Infinity Optional mathematics book 9 Final for CTP 2077

(v) Is cos2q + sin2q = 1 a trigonometric identity?
A mathematical statement which is always true for every condition or for any values
of its variable is known as an identity. Some examples of algebraic identities are :

(i) (x + 2)2 = x2 + 4x + 4, (ii) (x – 5) (x + 5) = x2 – 25

The mathematical statement which is true only for particular values of variable is
called an equation. x + 3 = 5, x2 – 9 = 0 etc. are equations because the first relation is
true only for x = 2 and second relation is true for x = ± 3. sin2θ + cos2θ = 1, secθ cosθ
= 1, tan2θ = sec2θ – 1 etc, are the trigonometric identities because these relations

are true for any value of θ.

WORKED OUT EXAMPLES

1. Prove the followings identities:

(a) sinθ.cotθ = cosθ

(b) tan2θ . cos2θ + cot2θ . sin2θ = 1 (c) cot2θ – cos2θ = cot2θ . cos2θ

(d) 1 – cos4A = 1 + 2cot2A
sin4A

(e) (asinA + bcosA)2 + (bsinA – acosA)2 = a2 + b2

(f) 1 – cosθ = cosec q – cotq
1 + cosθ

(a) sinθ.cotθ = cosθ

Solution:

(a) L.H.S. = sinθ . cotθ

= sinθ. cosθ = cosθ = R.H.S. proved.
sinθ

(b) tan2θ . cos2θ + cot2θ . sin2θ = 1

L.H.S. = tan2θ . cos2θ + cot2θ . sin2θ

= sin2θ .cos2θ + cos2θ . sin2θ
cos2θ sin2θ

= sin2θ + cos2θ

= 1 = R.H.S. proved.

Infinity Optional Mathematics Book - 9 201

(c) cot2θ – cos2θ = cot2θ . cos2θ

L.H.S. = cot2θ – cos2θ

= cos2θ – cos2θ
sin2θ

= cos2θ – cos2θ .sin2θ
sin2θ

= cos2θs(1in–2θsin2θ)

= cos2sθin. 2cθos2θ
= cos2θ . cot2θ

= R.H.S. proved.

(d) 1 – cos4A = 1 + 2cot2A
sin4A

L.H.S. = 1 – cos4A
sin4A

= sin14A – cos4A
sin4A

= cosec4A – cot4A

= (cosec2A – cot2A) (cosec2A + cot2A)
= 1(1 + cot2A + cot2A)
= 1 + 2cot2A

= R.H.S. Proved

(e) (asinA + bcosA)2 + (bsinA – acosA)2 = a2 + b2

L.H.S. = (asinA + bcosA)2 + (bsinA – acosA)2

= a2sin2A + 2absinA.cosA + b2cos2A + b2sin2A – 2absinA.cosA + a2cos2A

= a2sin2A + a2cos2A + b2sin2A + b2cos2A

= a2(sin2A + cos2A) + b2(sin2A + cos2A)

= a2 + b2 = R.H.S. proved.

202 Infinity Optional Mathematics Book - 9

(f) 1 – cosθ = cosec q – cotq
1 + cosθ

L.H.S. = 1 – cosθ = 1 – cosθ × 1 – cosθ
1 + cosθ 1 + cosθ 1 – cosθ

= (1 – cosθ)2 = (1 – cosθ)2
1 – cos2θ sin2θ

= 1 –sincoθsθ = = 1 θ – cos θ = cosecθ – cotθ =R.H.S. proved.
sin sin θ

(g) sin2θ . tan2θ + cos2θ . cot2θ = tan2θ + cot2θ – 1

L.H.S. = sin2θ . tan2θ + cos2θ . cot2θ

= sin2θ . sin2θ + cos2θ . cos2θ = sin4θ + cos4θ
cos2θ sin2θ cos2θ sin2θ

= sin6θ + cos6θ = (sin2θ)3 + (cos2θ)3
sin2θ.cos2θ sin2θ.cos2θ

= (sin2θ + cos2θ)3 – 3sin2θ.cos2θ(sin2θ + cos2θ)
sin2θ.cos2θ

= 1 –si3ns2iθn.2cθo.sc2oθs2θ

= sin2θ1.cos2θ – 3sin2θ.cos2θ
sin2θ.cos2θ

= cosec2θ . sec2θ – 3
= (1 + cot2θ) (1 + tan2θ) – 3
= 1 + cot2θ + tan2θ + tan2θ . cot2θ – 3
= tan2θ + cot2θ + 1 + 1 – 3
= tan2θ + cot2θ – 1

= R.H.S. proved.

(h) sin2a . sec2β + tan2β . cos2a = sin2α + tan2β

L.H.S. = sin2a . sec2β + tan2β . cos2a

= sin2a (1 + tan2β) + tan2β (1 – sin2a)

= sin2a + sin2a .tan2β + tan2β – sin2a . tan2β

Infinity Optional Mathematics Book - 9 203

= sin2a + tan2β
= R.H.S. proved.

(i) (1 + cotθ + cosecθ) (1 + cotθ – cosecθ) = 2cotθ

L.H.S. = (1 + cotθ + cosecθ) (1 + cotθ – cosecθ)

= (1 + cotθ)2 – (cosecθ)2

= 1 + 2cotθ + cot2θ – cosec2θ

= 1 + 2cotθ – 1 [ cosec2q – cot2q = 1]

= 2cotθ

= R.H.S. proved.

(j) (2 – cos2A) (1 + 2cot2A) = (2 + cot2A) (2 – sin2A)
L.H.S. = (2 – cos2A) (1 + 2cot2A)

= (2 – cos2A) 1 + 2cos2A
sin2A

= (2 – cos2A) sin2A + 2cos2A
sin2A

= (2 – cos2A) (sin2A + 2cos2A)
sin2A

= sin22A – cos2A (sin2A + 2 – 2sin2A)
sin2A

= (2cosec2A – cot2A) (2 – sin2A)

= (2 + 2cot2A – cot2A) (2 – sin2A)

= (2 + cot2A ) (2 – sin2A)

= L.H.S. proved.

(k) sseeccAA – tanA + 1 = 1 + secA + tanA
– tanA – 1 1 – secA – tanA

L.H.S. = secA – tanA + 1
secA – tanA – 1

= ((sseeccAA – tanA) + (sec2A – tan2A)
– tanA) – (sec2A – tan2A)

= ((sseeccAA––ttaannAA))+–((sseeccAA+– ttaannAA)) (sec A – tan A)
(secA + tanA)

204 Infinity Optional Mathematics Book - 9

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

= 11 +– secA + tanA
secA – tanA

= R.H.S. proved.

