cos2 6 + 4Sin2 3 + Cot2 4

4. Prove that the following.
p p p p
i) Cos2 2 + Sin2 4 + 1 Tan2 4 = Sin2 2
2

ii) 11 – Tan30° =2– 3 iii) 11+–CSoins3r6r = 7 + 4 3
+ Tan30° PRIME Opt. Maths Book - IX

214

iv) CCoot6t300°.°C–otC3o0t°60+°1 = Tan60° v) Sin30°Cos60° + Cos30°Sin60° = 1

5. If A = 0°, B = 30°, C = 45°, D = 60°, find the value of followings.

i) SinA + 2CosB + CotC + TanD ii) 3 TanB + CotC – 2 3 CotD

iii) Sin2B – Cos2C + Tan2D iv) 3Cot2D – 2 Cos2B + 4Sin2C
v) Cos(A + B) + Cos(D – B) 3

6. Find the value of A from the followings.

i) 2CosA – 1 = 0 ii) 2SinA = 3

iii) 3 TanA – 1 = 0 iv) CosA – 1 = 0

v) CotA – 3 = 0

7. PRIME more creative questions.
i) If 3Tanq – 3 = 0, find the value of ‘q’.

ii) Evaluate : 13 + 7 + 1 + 2 3 Sin60°

iii) Prove that : 1 – Sin60° = 1 – Tan30°
1 – Sin30° 1 + Tan30°

iv) Geometrically prove that Sin60° = 3
2

v) Geometrically prove that Cos45° = 1
2

Answer

1. Show to your teacher.

2. i) 1 ii) 1 iii) 3 iv) 1 v) 3

3. i) 3 3 ii) 6 iii) 4 iv) 3 v) 4 14
4 4

5. i) 2 3 + 1 ii) 0 iii) 2 3 iv) 2 12 v) 3
4

6. i) 60° ii) 60° iii) 30° iv) 0° v) 30°

7. i) 30° ii) 4

PRIME Opt. Maths Book - IX 215

5.8 Trigonometric ratio of complementary angles and solution of
right angled triangle

Let us consider that DPQR is a right angled triangle where,
\ Q = 90°
\ R = q (refrence angle)
\ \ P = 90° – q (Complementary angle of q)

Then, for the refrence angle (90° – q) P
90° – q
Sin(90° – q) = p QR = Cosq
h = PR Q

Cos(90° – q) = b PQ = Sinq
h = PR

Tan (90° – q) = p QR = Cotq q
b = PQ R

Cosec(90° – q) = h = PR = Secq
p QR

Sec(90° – q) = h = PR = Cosecq
b PQ

Cot (90° – q) = b = PQ = Tanq
p QR

Worked out Examples

1. Find the value of : Sin25° – Cos65°
Solution:
Here, Sin25° – Cos65°
= Sin25° – Cos(90° – 25°)
= Sin25° – Sin25°
= 0

2. Prove that : Cos55° + Sin65° = Sin35° + Cos25°
Solution :
L.H.S. = Cos55° + Sin65°
= Cos(90° – 35°) + Sin(90° – 25°)
= Sin35° + Cos25°
= R.H.S. proved

3. Prove that : Sin(90° – q) . Cosq . 1 = Sinq
Solution : Cosq Sin(90° – q) Sec(90°
– q)

L.H.S. = Sin(90° – q) . Cosq . 1
Cosq Sin(90° – Sec(90°
q) – q)

= Cosq × Cosq × 1
Cosq Cosq Cosecq

216 PRIME Opt. Maths Book - IX

= 1
Cosecq

= Sinq
= R.H.S. proved

4. Find the side QR of right angled DPQR. P
15 cm
Solution :

In rt. \ ed DPQR,

\ Q = 90°,

\ R = 60° (refrence angle) 60°
R
PR = 15 cm, Q

QR = ?

We have,

Cos60° = b = QR
h PR

or, 12 = QR
15

or, 2QR = 15

\ QR = 7.5 cm

5. If an electric pole is tied with a rope of length 20m where the rope made 60° angle

with the ground. How high does the rope from the ground? A
Solution:

Let, AB be the height of a pole. 20m ?
AC be the length of the rope.
Given :

\ B = 90° C 60° B
\ C = 60°

AC = 20 m

AB = ?

Now, In rt. \ ed DABC,
p
Sin60° = h = AB
AC

or, 3 = AB
2 AC

or, 2AB = 20 3

or, AB = 20 3
2

\ AB = 17.32 m

PRIME Opt. Maths Book - IX 217

Exercise 5.8

1. i) What is the complementary angle of 70°.

ii) Express Sin25° in terms of Cos as the complementary angle.
p
iii) Write down the equivalent of Cos 8 as the complementary angle.

iv) Find the value of Cotq.Cot(90° – q)

v) Find the value of Tan25°Tan65°.

2. Simplify :

i) Sin10° – Cos80° ii) Cos70° – Sin20° – Sin40° + Cos50°

iii) Cos(90° – q).Tan (90° – q).Cosec (90° – q)
iv) Cot(9C0o°s–eqc()9. S0e°c–(9q0)° – q)
v) Tan9°. Tan18°.Tan72°.Tan81°

3. Prove that the followings.

i) Sin34° = Cos56° ii) Cot17° = Tan 73°

iii) Cos40° + Sin55° = Sin50° + Cos35° iv) Sin74° – Cos37° = Cos16° – Sin53°

v) Sec80° + Cosec40° = Cosec10° + Sec50°

4. Prove that the followings.

i) Sin(90° – q).Cosq + Cos(90° – q).Sinq = 1

ii) Secq.Cosec(90° – q) – Cot (90° – q).Tanq = 1

iii) Tanq.Tan(90° – q) + Sinq.Sec(90° – q) = 2

iv) Sin Sini . Cos (90° – i) = Tan2q
(90° – Cosi
i)

Cos (90° – i) Tan (90° – i)
v) Sec (90° – i) . Cot (90° – i) . Cosec(90° – q) = Cosq

5. Find the side AB from the following right angled DABC.
i) A ii) A

10cm 60° 12cm

B 30° B C
iii) C A
30°
A iv) 40cm

C 45° B B C
218 20cm PRIME Opt. Maths Book - IX

v) A 60° D

25cm B
C

PRIME more creative questions:
6. Find the followings.

i) The angle made by ladder 20m long to the ground at the foot of the ladder is 30°

which is taken against a wall. Find the height of the wall.

ii) An electric pole of height 16ft is tied with a metallic rope where the rope makes
an angle of 60° with the ground, find the length of the rope.

iii) A pole of height 12ft forms the shadow during the sun’s altitude of 45°, find the
length of the shadow of the pole.

iv) Find the height of the temple given in diagram.

?

60°

2m

20 3m

v) If the kite is at a height of 102m from the ground, ?
find the length of the string used in the 30°

kite as given in diagram. 102m



2m

Answer

1. i) 20° ii) Cos65° iii) Sin 3p iv) 1 v) 1
8 v) 1
2. i) 0 ii) 0 iv) 1 v) 50cm
4. i) 5cm ii) 6cm iii) 1 iv) 80cm v) 200m
5. i) 10m ii) 32ft iv) 62m
iii) 20cm 219

iii) 12ft

PRIME Opt. Maths Book - IX

5.9 Trigonometrical ratios of additional angels:

Other than standard angles some of the angles lie in the four quadrants like 90°-A, 90°+A,
180°-A, 180°+A, 270°-A, 270°+A, 360°-A, 360°+A and (-A).
Here, we discuss the trigonometric ratios of those angles and their application in different
identities.

Y

90° + A 90° – A
180° – A
X’ A X
180° + A –A

Y’

• First quadrant : P
OP, PM and OM all are positive in first quadrant due to which all A
OM
trigonometric ratios are positive.

• Second quadrant : Y

p = PM = +ve and b = OM = -ve, OP = +ve P

It makes, Cosecq = +ve X’ M A q X
p O

Sinq = h = +ve ,

Cosq = b = -ve , Secq = -ve
h Cotq = -ve

Tanq = p = -ve , Y’
b Y

• Third quadrant :

p = PM = -ve

b = OM = -ve q
O
It makes, p X’ M X
h P Y’
Sinq = = -ve , Cosecq = -ve
Secq = -ve
Cosq = b = -ve , Cotq = +ve
h

Tanq = p = +ve ,
b

220 PRIME Opt. Maths Book - IX

• Fourth quadrant : Y

p = PM = -ve

b = OM = +ve q MX
O
It makes, p X’
h
Sinq = = -ve , Cosecq = -ve
Secq = +ve
Cosq = b = +ve , Cotq = -ve P
h Y’

Tanq = p = -ve ,
b

i. Trigonomertical ratios of (90°-A) Y
A
Let, XX’ and YY’ are intersected at the center of a circle ‘O’. O Q 90°–A
P
A revolving line OP makes an angle ‘A’ with y-axis. Y’
NMX
ie. \ POX = A
P
\ YOQ = A

\ QOX = 90° – A X’
Draw PM^OX and QN^OX.

Then, In right angled DPOM & DQON.

\ OQN = \ YOQ = A [Alternative angles]

\ \ QON = 90° – A

i. \ PMO = \ QNO = 90°

ii. OM = QN

iii. OP = OQ = radii of same circle.

\ DPOM b DQNO (By AAS axiom)

\ PM = ON and OM = QN.

Then, In right angled DQON.

Sin (90°-A) = QN = OM = CosA
OQ OP

Cos (90°-A) = ON = PM = SinA
OQ OP

Tan (90°-A) = QN = OM = CotA
ON PM

Cosec (90°-A) = OQ = OP = SecA
QN OM

Sec (90°-A) = OQ = OP = CosecA
ON PM

Cot (90°-A) = ON = PM = TanA
QN OM

ii. Trigonomertical ratios of (90°+A).

Let, A revolving line OP makes an angle ‘A’ with OX and again ‘A’ with OY at the position
of OQ.

ie. \ POX = A
\ QOY= A
\ QOX = 90°+A

PRIME Opt. Maths Book - IX 221

Draw the perpendiculars, Y
PM^OX and QN^OX’. Q
\ \ OQN = \ QOY = A
Now, In DPMO and DQNO, A A 90°A+A P
i. \ PMO = \ QNO = 90°
ii. \ PAM = \ OQN = A [Alternative angles] X’ N O MX
iii. OP = OQ = radii of same circle.
\ DPOM ≅ DQNO (By AAS axiom)
\ PM = –ON and OM = QN

Then, In right \ ed DQNO, Y’

Sin (90° + A) = QN = OM = CosA
OQ OP

Cos (90° + A) = ON = –PM = –SinA
OQ OP

Tan (90° + A) = QN = OM = –CotA
ON –PM

(and so on for others)

iii. Trigonomertical ratios of (180°-A)

Let, A revolving line OP makes an angle of ‘A’ with OX and again ‘A’ with OX’ at the

position of OQ. Y
ie. \ POX = A

\ QOX’= A QP

\ \ QOX = 180°-A 180°–A

Draw the perpendiculars, X’ N AA MX
PM^OX and QN^OX’
O

Now, In right angled DPMO and DQNO, Y’
i. \ PMO = \ QNO = 90°
ii. \ POM = \ QON = A
iii. OP = OQ = radii of same circle.
\ DPMO ≅ DQNO (By AAS axiom)
\ PM = QN and OM = –ON.

Then, In right angled DQNO,

Sin (180°–A) = QN = PM = SinA
OQ OP

Cos (180°–A) = ON = –OM = –CosA
OQ OP

Tan (180°–A) = QN = PM = –TanA
ON –OM

(and so on for the others)

222 PRIME Opt. Maths Book - IX

iv. Trigonomertical ratios of (180°+A)

Let, A revolving line OP makes an angle ‘A’ with the line OX and again ‘A’ with OX’ at

the position of OQ. Y
ie. \ POX = A

\ QOX’ = A P

\ \ QOX = 180°+A 180°–A
Draw the perpendiculars,
PM^OX and QN^OX’ X’ N A
Now, In DPMO and DQNO, A O MX

i. \ PMO = \ QNO = 90°

ii. \ PM = \ QON = A Q
iii. OP = OQ = radii of same circle.

\ DPMO ≅ DQNO (By AAS axiom) Y’
\ OM = ON and PM = -QN.

Then, In right angled DQNO,

Sin (180°+A) = QN = –PM = -SinA
OQ OP

Cos (180°+A) = ON = –OM = -CosA
OQ OP

Tan (180°+A) = QN = –PM = TanA
ON –OM

(and so on for the others)

v. Trigonomertical ratios of (270°-A)

Let, A revolving line OP makes an angle of ‘A’ with OX and again ‘A’ with OY’ at the

position X’ of OQ. Y
ie. \ POX = A

\ QOX’ = A P
\ \ QOX = 270°-A
Draw the perpendiculars, 270°–A
PM^OX and QN^OX’
\ \ OQN = \ QOY’ = A X’ N A MX
O

Now, In DPMO and DQNO, AA
i. \ PMO = \ QNO = 90°

ii. \ POM = \ NQO = A Q
iii. OP = OQ = radii of same circle. Y’
\ DPMO ≅ DQNO (By AAS axiom)

\ PM = –ON and OM = –QN.