(l) 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) = cosec A + cot A
1 – cosecA + cotA

= si1nA + cosA
sinA

= 1 +sincoAsA

= R.H.S. proved.

(m) secA 1 tanA – 1 = 1 – secA 1 tanA
– cosA cosA –

L.H.S. = secA 1 tanA – 1
– cosA

= sseecc2AA – tan2A – secA
– tanA

= (secA – tanA) (secA + tanA) – secA
(secA – tanA)

= secA + tanA – secA

= tanA

R.H.S. = 1 – secA 1 tanA
cosA +

= secA – sec2A – tan2A
secA + tanA

Infinity Optional Mathematics Book - 9 205

= secA – (secA – tanA) (secA + tanA)
(secA + tanA)

= secA – secA + tanA

= tanA

\ L.H.S. = R.H.S. proved

(n) 1 + (cota.secb)2 = 1 + (cosa.tanb)2
1 + (cota.secθ)2 1 + (cota.tanθ)2

L.H.S. = 11 + (cota.secb)2
+ (cota.secθ)2

= 1 + cot2a.sec2b
1 + cot2a.sec2q

= 11 ++ cot2a(1 + tan2b)
cot2a(1 + tan2θ)

= 11 ++ cot2a + cot2a.tan2b
cot2a + cot2a.tan2θ

= ccoosseecc22aa + cot2a.tan2b
+ cot2a.tan2θ

1 + cos2a .tan2b 1 + cos2a .tan2b
sin2a sin2a sin2a
= cos2a .tan2q =
1 sin2a 1 + cos2a .tan2q
sin2a + sin2a

= 11 ++ (cosa tanb)2 = R.H.S. proved.
(cosa .tanq)2

Exercise 5.6

Section 'A'

1. Multiply the followings :
(a) (sinθ + cosθ) (sinθ – cosθ)
(b) (sinA + cosA) (sin2A – sinA.cosA + cos2A)
(c) (5tanθ – 3cosθ) (7tanθ – 2cosθ) (d) (1 + tanθ) (1 – tanθ) (1 + tan2θ)
(e) (2sin2θ + cos2θ) (3sin2θ – 2cos2θ)

2. Simply the followings :
(a) 5sinA + 3sinA – 7cosA – 2sinA – 4cosA

206 Infinity Optional Mathematics Book - 9

(b) 3sin2θ + 7sinθ – 2sin2θ – sinθ
(c) 3tan2A + 4sin2A – 7(tan2A – 2sin2A) – 2
(d) (sinA – cosA) (sinA + cosA) (sin2A – 2sinA.cosA + cos2A)

(e) (sinA – secA)2 – (sinA + secA)2 (f) cosA + cosA
1 – sinA 1 + sinA
1 1 tanA (h) 1 – t1anA tan A
(g) secA – tanA – secA + + tanA – 1

3. Factorise :

(a) sin2A – cos2A (b) cos4θ – cos2θ. tan2θ

(c) sec4θ – cos4θ (d) 4sin2θ – 5sinθ – 6

(e) cos6θ – sin6θ (f) sin4θ + 4cot4θ

4. Prove the following.

(a) tan A. cos A = sin A. (b) sin A. sec A = tan A
sin θ . cosec θ
(c) cosec θ . cos θ. tan θ = 1
5. Prove the following identities : (d) tanθ . cosθ = cosec θ

(a) (1+ cot2θ) sin2θ = 1

(b) (1 + cot2θ) (1 – cos2θ) = 1

(c) tanα + cotα = secα . cosecα

(d) cosθ + sinθ. tanθ = secθ

(e) tan2θ – cot2θ = sec2θ (1 – cot2θ)

((fg)) sceoccso4eAsce2–θc2sθ–e1c2A= cos2θ + tan2A
= tan4A

(h) (sinA + cosA)2 + (sinA – cosA)2 = 2

(i) sin4A + 2sin2A . cos2A + cos4A = 1

(j) secA 1 – cos2A = tanA

(k) cosθ 1 + cot2θ = cosec2θ – 1

(l) sin6A + cos6A = 1 – 3sin2A.cos2A

(m) (sinA – cosecA)2 + (cosA – secA)2 = cot2A + tan2A – 1

(n) 1 – 2sinA.cosA = sinA – cosA

Section 'B'

6. Prove the following:

(a) 1 – 1 = 1
sin2A tan2A

(b) sin A – cos A = 1
cosecA sec A

Infinity Optional Mathematics Book - 9 207

(c) cosec2 q – 1 = cos2 q
cosec2q

(d) 1 = sec A + tan A
sec A–tan A

(e) 1 – tan2A = cot2A – 1
1 + tan2A cot2A + 1

(f) 1 – 1 + 1 A = 2 tan A × sec A
1 – sinA sin

(g) 1 + tan q = cos q + sin q
1 – tan q cos q – sin q

(h) sin A – sin B + cos A – cos B =0
cos A + cos B sin A + sin B

(i) cot2q = cos2q
1 + cot2 q

(i) sin A + cos A + sin A – cos A = 2
sin A + cos A sin A + cos A sin2A + cos2 A

(k) 1 + sin4A = 2 sec2A – 1
cos4 A

(l) 1 – 1 = 2 (1 + tan2A)
1 + sin A 1 – sin A

(m) 1 + sin q = (sec q + tan q)2
1 – sin q

(n) 1 + tanq 2= 1 + tan2q
1 + cotq 1 + cot2 q

(o) tanq – sinq = 2 cot q
secq – 1 1 + cosq

(p) 1 cos A + 1 + sin A = 2 sec A
+ sin A cos A

(q) cos2A + sin2A =1
1 – tan2 A 1 – cot2 A

(r) 1 + cosq = tan2q
1 – cosq (secq–1)2

7. Prove the followings:

(a) 1 – sinθ = secθ – tanθ
1 + sinθ

(b) 1 + cosθ = cosecq + cotq
1 – cosθ

208 Infinity Optional Mathematics Book - 9

(c) 1 – cosθ = sin θ
1 + cosθ 1 + cos θ

(d) 1 – tan2A = 1 + tanA
1 – tan2A cot A–1

(e) 1 + sinq – 1 – sinq = 2 tanq
1 – sinq 1 + sinq

(f) 1 + cosq – 1 – cosq = 2 cot q
1 – cosq 1 + cosq

8. Prove the following

(a) secA – tanA + 1 = 1 – sinA
sec A + tanA + 1 cosA

(b) cosecA – cotA – 1 = 1 – cosA
1 – cosec A + cot A sinA

(c) 1 – sinq + cosq = 1 + cosq
sinq + cosq – 1 sinq

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

secA – tanA + 1 1 + sec A + tan A
(e) sec A – tanA– 1 = 1 – sec A – tan A

1 – cosec A + cot A cosec A + cotA –1
(f) 1 + cosec A – cot A = cosec A + cotA + 1

Section 'C'

9. Prove the following identities :
(a) (1 + sinA + cosA)2 = 2(1 + sinA) (1 + cosA)
(b) (sinA + cosecA)2 + (cosA + secA)2 = tan2A + cot2A + 7
(c) (1 + cotA)2 + (1 + tanA)2 = (secA + cosecA)2
(d) (cotθ – 3) (3cotθ – 1) = 3cosec2θ – 10cotθ

(e) cosA(2tanA + 1) (tanA + 2) = 2secA + 5sinA

(f) (1 + cotA – cosecA) (1 + tanA + secA) = 2
(g) (3 – 4sin2θ) (1 – 3tan2θ) = (3 – tan2θ) (4cos2θ – 3)
(h) (sin6A – cos6A) = (1 – 2cos2A) (1 – sin2A.cos2A)
(i) sin8α – cos8α = (2sin2α – 1) (1 – 2sin2α . cos2α)

(j) tanA + cotA = 1 + secA . cosecA
1 – cotA 1 – tanA

Infinity Optional Mathematics Book - 9 209

(k) 1 – 1 cotA = 1 cotA – 1
sinA cosecA + cosecA – sinA

(l) 1 + 3sinθ – 4sin2θ = (1 + 2sinθ)2 (1 – sinθ )

(m) 1 + (cosecA. tanB)2 = 1 + (cotA. sinB)2
1 + (cosecC.tanB)2 1 + (cotC.sinB)2

(n) cosθ + cosa = cosθ + cosa
sinθ + cosa sina – cosq sinθ – cosa sina + cosq

(o) (1 + sinα – cosα)2 + (1 – sinα + cosα)2 = 4(1 – sinα . cosα)

(p) (sin A + cos A)3 = 3 (sin A + cos A) – 2(sin3A + cos3A)

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

(r) (cosec θ – sinθ) (sec θ – cosθ) (tanθ + cotθ) = 1

(s) sin6A + cos6A = 1 – 3 sin2A. cos2A.

(t) sec4A (1 – sin4A) – 2tan2A = 1

(u) secθ (1 – sinθ) (sec θ + tanθ) = 1

(v) (secθ + tan θ – 1) (sec θ – tan θ + 1) = 2 tanθ
(w) 2(cos6θ + sin6θ) – 3(sin4θ + cos4θ) + 1 = 0

210 Infinity Optional Mathematics Book - 9

Trigonometrical Ratios of Any Angle Y

In the adjoining figure, the two straight lines XX' and P
YY' intersect at O. These two lines divide the entire + ve
plane into four equal parts called quadrants. O – ve X

XOY is called first quadrant, YOX' is called second X' P'
quadrant, X'OY' is called third quadrant and XOY' is
called fourth quadrant. In first quadrant, the range of
angle is 0º to 90º. In second quadrant, the range of the
angle is 90º to 180º. In third, 180º to 270º and the fourth
quadrant, the range of the angle is 270º to 360º.

If a revolving line OP revolves about O in the anticlockwise direction, the angle
formed is taken as positive where as when OP revolves about O in the clockwise
direction, the angle formed is taken as negative.

Sign of Trigonometrical Ratios of Angles in the Quadrants.

(i) When the revolving line OP lies in the first quadrant

Let OP be a revolving line Y

Let(x, y) be the coordinates of the point P1 When the P1(x, y)
revolving line OP reaches at P1 of first quadrant.
r
Form P1, P1M^OX is drawn. θ
Since the point P1 lies on the first quadrant, X' O MP X
OM = x = +ve

PM = y = +ve

OP1 = r = +ve

For, ∠MOP1 =θ Y'

sinθ = P1M = y = +ve
OP1 r

cosθ = OM = x = +ve
OP1 r

tanθ = P1M = y = +ve
OM x

Infinity Optional Mathematics Book - 9 211

cosecθ = OP1 = r = +ve
P1M y

secθ = OP1 = r = +ve
OM x

cotθ = OM = x = +ve
P1M y

Hence, all the trigonometrical ratios of an angle in the first quadrant are

positive. Y
(ii) When the revolving line OP lies in the second quadrant

Let OP be a revolving line. Let (x, y) be the P2(x, y) X
coordinates of the point P2 when the revolving line θ P
OP reaches at P2 of the second quadrant. From P2,
P2M^OX' is drawn. X' M O

Since P2 lies in the second quadrant. Y'
OM = x = –ve

P2M = y = +ve
OP2 = r = +ve
P2M
Now, sinθ = OP2 = y = +ve
r

cosθ = OM = x = –ve
OP2 r

tanθ = P2M = y = –ve
OM x

cosecθ = OP2 = r = +ve
P2M y

secθ = OP1 = r = –ve
OM x

cotθ = OM = x = –ve
P2M y
Y
Hence, in the second quadrant sinθ and cosecθ are

positive and the rest are negative.
(iii) When the revolving line OP lies in the third quadrant
Mθ
Let P3(x, y) be the position of the point P of the X' O PX
revolving line OP in the third quadrant.

Here, OM = x = –ve P3(x, y)
Y'
M P3 = y = –ve
OP3 = r = +ve

Now, sinθ = P3M = y = – ve
OP3 r

212 Infinity Optional Mathematics Book - 9

cosθ = OM = x = – ve
OP3 r

tanθ = MP3 = y = +ve
OM x

cotθ = OM = x = +ve
P3M y

cosecθ = OP3 = r = –ve
MP3 y

secθ = OP3 = r = –ve
OM x

Hence, in the third quadrant, tanθ and cotθ are positive and the rest are

negative. Y

(iv) When the revolving line OP lies in the four quadrants

Let P4(x, y) be the coordinates of the point P4. O M X
When the revolving line OP reaches at P4 of the P
fourth quadrant, then X'

θ

OM = x = +ve P4(x, y)

M P4 = y = –ve Y'
OP4 = r = +ve

Now, sinθ= MP4 = y = –ve cosecθ = OP4 = r = –ve
OP4 r MP4 y

cosθ = OM = x = + ve secθ = OP4 = r = +ve
OP4 r OM x

tanθ = MP4 = y = – ve cotθ = OM = x = –ve
OM x OP4 y

Hence, in the fourth quadrant, cosθ and Y
secθ are positive and the rest are negative.
2nd quadrant 1st quadrant
The sign of the six trigonometrical ratios sin and cosec (+ve) All (+ve)
of an angle of any magnitude can be
remembered from the adjoining diagram. X' S AX
TO C
Note: This rule is popularly known as CAST rule.