Then, In right angled DQNO,

Sin (270°-A) = QN = –OM = –CosA
OQ OP

Cos (270°-A) = ON = –PM = –SinA
OQ OP

Tan (270°-A) = QN = –OM = CotA
ON –PM
(And so on for the others)

PRIME Opt. Maths Book - IX 223

vi. Trigonomertical ratios of (270°+A)

Let, A revolving line OP makes an angle ‘A’ with OX and again ‘A’ with OY’ at the

position of OQ. Y

ie. \ POX = A

\ QOY’ = A P
\ \ QOX(reflex) = 270°+A
Draw the perpendiculars, 270°–A AN X
X’ O M
PM^OX and QN^OX
AA
\ \ OQN = \ QOY’ = A

Now, In right angled DPMO and DQNO, Q
i. \ PMO = \ QNO = 90° Y’
ii. \ POM = \ OQN = A
iii. OP = OQ = radii of same circle.
\ DPMO ≅ DQNO (By AAS axiom)
\ PM = ON and OM = –QN.

Then, In right angled DQNO,

Sin (270°+A) = NQ = –OM = –CosA
OQ OP

Cos (270°+A) = ON = –PM = SinA
OQ OP

Tan (270°+A) = QN = –OM = –CotA
ON PM

(And so on for the others)

vii. Trigonomertical ratios of (360°-A)

Let, A revolving line OP makes an angle ‘A’ with OX at the position of OQ.

ie. \ POX = A Y

\ QOX = A P
\ \ QOX (reflex) = 360°-A

Draw the perpendiculars, 360°–A
PM^OX and QN^OX
\ \ OQN = \ QOY’ = A X’ A M X
OA N

Now, In DPMO and DQNO, A

i. \ PMO = \ QNO = 90° Q

ii. \ POM = \ QON = A

iii. OP = OQ = radii of same circle. Y’

\ DPMO ≅ DQNO (By AAS axiom)

\ PM = –QN and OM = –ON.

Then, In right angled DQON,

Sin (360°-A) = QN = –PM = –SinA
OQ OP

224 PRIME Opt. Maths Book - IX

Cos (360°-A) = QN = OM = CosA
OQ OP

TanA (360°-A) = QN = –PM = –TanA
ON OM

(And so on others can be calculated)

viii. Trigonomertical ratios of (360°+A)

Let, A revolving line OP makes an angle ‘A’ with OX Y

and again ‘A’ with OX at the position of OQ.

ie. \ POX = A P, Q
A MX
\ QOX = A 360°+A
\ \ QOX (reflex) = 360°+A

X’ O
Here, OP and OQ lines at same place (coincide) so far

both angles ‘A’ and 360°+A, trigonometric ratios will

be same as follows, Q
Y’
Sin(360°+A) = RM = PM = SinA
OP OP

Cos(360°+A) = OM = OM = CosA
OQ OP

Tan(360°+A) = QM = PM = TanA
OM OM

(And so on for the others)

ix. Trigonomertical ratios of (-A)

Let, A revolving line OP makes an angle ‘A’ with OX and again it makes an angle (-A)

with OX at the position of OQ. Y
ie. \ POX = A (anti-clockwise +ve)

\ QOX = -A (clockwise -ve) P
Draw the perpendiculars,

PM^OX and QN^OX

\ \ OQN = \ QOY’ = A X’ O A N X
Now, In DPMO and DQNO, –A M

i. \ PMO = \ QNO = 90°

ii. \ POM = \ QON = A Q
iii. OP = OQ = radii of same circle.

\ DPMO ≅ DQNO (By AAS axiom) Y’
\ PM = -QN and OM = -ON.

Then, In right angled DQNO,

Sin (-A) = QN = –PM = – SinA
OQ O

Cos(–A) = ON = OM = CosA
OQ O

Tan(–A) = QN = –PM = –TanA
ON OM

PRIME Opt. Maths Book - IX 225

Cosec(–A) = OQ = OP = –CosecA
QN –PM

Sec(–A) = OQ = OP = SecA
ON OM

Cot(–A) = ON = OM = –CotA.
QN –PM

x) Generalization of the angles modified from standard angles as above:
Sign rule (CAST RULE)
Above derivation of the different forms of angles belonging to the quadrants, we

Conclude that:
1st quadrent → All trigonometrical ratios are positive
2nd quadrent → only Sin and Cosec are positive(S)
3rd quadrent → only Tan and Cot are positive (T)
4th quadrent → only Cos and Sec are positive (C)

Y

(90° + A, 180° – A) S (360° + A, 90° – A)
2nd Quadrent (Sin + Cosec) 1st Quadrent (All)
A

X’ O X

T C
3rd Quadrent (Tan + Cot) 4th Quadrent (Cos + Sec)

(180° + A, 270° – A) Y’ (360° – A, 270° + A)

5.9.1 Multiple of 90° rule:

Any kind of angles can be derived in the form of (90° × n + A) where n is any constant

(Natural number).
• If ‘n’ being even, the trigonometrical ratios will not be changed.
• If ‘n’ being odd, the trigonometrical ratios will be changed as,

Sin → Cos Cot →Tan Cos → Sin
Cosec → Sec
Tan → Cot Sec → Cosec

Examples: Y
Sin(270° + A)
= Sin(90° × 3 + A) X’ SA X
= Sin (90° × odd + A) TC
It belongs to 4th quadrent.

= – CosA Y’

Y

Tan(180° + A) = Tan(90° × 2 + A) X’ SA X
= Tan(90° × even + A) TC
It belongs to 3rd quadrent.
= TanA
Y’



226 PRIME Opt. Maths Book - IX

Cos 120° = Cos(90° × 1 + 30°) Y

= Cos(90° × odd + 30°) SA
TC
It belongs to 2nd quadrent. X’ X

Tan 780° = –Sin30° Y’

=- 1
2

= Tan(90° × 9 – 30°) Y
Cos 150° = Tan (90° × odd – 30°)
It belongs to 1st quadrent. X’ SA X
= –sin30° TC
=3
Y’

Cot 150° = Cos(90° × 1 + 60°) Y
= Cos(90° × odd + 60°)
It belongs to 2nd quadrent. X’ SA X
= Cot30° TC
=3
Y’

= Cot(90° × 2 - 30°) Y
= Cot (90° × even + 60°)
It belongs to 2nd quadrent. X’ SA X
= -Cot 30° TC
=- 3
Y’

Worked out Examples

1. Find the value of Sin(–1140°)

Solution:

Sin (-1140) = –Sin 1140

= –Sin(90° × 12 + 60°)

= –Sin 60°”

=3

2. Prove that: Cos 55° + Sin 75° = Cos 15° + Sin 35°
Solution:
L.H.S = Cos 55° + Sin 75°
= Cos(90° × 1 – 35)+ Sin (90° × 1 – 15°)
= Sin 35° + Cos15°
= R.H.S

PRIME Opt. Maths Book - IX 227

3. Prove that: Sin 420° Cos (–390°) + Cos 300. Sin(–330) = 1

Solution:

L.H.S = Sin 420° . Cos(–390) + Cos 300° . Sin(–330)

= Sin(90° × 4 + 60°) . Cos(90° × 4 + 30) + Cos(90° × 3 + 30°) [\Cos(–q) = Cosq]

= Sin 60° . Cos30° + Sin 30° (Cos 60°)

= 3 × 3 + 1 × 1
2 2 2 2

= 3 + 1
4 4

= 4
4
=1

= R.H.S.