3rd quadrant 4th quadrant

tan and cot (+ve) cos and sec (+ve)

Y'

Infinity Optional Mathematics Book - 9 213

Trigonometrical Ratios of Negative Angle

Let OP be a revolving line.
Let P1(x, y) be the coordinates of the point P1. When the revolving line OP makes an
angle θ with X-axis in anti clock wise direction

Y

i.e. ∠P1OP = θ = P1M = y P1(x, y)
Then, sinθ OP1 r

cosθ = OM = x X' O M X
OP1 r –θ P

tanθ = P1M = y P2(x, –y)
OM x

cosecθ = OP1 = r Y'
P1M y

secθ = OP1 = r
OM x

cotθ = OM = x
P1M y

Let P2(x, –y) be the image of P1(x, y) when the reflection takes place about
X-axis.

Then ∠POP2 = – θ MP2 –y
OOMP2 r
Now, sin(– θ) = OP2 = = – sin q
cos(– θ) =
= x = cosθ
r

tan(– θ) θ=) OM=MPMO22 PP=22– r y = – tanθ
cosec(– =
r = – cosecθ
y

sec(– θ) = OP2 = r = secθ
OM x

cot(– θ) OM = x = – cotθ
= MP2 –y

Hence the trigonometrical ratios of negative angle is written as follows :

sin(– θ) = – sinθ

cosec(– θ) = – cosecθ

cos(– θ) = cosθ

sec(– θ) = secθ

tan(– θ) = – tan θ

cot(– θ) = – cotθ
214 Infinity Optional Mathematics Book - 9

Trigonometrical Ratios of Complementary Angle (90º – θ)
Let O be the centre of circle and r be its radius.

Let (x, y) be the coordinates of the ∠pPoi1nOtPP=1 when the Y
revolving line OP reaches at P1 Let θ.
P1(x,y)

Draw P1M ^OX. r 90° – θ
Then ∠OP1M = 90º – θ. θ
X' O M P X

Here, OM = x, P1M = y and OP1 = r.
Now, sinθ
cosθ = P1M = y
OP1 r
tanθ
= OM = x
OP1 r

= P1M = y
OM x

cosecθ = OP1 = r
P1M y

secθ = OP1 = r
OM x

cotθ = OM = x
P1M y

Again, OM = x = cosθ
sin(90° – θ) = OP1 r

cos(90° – θ) = O1M = y = sinθ
OP1 r

tan(90° – θ) = OM = yx = cotθ
O1M

cosec(90° – θ) = OP1 = xr = secθ
OM

sec(90° – θ) = OP = r = cosecθ
O1M y

cot(90° – θ) = P1M = y = tanθ
OM x

Hence the trigonometrical ratios of (90º – θ) are as follows :
sin(90º – θ) = cosθ

Infinity Optional Mathematics Book - 9 215

cos(90º – θ) = sinθ
tan (90º – θ) = cotθ
cosec(90º – θ) = secθ
sec (90º – θ) = cosecθ
cot (90º – θ) = tanθ

Trigonometrical Ratios of (90º + θ)

Let O be the centre of circle and r be the radius. Y

2(– y, x) + θ) P1(x, y)
(90°
Let (x, y) be the coordinates of the point P1 when the P
revolving line OP reaches at P1. Let ∠P1OP = θ.
X' θ PY
Y -coordinate of P1 y O
Then, sinθ = Radius OP1 = r

cosθ = X -coordinate of P1 = x
Radius OP1 r
Y'

tanθ = Y -cocrdinate of P1 = y
X-coordinate of P1 x

cosecθ = Radius OP1 = r
Y -coordinate of P1 y

secθ = Radius OP1 P1 = r
Y -coordinate of x

cotθ = X -cocrdinate OP1 = x
Y -coordinate of P1 y

Hence, P2(– y, x) is the image of P1 after rotation through + 90º about the origin O.
Then ∠P2OY = θ and ∠P2OP = 90º + θ.

Now, sin(90°­ + θ) = Y -coordinate of P2 = xr = cos θ
Radius of OP2

cos(90°­ + θ) = X -coordinate of P2 = –y = – sinθ
Radius of OP2 r

tan(90°­ + θ) = Y -cocrdinate of P2 = –xy = – cotθ
cosec(90°­ + θ) = X -coordinate of P2

Radius of OP2 = xr = secθ
Y -coordinate of P2

sec(90°­ + θ) = Radius of OP2 = r = – cosecθ
cot(90­° + θ) Y -coordinate of P2 –y = – tanθ

= X -cocrdinate of P2 = –xy
Y -coordinate of P2

216 Infinity Optional Mathematics Book - 9

Hence, the trigonometrical ratios of (90º + θ) are as follows :
sin(90º + θ) = cosθ
cos(90º + θ) = – sinθ
tan (90º + θ) = – cotθ
cosec(90º + θ) = secθ
sec (90º + θ) = – cocecθ
cot (90º + θ) = – tanθ

Trigonometrical Ratios of (180° – θ)

Let O be the centre of circle and r be its radius. Let (x, y) be the coordinates of the

point P1 when the revolving line OP reaches at P1. Y
Let ∠P1OP = θ P 2(–(y1,8x0) ° – θ)
Then,
sinθ = yr x
cosθ = r P1(x, y)

tanθ = xy cosecθ = r X' θθ x
y O P
secθ = xr x
cotθ = y

Y'