4. Prove that: Cos r + Cos 5r + Cos 11r + Cos 15r =0
16 16 16 16
Solution:
r 5r 11r 15r
L.H.S = Cos 16 + Cos 16 + Cos 16 + Cos 16

= Cos r + Cos 5r + Cos a 16r - 5r k + Cos a 16r - r k
16 16 16 16

= Cos r + Cos 5r + Cos ar - 5r k + Cos `r - r j
16 16 16 16

= Cos r + Cos 5r + Cos a90°×2 - 5r k + Cos `90°×2– r j
16 16 16 16

= Cos r + Cos 5r - Cos 5r - Cos r
16 16 16 16

=0
= R.H.S

5. Tan r Tan 3r Tan 5r Tan 7r Tan 9r =1
20 20 20 20 20
Solution:
r 3r 5r 7r 9r
L.H.S = Tan 20 Tan 20 Tan 20 Tan 20 Tan 20

= Tan r Tan 3r Tan 45°.Tan a 10r – 3r k .Tan a 10r – r k
20 20 20 20

= Tan r Tan 3r × 1 × Tan a r – 3r k .Tan ` r – r j
20 20 2 20 2 20

= Tan r Tan 3r .Tan a90°×1 – 3r k .Tan `90°×1 – r j
20 20 20 20

= Tan r .Tan 3r + 3r .Cot r
20 20 20 20

= (Tan 2p0 .Cot 2p0 ) (Tan 3p .Cot 3p )
20 20

= 1 × 1

= 1

= R.H.S

228 PRIME Opt. Maths Book - IX

6. Prove that: Sin(90° + A).Tan(180° – A).Cosec(360° – A) = –1
Solution: Cos(180° – A).Tan (180° – A).Cosec (360° – A)

L.H.S = Sin(90° + A).Tan(180° – A).Cosec(360° – A)
Cos(180° – A).Tan (180° – A).Cosec (360° – A)

= Sin(90° × 1 + A).Tan(90° × 2 – A).Sec(90° × 3 – A)
Cos(90° × 2 – A).Tan (90° × 2 + A).Cosec (90° × 4 + A)

= CosA(–1TanA).(–CosecA)
(–CosA).TanA.CosecA

= – CosA.TanA.CosecA
CosA.TanA.CosecA

= -1

= R.H.S

7. Prove that: Sin2 r + Sin2 3r + Sin2 5r + Sin2 7r =2
8 8 8 8
Solution:
p
L.H.S = Sin2 8 + Sin2 3r + Sin2 5r + Sin2 7r
8 8 8
4r + r
= Sin2 p + Sin2 3r + Sin2 a 8 k + Sin2 a 4r + 3r k
8 8 8
r r r 3r
= Sin2 p + Sin2 3r + Sin2 ` 2 + 8 j + Sin2 a 2 + 8 k
8 8
r 3r
= Sin2 p + Sin2 3r + Sin2 `90°×1 + 8 j + Sin2 a90°×1 + 8 k
8 8

= Sin2 p + Sin2 3r + Cos2 p + Cos2 3r
8 8 8 8

= (Sin2 p + Cos2 p ) + (Sin2 3r + Cos2 3r )
8 8 8 8

= 1+1

= 2

= R.H.S

8. Find the value of ‘x’ from the followings equation:

xSec(180° – A) Cosec(270° + A) – Sin(90° + A). Cos(360° – A) = Cos(270° – A) Sin(180°

+ A) + x Tan(270° + A). Tan(180° – A)

or, x Sec(90° × 2 – A). Cosec(90× 3 + A) –Sin(90° × 1 + A). Tan(90° × 4 – A) =

Cos(90× 3 – A). Sin(90° × 2 + A) + x Tan(90° × 3 + A). Tan(90° × 2 – A)

or, x (–SecA). (–SecA) – CosA . CosA = (–SinA) (–SinA) + x(–CotA). (–TanA)

or, x Sec2A – Cos2A = Sin2A + x × 1

or, x (Sec2A – 1) = Cos2A + Sin2A

or, x= 1
Tan2 A

\ x = Cot2A.

PRIME Opt. Maths Book - IX 229

9. Find the value of ‘A’ from SinA = Cos2A.
Solution:
SinA = Cos2A
or, SinA = Sin(90° – 2A)
or, A = 90° – 2A
or, 3A = 90°

90° 30°
or, A = 3

\ A = 30°

Exercise 5.9

1. i) Show the sign rule in a quadrant.
ii) Write down the conditional of trigonometric ratio changed or not according to

CAST rule in a quadrant.
iii) Simplify Sin(270° – q) by showing in quadrant.
iv) Simplify Tan(180° – q) by showing in quadrant.
v) Simplify Cos(540° – q) by showing in quadrant.

2. Find the value of the following: ii) Cos(-780°)
i) Sin 240° iv) Cot(-1200°)
iii) Tan 960°
v) Cosec 1830°

3. Find the value of the followings:

i) Sin2120° + Cos2150° – Tan2135°

ii) 34 Tan2150°- 1 Cos 180° + Tan 135°
2
1
iii) 2Cos2135° + Sin2150° + 2 Cos 180° + Tan 135°

iv) Sin2135° Cos2120° - Sin2120°Cos2135°

v) Sin2 r + Sin2 3r + Sin2 5r + Sin2 7r
4 4 4 4

4. Find the value of the followings.
i) Cos2150° + Sin2120° + Sin2150° + Cos2120°
ii) Sin 300°.Cos(-330°) + Cos 420°.Sin(-390°)
iii) Cos 330° Sin 300° – Sin 330° Cos 660° 2Sin 750°
iv) Cos 510° Sin 600° – Cos 420° Sin (–630°)
v) Sin570°Cos420° + Cos750°Sin600°

5. Prove that the following:
i) Tan 32° + Cos 53° – Cosec 70° = Sin 37° + Cot 58° – Sec 20°
ii) Tan 9° Tan 27° Tan 45° Tan 63° Tan 81° = 1

230 PRIME Opt. Maths Book - IX

iii) Cos 112° + Sin 74° + Cos 68° – Sin 106° = 0
iv) Cos 306° + Cos 234° + Cos 162° + Cos 18° = 0
v) Cos 20° + Cos 40° + Cos 140° + Cos 160° = 0
vi) Cot 9° Tan 27° – Tan 81° Cot 63° = 0