Let P2 be the image of P1 after reflection through Y-axis. Then the image of P2 is
(– x, y), ∠P2OX' = θ and ∠P2OP = 180º – θ.
Now,
sin(180º – θ) = y = sinθ

r
cos(180º – θ) = – x = – cosθ

r
tan (180º – θ) = y = – tanθ

–x
cosec(180º – θ) = r = cosecθ

y
sec (180º – θ) = r = – secθ

–x
cot(180º – θ) = – x = – cotθ

y
Hence the trigonometrical ratios of (180º – θ) are as follows :
sin(180º – θ) = sinθ
cos(180º – θ) = – cosθ
tan (180º – θ) = – tanθ
cosec(180º – θ) = cosecθ
sec (180º – θ) = – secθ
cot(180º – θ) = – cotθ
Trigonometrical Ratios of (180º + θ)

We know that, (180º + θ) = 90º + (90º + θ)

Infinity Optional Mathematics Book - 9 217

So, sin(180º + θ) = sin{90º + (90º + θ)} = cos(90 + θ) = – sin θ

cos(180º + θ) = cos{90º + (90º + θ)} = – sin(90 + θ) = – cosθ

tan(180º + θ) = tan{90º + (90º + θ)} = – cot(90 + θ) = tanθ

cosec(180º + θ) = cosec{90º + (90º + θ)} = sec(90 + θ) = – cosecθ

sec(180º + θ) = sec{90º + (90º + θ)} = cosec(90 + θ) = – secθ

cot(180º + θ) = cot{90º + (90º + θ)} = – tan(90 + θ) = cotθ.

Trigonometrical Ratios of (270º – θ)

We know that, (270º – θ) = 180º + (90º – θ)
So, sin(270º – θ) = sin{180º + (90º – θ)} = –sin(90 – θ) = –cosθ
cos(270º – θ) = cos{180º + (90º – θ)} = – cos(90 – θ) = –sinθ
tan(270º – θ) = tan{180º + (90º – θ)} = tan(90 – θ) = cotθ
cosec(270º – θ) = cosec{180º + (90º– θ)} = – cosec(90 – θ) = –secθ
sec(270º – θ) = sec{180º + (90º – θ)} = – sec(90 – θ) = –cosecθ
cot(270º – θ) = cot{180º + (90º – θ)} = cot(90 – θ) = tanθ.

Trigonometrical Ratios of (270º + θ)

We know that, (270º + θ) = 180º + (90º – θ)
So, sin(270º + θ) = sin{180º + (90º + θ)} = – sin(90 + θ) = – cosθ
cos(270º + θ) = cos{180º + (90º + θ)} = – cos(90 + θ) = sinθ
tan(270º + θ) = tan{180º + (90º + θ)} = tan(90 + θ) = – cotθ
cosec(270º + θ) = cosec{180º + (90º+ θ)} = – cosec(90 + θ) = – secθ
sec(270º + θ) = sec{180º + (90º + θ)} = – sec(90 + θ) = cosecθ
cot(270º + θ) = cot{180º + (90º + θ)} = cot(90 + θ) = – tanθ.

General rule for finding the trigonometrial ratios of any angle with
referred to 90º.

The trigonometric ratios of any angle can be expressed interms of acute angle by
expressing the angle in the form of (n × 90º ± θ) and the trigonometric ratios is
determined by the following steps.

Step 1 : First express the given angle in the form (n × 90º ± θ) where n is a whole
number.

Step 2 : Find the quadrant in which the angle terminates.

Step 3 : The sign of the value of the trigonometric ratios is determined according to
the quadrant where the angle terminates.

Step 4 : If n is even in the angle (n × 90º ± θ), there will be no change in the
trigonometric ratios.

For example : SA
(i) sin(180º + θ) = sin(90 × 2 + θ) = – sinθ TC
(ii) cos(720º+ θ) = cos(90 × 8 + θ) = cosθ
(iii) tan1860º = tan(90 × 20 + 60º) = tan60º = 3

218 Infinity Optional Mathematics Book - 9

Step 5 : In n is odd in the angle (n × 90º ± θ), then the trigonometrical ratios
change from sine to cosine, cosine to sine, tangent to cotangent, cotangent
to tangent, cosecant to secant and secant to cosecant

For example :
(i) sin(90 + θ) = sin(90 × 1 + θ) = cosθ
(ii) tan480º = tan(90 × 5 + 30) = – cot30º = – 3
(iii) cos1035º = cos(90 × 11 + 45) = sin45º = 1

2

The numerical values of the angles 0º, 30º, 45º, 60º, 90º, 120º, 135º, 150º and 180º are
given below:

Angles → 0º 30º 45º 60º 90º 120º 135º 150º 180º
Ratios ↓

sin 011 31 31 10

2 22 2 22

cos 1 31 1 0 – 1 –1 – 3 –1
2 2 2
2 22

Tan 0 1 1 3∞ –3 –1 1 0
Cosec
33

∞2 22 1 2 22 ∞
3
3

Sec 12 22 ∞ – 2 – 2 – 2 –1

33

Cot ∞ 3 1 1 0 – 1 – 1 – 3 ∞
33

WORKED OUT EXAMPLES

1. Find the complementary angle of 35º.

Solution: The complementary angle of 35º = 90º – 35º = 55º

2. Find the supplementary angle of 105º.

Solution : The supplementary angle of 105º = 180º – 105º = 75º

3. Prove the followings :

(a) cos80º – sin70º = sin10º – cos20º
(b) tan 10º . tan35º . tan55º . tan80º = 1
(c) sin40º . cos50º + cos50º . sin40º = 1
(d) sin112º + cos106º + cos74º – sin68º = 0

Infinity Optional Mathematics Book - 9 219

Solution :

(a) L.H.S. = cos80º – sin70º

= cos(90º – 10º) – sin(90º – 20º)

= sin10º – cos20º

= R.H.S. proved.

(b) L.H.S. = tan10º . tan35º . tan55º . tan80º

= tan10º . tan35º . tan(90º – 35º) . tan(90º – 10º)

= tan10º . tan35º . cot35º . cot10º

= tan10º . cot10º . tan35º . cot35º

=1×1

=1

= R.H.S. proved

(c) L.H.S. = sin40º . cos50º + cos50º . sin40º

= sin(90º – 50°) . cos50º + cos(90º – 50º) . sin50º

= cos50º . cos50º + sin50º . sin50º

= cos250º + sin250º

=1

= R.H.S. proved.