6. Prove that the followings:
Sec (360°–i) Cos (180°–A) Sin (90° + A)

i) Tan (270° + i) × Cot (360°–A) × Cos (270° – A) = TanA

ii) TSainn((118800°°–+ii)) . Cos (360° + i) . Cot (–i) = – Sinq
Sin (360°–i) Tan (90° + i)

iii) STiann(1(18800°°+–AA)) . Cot (270°–A) . Cos (360°–A) = SinA
Tan (90°–A) Sin (–A)

iv) TanC(o1t(8207°0–°q–).qC)o.Ct(o9t(01°8–0q°)–.Cqo)s.S(3in60(–°q+) q) = –Tanq

v) Sin(90°+q).Cos(180°–q).Tan(360°–q) = Sin2(90°–q).Cot(270°–q)

7. Prove that the followings:

i) Cos102° + Cos126° + Sin36° + Sin12° = 0.
p
ii) Cos 8 + Cos 3r + Cos 5r + Cos 7r = 0.
8 8 8

iii) Sin 2r + Sin 3r – Sin 10r – Sin 11r = 0.
13 13 13 13

iv) Cos 3r + Cos 5r + Cos 11r + Cos 13r = 0.
16 16 16 16

v) Cos 24° + cos 48° + Cos 132° + Cos 156° = 0

8. Prove that the following:
r 3r 5r 7r
i) Sin2 8 + Sin2 8 + Sin2 8 + Sin2 8 =2

ii) Cos2 r + Cos2 3r + Cos2 5r + Cos2 7r =2
16 16 16 16

iii) Tan12° Tan36° Tan84° Tan108° = 1
r 3r 5r 7r 9r
iv) Cot 20 Cot 20 Cot 20 Cot 20 Cot 20 = 1.

v) Sin2 3r + Sin2 5r + Sin2 11r + Sin2 13r = 2.
32 32 32 32

9. Find the value of ‘x’ from the followings:
p p p p p
i) x Sin 4 Cos 4 . Tan 4 – Tan2 4 + Cos2 4 = 0

ii) x Cotq. Tan(270° + q) = Tan (90° + q). Cot (180 – q) + x Sec(90° + q). Cosec(180° – q)

iii) Tan(270° + A) Sin(180° + A). Cosec(270° – A) = x Sin (180° + A) – Cos2(90° + A)

Tan2(450° + A)

iv) 2Cot120°. Tan135° Cos180° – x sin120° Cos180° = xTan150°

v) x Cos(90° + q). Cos(90° – q) + Tan(180° – q). Cot(90° + q) = Sin(180° – q). Sin(360° + q)

PRIME Opt. Maths Book - IX 231

10. Find the value of the angle ‘A’ from the followings:

i) 3 TanA + 1 = 0 ii) 4CoaA + 2 = 0

iii) SinA + CosA = 0 iv) Sin2A + SinA = 0

v) Cot3A + TanA = 0

11. PRIME more creative questions: 2r 5r
2r 5r 7 7
i) Prove that : 1 + Cos 7 Cos 7 = Sin Sin .
ii) If A, B and C are the
angles of a triangle, prove that: Tan a A + B k = Cot C .
2 2

iii) If A, B, C and D are the angles of a quadrilatral, prove that: Cos a A + B k +
2
C +D
Cos a 2 k = 0.

iv) If P, Q, R and S are the angles of a quadrilateral, prove that CosA + CosB + CosC +

CosD = 0.
v) If x Cos(180° + q). Cot(270° + q) = Sin(90° + q) + Cosec(270° – q), find the value

of ‘x’.

Answer

1. Show to your teacher.

2. i) c– 3 m ii) 12 iii) 3 iv) 13 v) 2
2 v) 2
v) – 1
3. i) 1 ii) – 14 iii) – 14 iv) – 14 v) Tan2q
2 v) 45°

4. i) 2 ii) –1 iii) 1 iv) 1 14
2

9. i) 3 ii) Cot2q iii) SinA iv) 4
5

10. i) 150° ii) 120° iii) 135° iv) 180°

11. v) –Tanq

232 PRIME Opt. Maths Book - IX

5.10 Compound Angle

(A + B) and (A – B) are taken as the compound angle of the angles ‘A’ and ‘B’ . Here, we
discussed about the trigonometrical ratios of such compound angles A + B, A – B, A + B +
C, A – B – C etc.

5.10.1 Trigonometrical ratios of (A + B) Z

Let, a revolving line OY makes and angle A with OX Q
N
and OZ makes an angle B with OY.

i.e. ∠YOX = A, ∠ZOY = ∠B P
A
Taking a point p on OZ, Draw the perpendiculars Y
R X’
PM^OX, PQ^OY, QN^OX & QR^PM
B 233
where, A
M
∠RQO = ∠YOX = A

∠PQR = 90° – A

∠RPQ = 90° – ∠PQR

= 90° – (90° – A) = A. O

Then,

In right angled DPMO, ∠POM = A + B

i) Sin(A + B) = PM
OP

= PR + RM
OP

= PR + QN
OP OP

= PR PQ + QN OQ
PQ × OP OQ × OP

= CosA × SinB + SinA × CosB

\ Sin(A+B) = SinA.CosB + CosA.SinB

ii) Cos(A+B) = OM
OP

= ON – MN
OP

= ON – RQ
OP OP

= ON × OR – RQ PQ
OR OP PQ × OP

\ Cos(A+B) = CosA.CosB – SinA.SinB

iii) Tan(A+B) = Sin (A + B) = SinA.CosB + CosA.SinB
Cos (A + B) CosA.CosB – SinA.SinB

Dividing each terms by CosA.CosB = SinA.CosB + CosA.SinB
PRIME Opt. Maths Book - IX CosA.CosB

CosA.CosB – SinA.SinB
CosA.CosB

Tan(A+B)= TanA + TanB
1 – TanA.TanB

Cot(A+B) = Cos (A + B)
Sin (A + B)

CosA.CosB +SinA.SinB

= SinA.SinB
SinA.CosB + CosA.CosB

SinA.SinB

\ Cot(A+B) = CotA.CotB – 1
CotB + CotA

5.10.2 Trigonometrical ratios of (A – B)

Sin(A – B) = SinACosB – CosASinB

Cos(A – B) = CosACosB + SinASinB

Tan(A – B) = TanA – TanB
1 + TanA.TanB

Cot(A – B) = CotA.CotB + 1
CotB – CotA

5.10.3 Trigonometrical ratios of A + B + C

i) Sin(A + B + C)
= Sin{(A + B) + C}
= Sin(A + B)CosC + Cos(A + B)SinC
= (SinACosB + CosASinB) CosC + (CosACosB – SinASinB)SinC
= SinACosBCosC + CosASinBCosC + CosACosBSinC – SinASinBSinC

ii) Cos(A + B + C)
= Cos{(A + B) + C}
= Cos(A + B)CosC – Sin(A + B)SinC
= (CosACosB – SinASinB)CosC – (SinACosB + CosASinB)SinC
= CosACosBCosC – SinASinBCosC – SinACosBSinC – CosASinBSinC

iii) Tan(A + B + C)

= Tan {(A + B) + C}
Tan(A + B) + TanC
= 1 – Tan(A + B).TanC

TanA + TanB + TanC
1 – TanA.TanB
=
TanA + TanB
1– 1 – TanA.TanB TanC

= TanA + Tan B + TanC – TanA.Tan B.TanC
1 – TanA.TanB – TanB.TanC – TanC.TanA

234 PRIME Opt. Maths Book - IX

iv) Cot(A + B + C)

= Cot{(A + B) + C}

= Cot(A + B).CotC –1
CotC + Cot(A + B)

CotA.CotB – 1 × CotC – 1
CotB + CotA
=
CotA.CotB – 1
CotC + CotB + CotA

= CotA.CotB.CotC. – CotA – CotB – CotC
CotA.CotB + CotB.CotC + CotC.CotA – 1

Worked out Examples

1. Find the value of Sin15° without using calculator.

Solution :

Sin15° = Sin(45° – 30°)

= Sin45°Cos30° – Cos45°Sin30°

= 1× 3 – 1 × 1
2 2 2 2

= 3 – 1
22

2. If SinA = 3 , CosB = 12 , find the value of Sin(A + B).
5 13
Solution :

SinA = 3 , CosB = 12
5 13

\ CosA = 1 – Sin2 A , SinB = 1 – Cos2 B

= 1 – a 3 2 = 1 – a 11 2
5 13
k k

= 4 = 5
5 13
Then,

Sin(A+B) = SinA.CosB + CosA.SinB

= 3 × 12 + 4 × 5
5 13 5 13

= 36 + 20
65

= 56
65

3. If A + B = 45°, prove that (CotA – 1)(CotB – 1) = 2

Solution: Taking, A + B = 45°

or, Cot(A + B) = Cot45°

or, CotA.CotB – 1 =1
CotB + CotA

or, CotA.CotB – 1 = CotB + CotA

or, CotA.CotB – CotB – CotA = 1

PRIME Opt. Maths Book - IX 235

Adding ‘1’ on both sides,
or, CotA.CotB – CotB – CotA + 1 = 1 + 1
or, CotB(CotA – 1) –1(CotA – 1) = 2
or, (CotA – 1) (CotB – 1) = 2
L.H.S. = R.H.S. Proved

4. Prove that the followings.

i) Tan7A – Tan4A – Tan3A = Tan7A.Tan4A.Tan3A

ii) Cot20°.Cot50° + Cot50°.Cot110° + Cot110°.Cot20° = 1

Solution:

i) We have,

7A = 4A + 3A

or, Tan(7A) = Tan(4A + 3A)
Tan4A + Tan3A
or, Tan7A = 1 – Tan4A.Tan3A

or, Tan4A + Tan3A = Tan7A – Tan7A.Tan4A.Tan3A

or, Tan7A.Tan4A.Tan3A = Tan7A – Tan4A – Tan3A

\ Tan7A – Tan4A – Tan3A = Tan7A.Tan4A.Tan3A.

L.H.S. = R.H.S.

ii) We have,

or, 20° + 50° + 110° = 180°
or,
20° + 50° = 180° – 110°
or,
Cot(20 + 50) = Cot(180° – 110°)
or, Cot20°. Cot50° – 1
or, Cot50° + Cot20° = – Cot110°

Cot20°.Cot50° – 1 = – Cot50°.Cot110° – Cot110°.Cot20°

Cot20°.Cot50° + Cot50°.Cot110° + Cot110°.Cot20° = 1

L.H.S. = R.H.S.

5. If an angle A is divided into two parts a and b that bTana = aTanb, prove that Sin(a

– b) = a – b SinA.
a + b

Solution : aTanb aSinbCosa
Tana SinaCosb
Here, bTana = aTanb or, b= =
Then,

R.H.S. = a – b SinA
a + b

a– Sinb.Cosa
Sina.Cosb
= SinA
aSinb.Cosa
a+ Sina.Cosb

= a(Sina.Cosb – Sinb.Cosa) SinA
a(Sina.Cosb + Sinb.Cosa)

= Sin(a – b) Sin(a + b) Sin(a – b) = L.H.S. proved
Sin(a + b)

236 PRIME Opt. Maths Book - IX

Exercise 5.10

1. i) Write down the formula of Sin(A + B).

ii) Write down the formula of Cos(a – b).

iii) Write down the formula of Tan(A + B).

iv) Prove that Cot(A + B) = CotA.CotB – 1
CotB + CotA

v) Prove that Tan(a + b) = Tana – Tanb
1 + Tana.Tanb

2. i) If SinA = 4 , SinB = 5 , find the value of Cos(A + B).
5 13

ii) If CosA = 1 , CosB = 3 , find the value of Sin(A – B).
2 2

iii) If 3TanA = 4, and TanB = 5 find the value of Tan(A + B)
12

iv) If Cota = 3 and Cotb = 2, find the value of Tan(a – b).

v) If Tan(A + B) = 1 and Tan A = 1 , find the value of TanB.
2

3. Find the value of the followings with out using calculator.

a) i) Sin 75° ii) Cos 15° iii) Tan 105°

iv) Cos 285° v) Sin 555°

b) i) Tan 15° + Cot 15° ii) Sin 15° + Sin 75°

iii) Cos 105° + Cos 165° iv) Sin 165° + Sin 75°

v) Cos 195° + Sin 285°

4. Prove that the following using given condition.

a) i) If A + B = 45°, prove that : CotA.CotB – CotA – CotB = 1

ii) If A + B = 45°, prove that : (1 + TanA)(1 + TanB) = 2

iii) If a + b = 225°, Prove that : (Cota – 1)(Cotb – 1) = 2
p
iv) If A + B = 4 , prove that : TanA(TanB + 1) = 1 – TanB

v) If a + b = p , prove that : Cotb (Cota – 1) =1
4 (Cota + 1)

b) i) If A + B + C = 180°, prove that : TanA + TanB + TanC = TanA.TanB.TanC

ii) If a + b + g = p, prove that : Cota.Cotb + Cotb.Cotg + Cotg.Cota = 1

iii) If A + B + C = p, prove that : Tan A .Tan B + Tan B .Tan C + Tan. C
Tan A = 1 2 2 2 2 2

2
iv) If A + B + C = p and CosA = CosB.CosC, prove that : TanA = Tan B + TanC.

v) If A + B + C = p and CosA = CosB + CosC, Prove that : CotB.CotC = 1
2

c) i) If TanA = 3 , TanB = 1 , prove that A + B = 45°.
4 7

ii) If TanA = m , Tan B = 1 , prove that A + B = rc .
m+1 2m + 1 4

PRIME Opt. Maths Book - IX 237

iii) If Cota = 11 and Cotb = 6 , prove that a + b = rc .
5 4

iv) If Sina = 1 & Cosb = 2 , show that a + b = 45°.
10 5

v) If Cosq = 4 & Cosf = 7 , show that q – f = rc .
5 52 4

5. Prove that the followings:
a) i) Sin(A + 45°) = 1 (SinA + CosA)

2

ii) Cos(A – 45°) = 1 (CosA – SinA)
2

iii) Sin(A + 45°) – Sin(A – 45°) = 2 CosA.