(d) L.H.S. = sin112º + cos106º + cos74º – sin68º

= sin(180º – 68º) + cos(180º – 74º) + cos74º – sin68º

= sin68º – cos74º + cos74º – sin68º

=0

= R.H.S. proved

4. Prove the followings :

(a) tanθ + tan(180º – θ) + cot(270º + θ) + cot(270º – θ) = 0

(b) sin2135º + cos2150º – tan2225º = 1
4
(c) sin(– θ) . cos(90 + θ) + cos(– θ) . sin(90 + θ) = 1
1
(d) sin420º . cos570º + cos(– 660º) . sin(– 690º) = – 2

(e) sinc(2o7s(03°6–0°θ)–.θc)o.sc(o1s8θ0.°c–otθ()9.0t°a+nθ()3.6c0o°s–ecθ()90. °se–cθ()– θ) = 1

(f) sin  + sin 3 – sin 3 – sin 7 = 0
8 8 8 8
 3 3 7
(g) tan 8 . tan 8 . tan 8 . tan 8 = 1
Solution:

(a) L.H.S. = tanθ + tan(180º – θ) + cot(270º + θ) + cot(270º – θ)

= tanθ + (– tanθ) + (– tanθ) + tanθ

= tanθ – tanθ – tanθ + tanθ

=0

= R.H.S. proved.

220 Infinity Optional Mathematics Book - 9

(b) L.H.S. = sin2135º + cos2150º – tan2225º

= sin2(90 × 1 + 45º) + cos2(90 × 1 + 60º) – tan2(90 × 2 + 45º)

= cos245º + sin260º – tan245º

= 1 2 3 2 – (1)2
2
2 +

= 1 + 3 – 1
2 4

= 2 + 3 – 4 = 1
4 4

= R.H.S. proved.

(c) L.H.S. = sin(– θ) . cos(90° + θ) + cos(– θ) . sin(90° + θ)

= – sinθ . (– sinθ) + cosθ . cosθ S A
T C
= sin2θ + cos2θ

=1

= R.H.S. proved.

(d) L.H.S. = sin420º . cos570º + cos(– 660º) . sin(– 690º)

= sin(90 × 4 + 60) . cos(90 × 6 + 30º) + cos660º .

(– sin690°)

= sin60º (– cos30º) – cos(90 × 7 + 30º) . sin (90 × 8 – 30º)

= – 3 × 3 – sin30º (– sin30º)
2 2
–3 1 1
= 4 + 2 × 2

= –3 + 1
4 4

= –2 = –1
4 2
= R.H.S. proved.

(e) L.H.S. = sin(270° – θ) . cos (180° – θ) . tan (360° – θ) . sec (– θ)
cos(360° – θ) . cosθ . cot(90° + θ) . cosec(90° – θ)

= sin(90 × 3 – θ) . cos(90 × 2 – θ) . tan (90 × 4 – θ) secθ
cos(90 × 4 + θ) . cosθ . cot(90 × 1 + θ) . cosec(90 × 1 – θ)

= (– cosθ) . (– cosθ) . (– tanθ) . secθ
cosθ . cosθ . (–tanθ) . secθ

= 1.

= R.H.S. proved

(f) L.H.S. = sin  + sin 3 – sin 5 – sin 7
8 8 8 8

= sin – 7 + sin  – 5 – sin 5 – sin 7
8 8 8 8

Infinity Optional Mathematics Book - 9 221

= sin 7 + sin58 – sin58 – sin78
8

=0

= R.H.S. proved

(g) L.H.S. = tan  . tan 3 . tan 5 . tan 7
8 8 8 8

= tan  . tan 3 . tan  +  . tan  + 3
8 8 2 8 2 8

= tan  . tan 3 . – cot8 . – cot38
8 8

= tan8 . cot8 . tan38 . cot38

=1×1=1

= R.H.S. proved.

5. Find the value of x :

tan(90° + θ) . cot(180° – θ) + x . sec(90 + θ) . cosecθ = xcotθ . tan(90 + θ)

Solution :

Here, tan(90 + θ) . cot180º – θ)+ x . sec(90 + θ) . cosecθ = xcotθ . tan(90 + θ)

or, (– cotθ) . (– cotθ) + x . (– cosecθ) . cosecθ = xcotθ . (– cotθ)

or, cot2θ – xcosec2θ = – xcot2θ

or cot2θ = xcosec2θ – xcot2θ

or, cot2θ = x(cosec2θ – cot2θ)

or, cot2θ = x × 1

\ x = cot2θ.
6. If A, B and C are the angles of a triangle ABC, show that

sin(A + B) = sinC

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

A + B = 180º – C.

L.H.S. = sin(A + B)

= sin(180º – C)

= sinC = R.H.S. proved

222 Infinity Optional Mathematics Book - 9

Exercise 5.7

Section 'A'

1. Find the complementary angle of the following angles :

(a) 45º (b) 50º (c) 10º (d) 70º (e) 35º
2. Find the supplementary angle of the following angles :

(a) 70º (b) 100º (c) 135º (d) 19º (e) 170º
3. Prove the followings :

(a) sin72º + cos36º = cos18º + sin54º
(b) tan55º – cos25º = cot35º – sin65º
(c) sin10º . sin70º . sin85º = cos5º . cos20º . cos80º
(d) sin10º . cos75º . sec80º . cosec15º = 1
(e) sin95º + cos105º – tan115º = sin85º – cos75º + tan65º
(f) sin20º + cos40º – tan80º – sin160º – sin50º – tan100º = 0
(g) tan81º . tan63º . tan45º . tan27º . tan9º = 1
(h) sin40º . sin140º – cos40º . cos140º = 1
4. Find the value of the followings :

(a) sin240º (b) sin330º (c) sin420º
(d) cos150º (f) cos1020º
(g) cos(– 570º) (e) cos480º (i) tan1410º
(j) tan(–1395º) (l) sec(– 405º)
(h) tan690º
(m) cot(– 600º) (o) cot960º.
(k) cosec(– 585º)

(n) cosec 11
4

Section 'B'

5. Find the values of the followings :

(a) 6sin150º + 2cos120º + tan135º
(b) cos150º + sin120º + sin2150º + cos2120º
(c) sin2120º – cos2120º – sin2135º – tan2150º
(d) tan245º – 4sin260º + 2cos245º + sec180º + cosec135º
(e) 3stiann22112500°°–+ccoost22112305°°

Infinity Optional Mathematics Book - 9 223

6. Prove the followings :

(a) cos150º + tan2135º + cos180º + sin120º = 0

(b) 34 cos2150º – 1 sec2135º + 3tan2150º – 1 sin2120º = 3
2 3 4
1
(c) cos290º + 2sin2120º – 2 tan2135º + cot2150º = 4