iv) Sin(A + 30°) = 1 ( 3 SinA + CosA)
2

v) Cos( p + q) + Cos( p – q) = Cosq.
3 3

b) i) Tan (45° + a) = Cosa + Sina
Cosa – Sina

ii) Cot ( p – q) = Cosi + Sini
4 Cosi – Sini

iii) Cos(60° + A) + Cos(60° – A) = 0

iv) Sin( 2r + A) + Sin( 2r – A) – 3 CosA = 0
3 3
4Tani
v) Tan(45° + A) – Tan(45° – A) = 1 – Tan2 i

c) i) Sin(A + B) + Sin(A – B) = 2SinA.CosB
ii) Cos(A + B) + Cos(A – B) = 2CosA.cosB
iii) Sin(A + B) – Sin(A – B) = 2CosA.SinB

iv) Cos(A – B) – Cos(A + B) = 2SinA.SinB

v) Tan(A + B) – Tan(A – B) = 2Sec2A.TanB
1 – Tan2A.Tan2B
d) i)
ii) 1 – Tan13°Tan32° = Tan13° + Tan32°
iii) (1 + Tan22°) (1 + Tan23°) = 2
iv) (Cot20° – 1) (Cot25° – 1) = 2
v) Cot 100° + Cot 35° + Cot 100°.Cot35° = 1
e) i) (Tan55° – 1°) (Tan80° – 1) = 2
ii) Tan7A – Tan4A – Tan3A = Tan7A.Tan4A.Tan3A
iii) Tan70° – Tan50° – Tan20° = Tan70°.Tan50°.Tan20°
Cot10°.Cot15° – Cot15°.Cot25° – Cot25°Cot.10° = 1
iv)
CotA.Cot2A – CotA.Cot3A – Cot3A.CotA = 1
v) Sin25°.Cos20° + Cos25°.Sin20° = 1

f) i) 2
ii) Tan20° + Tan75° + Tan85° = Tan20°.Tan75°.Tan85°
Cot10°.Cot70° + Cot70°Cot100° + Cot100°Cot10° = 1
238
PRIME Opt. Maths Book - IX

iii) CCooss1100°° – Sin10° = Tan 35° iv) Cos14° + Sin14° = Cot31°
+ Sin10° Cos14° – Sin14°

v) Cos8° + Sin8° = Tan53°
Cos8° – Sin8°

g) i) Sin (A – B) = TanA – TanB ii) Sin2(45° + q) + Sin2(45° – q) = 1
CosA.CosB iv)
Sin2 A + Sin2 B
iii) Cot(45° + A).Cot(45° + A) = 1 Tan(A+B).Tan(A–B)= Cos2 B – Sin2 A

v) Sin3A + Cos3A = Sin5A
Sec2A Co sec 2A

h) i) Tan70° = 2Tan50° + Tan20° ii) Tan50° = 2Tan10° + Tan40°
iii) Cos18° – Sin18° = 2 Sin27° iv) Cos10° + Sin10° = 2 Sin55°
v) Tan(a + b) – Tana = Tanb

1 + Tan(a + b).Tana

6. PRIME more creative questions:
a) i) Prove that : Sin(A + B + C) = CosA.CosB.CosC(TanA + TanB + TanC –

TanA.TanB.TanC)
ii) Prove that : SinA.Sin(B – C) + SinB.Sin(C – A) + SinCSin(A – B) = 0
iii) Prove that : Tan(a – b) + Tan(b – g) + Tan(g – a) = Tan(a – b).Tan(b –

g). Tan(g – a)
iv) If A + B + C = p and CosA = CosB.CosC, prove that : TanA = TanB + TanC
v) If A + B + C = p and SinA = SinB.SinC, prove that : CotB + CotC = 1

b) i) Prove that : Sin2A + Sin2B + 2SinASinB.Cos(A + B) = Sin2(A + B)
ii)
iii) Prove that : Sin (A – B) + Sin (B – C) + Sin (C – A) =0
CosA.CosB CosB.CosC CosC.CosA

iv) An angle q is divided into A and B such that TanA.TanB = m:n, prove
v)
that : Sin(A – B) = ` m – n j .Sinq.
m + n Sin (a
+ b) k +1
If Tana = kTanb, prove that : Sin (a – b) = k –1

If 2TanA + CotA = TanB, prove that: 2Tan(B – A) = CotA.

7. Project work
Collects the formula used in trigonometry in a chart paper and present into the

classroom and paste at the project board.

Answer

1. Show to your teacher.

2. i) 16 ii) 12 iii) 1 1156 iv) – 17 v) 13
65

3. i) 6+ 2 ii) 6 + 2 iii) – (2 + 3) iv) 6– 2 v) 2– 6
4 4 4 4

PRIME Opt. Maths Book - IX 239

Co-ordinate Geometry

Unit Test - 1

Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. What is radian angle? Write down its value in degrees.

2. a. Express 35° 42’ 54” into centesimal measurement.

b. Prove that: nC,ofsin8pd+sCeoc2sA38–πTa+nC2Aos. 5π + Cos 7π =0
c. If m SinA = 8 8

3. a. Prove that: TanA + CotA = SecA.CosecA + 1
1 – CotA 1 – TanA

b. If x Cotq. Tan(90° + q) = Tan(270° + q). Cot(180° – q) – xSec(90° – q).
Cosec(360° – q), find the value of x.

4. If the angles of a triangle in degrees, grades and radians respectively are in the ratio
288: 280: p, find the angles of the triangle in degrees.

Unit Test - 2 Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. Write down the formula to find central angle of a circle with the usual meaning of the
symbols used.

2. a. Find the radius of a circle forms a central angles of 60° with an arc of 8.8 cm.

b. Prove that: 1 – Cos30° = 7 – 4 3
1 + Sin60°

c. Prove that: 1 – Cos4 A = 2Cosec2A – 1
Sin4 A

3. a. If 5Cosq + 12Sinq = 13, find Cotq. 5π 7π
p 3π 8 8
b. Prove that: Sin2 8 + Sin2 8 + Sin2 + Sin2 = 2.

4. Prove that: (3 – 4Sin2q) (1 – 3Tan2q) = (3 – Tan2q) (4Cos2q – 3)

240 PRIME Opt. Maths Book - IX

Unit 6 Vector Geometry

Specification Grid Table

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

No. of Questions 1 2 – 1 4 10 12
Weight 1 4 – 5

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

Objectives : At the end of the lesson
• Student are able to denote the vector and scalar with examples.
• Students are able to identify the types of vectors.
• Students are able to find the operation of vectors like addition, subtraction

multiplication by scholar, unit vector etc.
• Students are able to find magnitude and direction of vectors.
• Students are able to the simple geometry involving vector.