(d) sin135º . cos120º + cos135º . sin260º = – 5
4 2
1
(e) cos150º + sin120º + sin2150º + cos2120º = 2

7 Prove the followings :

(a) sin420º . cos390º + cos660º . sin480º = 3+ 3
4
(b) tan585º . cot405º + tan765º . cot675º = 0

(c) cos570º . sin150º – sin330º. cos390º = 0

(d) cos420º . sin390º – sin300º . cos(–330º) = 1

(e) sin780º . cos390º + sin(– 330º) . cos300º = 1

(f) sin225º . sin300º – cos315º . cos240º = 3 +1
22

(g) cos306º + cos234º + cos162º + cos18º = 0

8. Simplify :

(a) cot(90° – A) + tan(90° – A) – cosecA . secA
(b) tan2θ . sec2(90 – θ) – sin2θ . cosec2(90 – θ)
(c) cos(270º – A) . sec(180º – A) . sin(270º + A) . cos(450º – A)

(d) cos(s9i0n°A– A) × cosA A) . tanA A)
sin(90° – cot(90° –

(e) tsainn((118800°° – A) × cot(90° – A) × cot(– A)
+ A) tan(90° + A) cos(270° – A)

(f) ssiinn((19800°°+–θθ)) × sec(90° + θ) × cot(– θ) θ)
cot(180° + θ) cos(180° –

(g) sin(180ta°n+(2θ7).0c°o–t(θ1)8.0c°o–s(θ1)8.0t°a+n(θ9)0° + θ)

(h) ssiinn((19800°°––AA)) + tan(180° – A) – sec(90° + A)
tan(180° + A) cosec(180° – A)

(i) csoisn32((19800°°++AA)) × tan(360° – A) × cosec2(180° – A) × secA
sec2( – A) cos(180° – A) cosec(90° + A)

(j) tsainn((217800°° – A) × cot(270° – A) × cos(360° – A)
– A) tan(270° + A) sin(180° + A)

224 Infinity Optional Mathematics Book - 9

9. Prove the followings :

(a) cos  + cos 3 + cos 3 + cos 7 =0
8 8 8 8

(b) tan  + tan 3 + tan 5 + tan 7 =0
8 8 8 8

(c) cos2  + cos2 3 + cos2 5 + cos2 7 = 2
8 8 8 8

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

(e) cos2  + cos2 3 + cos2 5 + cos2 7 = 2
16 16 16 16

(f) tan  . tan 3 . tan 5 . tan 7 = 1
8 8 8 8

10. Find the value of x :

(a) 2cot120º – x sin120º . cos180º = tan150º
(b) tan2135º – sec260º = x . sin135º . cos45º . tan60º

(c) 3sin420º + x. cos120º . tan225º = x . cot120º

(d) xcosθ . cot(90° + θ) + sin(90° + θ) × cosec(90° – θ) = 0

(e) tan(90° + θ) . cot(180° – θ) = xcotθ . tan(90° + θ) – xsec(90° + θ). cosec θ.

(f) xsin(90°–θ).cos(90°–θ)–cos(180°–θ)cos(90°+θ) = sin(180°–θ) . sin(90° + θ)

11. If A, B, C are the angles of a triangle, show that

(a) sin(A + B) = sinC (b) cos AB = sin C
2+2 2

(c) tan(2B + 2C) + tan2A = 0 (d) cosec AC = sec B
2+2 2

(e) cosA + cosB + cosC + cos(A + B) + cos(B + C) + cos(C + A) = 0

12. Find the value of A from the following (0° ≤ A ≤ 90°).

(a) sinA = cos2A (b) sec7A = cosec3A (c) tan6A = cot3A

(d) sin5A = cos4A (e) cos7A = sin2A (f) sin A × cot A = sin 3A

13. If tan25º = P, show that: tan205° – tan115° = 1 + p2
tan245° + tan335° 1 – p2

Infinity Optional Mathematics Book - 9 225

5.3 Compound Angles

The sum or difference of two or more angles is called a compound angle. If A, B and
C are the three angles, then (A + B) or (A –B) or (A + B + C) or (A – B – C) etc are
the compound angles.

X'

Trigonometric ratios of (A + B) and (A – B)

Let XOX' and YOY' be X-axis and Y-axis respectively. R P (cosA, sinA)

Consider a unit circle (circle having radius 1 unit) B X'
A M
X' (1, 0)
O –B
with centre at origin O. Let the circle cuts X-axis

at M, then the coordinates of M are (1, 0). Let P be Q (cosB – sinB)

another point on the circumference of the circle such

that ∠POM = A. Then the coordinates of P are (cosA,

sinA).

Let R be another point on the circumference of the Note:

circle such that ∠ROP = B. Then ∠ROM = A + B. So, P (ON, PN)

coordinates of R are (cos (A + B), sin (A + B)). NO

Also take a point Q on the circumference of the circle cosA = ONOP= ON1
\ON = cosA
as shown in the figure making ∠QOM = ∠POR = B. SinA = PNPO= PN1
\ PN = sinA
But ∠QOM is subtended in –ve direction so ∠MOQ

= – B.

Hence, the coordinates of Q are (cos (–B), sin (–B))

= (cosB, – sinB).

RM and PQ are joined.

In the figure
PM/A, PR / B and MQ/ – B.
Now, PM + PR = PM + MQ
or, MPR = PMQ
\ chord RM = chord PQ. [Equal arcs cut of equal chords]
Now, RM2 = PQ2
or, [cos (A + B) – 1]2 + [sin (A + B) – 0]2 = (cosA ­– cosB)2 + (sinA + sinB)2
or, cos2 (A + B) – 2 cos (A + B) + 1 + sin2 (A + B) = cos2A – 2cosA . cosB + cos2B +

sin2A + 2sinA . sinB + sin2B)
or, 1 + 1 – 2 cos(A + B) = 1 + 1 – 2cosA . cosB + 2sinA.sinB
or, 2 – 2cos(A + B) = 2 – 2cosA . cosB + 2sinA . sinB

226 Infinity Optional Mathematics Book - 9

\ cos(A + B) = cosA . cosB – sinA . sinB ... (i)

Replacing –B by B in equation (i), We get
Cos(A – B) = cosA . cos(– B) – sinA . sin(– B)

\ cos(A – B) = cosA . cosB + sinA . sinB …(ii)

∵ [cos(– B) = cosB = Sin(– B) = – sinB]
Again, sin(A + B) = cos [90º – (A + B)]
= cos [(90º – A) – B]
= cos (90 – A) . cosB + sin (90 – A). sinB

\ sin(A + B) = sinA . cosB + cosA . sinB … (iii)

Similarly, sin(A­– B) = cos [90 – (A – B)] = cos [(90 – A) + B]

= cos (90 – A) . cosB – sin (90 – A) . sinB

\ sin(A – B) = sinA .cosB – cosA . sinB … (iv)

Again, Tan (A + B) = sin(A + B)
cos(A + B)

= sinA . cosB + cosA . sinB
cosA . cosB – sinA . sinB

= sinA . cosB + cosA . sinB
cosA . cosB

cosA . cosB – sinA sinB
cosA . cosB

sinA . cosB + cosA . sinB
cosA . cosB cosA . cosB
= [Dividing Nr andDr by cosA . cosB]
cosA . cosB sinA . sinB
cosA . cosB – cosA . cosB

= tanA + tan B
1 – tan A. tan B.