Materials Required:
• Chart paper.
• Graph paper.
• Flash card.
• Sample of position vector of a point in chart paper.
• Chart of types of vector and operation on vector.
• Chart of vector to find magnitude and direction.

PRIME Opt. Maths Book - IX 241

6.1 Vector and Scalar

Out of different physical quantities in our daily life, some of them have magnitude only
while some of the others have magnitude as well as direction. The physical quantities
which have only magnitude are the scalar quantities and which have magnitude as well
as direction are called vector quantities.

Vector:
The physical quantity which has magnitude as well as
direction is called vector. Eg : displacement, velocity, force,
acceleration, etc.

Scalar:
The physical quantity which has only magnitude is called scalar
quantity. Eg : distance, area, speed, density, mass, length etc.

6.1.1 Representation of vector & Scalar:

• Vector quantity is represented by using a directed line segment with arrow.
B

A , AB , a etc.

• Scalar quantity is represented by a line segment only.
A B, AB, a etc.

• The line segment having speci�ic direction is called directed line segment.

PQ

• Vector is represented by using two components ‘x’ and ‘y’ as

AB = e x - Component o
y - Component

Where, for a point A(x, y), components of OA are Y

(Projection of OA on x-axis) A(x, y)
MX
(Projection of OA on y-axis)

x – component = OM = x

y – component = AM = y

∴ OA = d x–Comp n = d yx n
y–Comp
O

242 PRIME Opt. Maths Book - IX

• If A(x1, y1) and B(x2, y2) are given vector AB is calculated as shown in diagram.
Y

N’ B (x2, y2)

(x , y ) y

1 1 P
NX
M’ A x

OM

Where,

Here,

AP = x – Comp. = MN = ON – OM = x2 – x1 (projection of AB on x-axis)
BP =y – Comp. = M’N’ = BN – PN = BN – AM = y2 – y1 (projection of AB on y-axis)
\ AB =
x - comp. = dx2 – x1 n
d - n y2 – y1
y
comp.

• Study the following graph having directed line segment.

Here, From graph we get,

AB = x - comp. = 3 B
dn dn
y - comp. 2 R

PQ = x - comp. = 3 A M
dn dn P A(4, 3)
y - comp. –4 3
N
RS = dx - comp. n = –4 S 4M X
y - comp. dn
–4 Y

MN = dx - comp. n = –3 Q O
y - comp. dn

5

6.1.2 Position Vector of a point:
In the vector OA , the initial point is the origin and final

point is A(4, 3).

Where,

x - component = OM = 4

y - component = AM = 3

And position vector of a point A = OA = 4
dn
3

The position vector of a point P(x, y) is the vector

OP where ‘O’ is the origin and OP = dx n .
y

PRIME Opt. Maths Book - IX 243

6.1.3 Types of vector Y
B(4, 4)
i) Column vector:

The vector whose components are written in a column

is called column vector.

In the diagram, A(1, 2)
O M3 N
AB = x - comp. = 4–1 = 3 X
dn dn dn
y - comp. 4–2 2

ii) Row vector : Y
The vector whose components are written in a row B(5, 7)

is called row vector. A(1, 2) C

In the given diagram, OM 4 N X

AB = (x - comp. y - comp)
= (4 6)

iii) Zero vector (Null)
The vector having magnitude zero is called zero

vector whose components are also zero.
i.e. AA = d0 n , BB = d0 n

00
The direction of zero vector is not fixed.

iv) Equal vectors:

Any two vector having same direction as well as equal magnitude are called equal

vectors.

Eg : A B P Q

a a

Here, AB = PQ = a

v) Negative vectors:

Any two vectors having equal magnitude but in opposite direction are called negative

vectors.

Eg: A B A B

Here, a –a

AB = a

BA = a (in opposite direction)

i.e. AB = a = – (– a ) = – BA

\ AB & BA are negative vector

i.e. AB = – BA

244 PRIME Opt. Maths Book - IX

vi) Co-intial vectors: B
A
The vectors having same initial point are called co-initial vectors.

Here, The vectors OA and OB have the same initial point ‘O’.

Hence, they are co-initial vectors.

But OA ≠ OB O

vii) Like vectors
Any two vectors having same direction are called like

vectors. The like vectors may be different in their
magnitude.

viii) Unlike vectors
Any two vectors having opposite direction are called

unlike vectors. The unlike vectors may be different in
their magnitude.

ix) Parallel vectors

Any two vectors which are in the form of multiplication

with a constant number to each other are called parallel

vectors.

i.e. a = mb and b = na .

They are like as well as unlike.

Example :

If p = d2n and q = d6 n ,
1 3

Here, q = d6n = 3 d2 n = 3p which is same as
3 1

a = mb for a ||b

\ p ||q is proved



Worked out Examples

1. If a point P(3, 4) is directed by a line segment from the origin, find OP and its

components.

Solution : Initial point is 0(0, 0)

Final point is P(3, 4)

\ x-component = x2 – x1 = 3 – 0 =3
= 4 – 0 =4
y - component = y2 – y1

Then, dx - comp n = 3
OP = y - comp cm
4

PRIME Opt. Maths Book - IX 245

2. If A(1, 2) and B(7, 10) are any two points, find components of AB & vector AB .
Solution :
Initial point is A(1, 2)
Final point is B(7, 10)
Then

x - component = x2 – x1 = 7 – 1 = 6
y - component = y2 – y1 = 10 – 2 = 8

\ AB = dx - comp n = 6
dn
y - comp 8

3. If A(–3, 2), B(1, 4), P(5, 1) and Q(1, –1) are the four points, prove that AB = –PQ .

Also prove that they are parallel to each others.

Solution :

The given points are A(–3, 2), B(1, 4), P(5, 1) and Q(1, –1)

Then,

AB = dx2 – x1n = d1 – (–3)n = d4n
y2 – y1 4–2
2

PQ = dx2 – x1n = d 1–5 n = d –4 n
y2
– y1 –1–1 –2 Again,
AB = – PQ is same as
Here, a = mb for a ||b

AB = d4 n = – d–4n = – PQ \ AB || PQ proved.

2 –2

\ AB = – PQ

4. If P(a, 3), Q(7, 1), R(5, 3), S(2, 5) and PQ = RS , find the value of ‘a’.
Solution:

Taking the points P(a, 3) & Q(7, 1)

PQ = dx2 – x1n = 7–a = 7–a
y2 – y1 cm cm
1–3
–2

Taking the points R(5, 3) & S(2, 5)

RS = dx2 – x1n = 2–5 = –3
y2 – y1 dn cm
5–3
2

Also, Taking

PQ = – RS

or, 7 – a = – –3
c m cm
–2 2

or, 7 – a = c 3 m
c m
– 2 –2

246 PRIME Opt. Maths Book - IX