\ tan (A + B) = tanA + tan B … (v)
1 – tan A. tan B.

Tan (A – B) = sin(A – B)
cos(A – B)

= sinA . cosB – cosA . sinB
cosA . cosB + sinA . sinB

sinA . cosB – cosA . sinB

= cosA . cosB [Dividing Nr andDr by cosA . cosB]

cosA . cosB + sinA . sinB

cosA . cosB

= 1 tanA – tan B . Similarly,
+ tan A. tan B.

cot(A + B) = cotA . cotB –1 … (vii)
cot A + cotB

Infinity Optional Mathematics Book - 9 227

cot(A – B = cotA . cotB + 1 ... (viii)
cotB – cotA

Compound angles formulae

Sin(A + B) = sinA . cosB + cosA . sinB Sin(A ­– B) = sinA . cosB – cosA . sinB

cos(A + B) = cosA . cosB – sinA.sinB cos(A – B) = cosA.cosB + sinA.sinB

tan(A + B) = tanA + tanB tan(A – B) = tanA – tanB
1 – tanA . tanB 1 + tanA . tanB

cot(A + B) = cotA . cotB – 1 cot(A – B) = cotA . cotB + 1
cotB + cotA cotB – cotA

WORKED OUT EXAMPLES

1. Find the value of :
(a) Sin 75° (b) tan 105° (c) cos(– 285°)

Solution: (b) Tan105º
(a) sin 75°

= sin (45º + 30º) = tan (60º + 45º)

= Sin45º. Cos30º + cos45º. sin30º = tan60° + tan45°
1 3 1 1 1 – tan60° . tan45°
= 2 . 2 + 2 . 2
3 + 1
= 3 +1 = 1– 3 .1
22
= 3 +1 × 1+ 3
1– 3 1+ 3

(c) cos (– 285º) = 1+2 3 +3
1–3
= cos285º 2(2 + )
= –2 3
= cos (90 × 3 + 15)
= sin 15° = ­–­ (2 + 3 )
= sin (45° – 30°)
= sin 45°. cos 30° – cos 45°. sin30°

= 1 . 3 – 1 . 1
2 2 2 2

= 3 –1
22
3 153,
2. If sinA = 5 and cosB = find the value of sin(A – B) and cos (A + B)

Solution : Here, sinA = 3
5

cosA = 1 – sin2A = 1– 3 2 = 54.
5
cosB = 5
13

228 Infinity Optional Mathematics Book - 9

sinB = 1 – cos2B = 1– 52 = 1132.
13

Now, sin(A – B) = sinA . cosB – cosA . sinB Cos (A + B) = cosA.cosB – sinA . sinB

= 3 . 5 – 4 × 12 = 4 × 5 – 3 × 12
5 13 5 13 5 13 5 13

= 15 – 48 = 20 – 36
65 65 65 65
P=ro­–v635e3 that: ­–6156
sin2(45° – A) + sin2(45° + 1 =
3. A) =

Solution: = sin2(45° – A) + sin2(45° + A)
L.H.S.
= (sin45°.cosA – cos45°.sinA)2 + (sin45°.cosA + cos45°.sinA)2

= 1 cosA – 1 sinA 2 + 1 cosA + 1 sinA 2
2 2 2 2

= 1 cos2A – 2 . 1 cosA . sinA + 1 sin2A + 1 cos2A + 2 . 1 cosA . sinA + 1 sin2A
2 2 2 2 2 2

= 1 (cos2A + sin2A) + 1 (cos2A + sin2A)
2 2
1 1
= 2 + 2

= 1 = R.H.S. proved

4. Prove that : sinA + sin(A + 60°) + sin(A – 60°) = 2sinA

Solution :

L.H.S. = sinA + sin(A + 60º) + sin(A – 60º)

= sinA + sinA . cos60º + cosA . sin60º + sinA . cos60º – cosA . sin60º

= sinA + 2sinA . cos60º
1
= sinA + 2sinA . 2proved
= 2sinA = R.H.s.

5. Prove that : tanA + cot2A = cosec2A

Solution :

L.H.S. = tanA + cot2A

= sinA + cos2A
cosA sin2A

= sin2A . sinA + cos2A . cosA

cosA . sin2A

= cos(2A – A)
cosA . sin2A
cosA
= cosA . sin2A

= cosec2A = R.H.S. proved

Infinity Optional Mathematics Book - 9 229

6. Prove that : cos8° + sin8° = tan53°
Solution : cos8° . sin8°

R.H.S. = tan53º

= tan(45° + 8°)

= sin(45° + 8°)
cos(45° + 8°)

= sin45° . cos8° + sin8° . cos45°
cos45° . cos8° – sin8°. sin45°
11

= 2 cos8° + 2 sin8°
1 cos8° – 1 sin8°
22

= cos8° + sin8° = L.H.S. proved
cos8° – sin8°
7. Prove that : sin35° + sin55° = 2 sin80°

Solution :

L.H.S.= sin35º + sin55º

= sin(45º – 10º) + sin(45º + 10º)

= sin45º.cos10º – cos45º.sin10º + sin45º.cos10º + cos45º.sin10º

= 2 sin 45°. cos 10°

=2× 1 cos10º
2

= 2 cos (90º – 80º)

= 2 sin80º

= R.H.S. proved

8. Prove that : tan35° + tan10° + tan35° . tan10° = 1

Solution :
We know that 35º + 10º = 45º
Now, tan(35º + 10º) = tan45º
or, 1t–atnan353°5+°.ttaann1100° ° = 1
or, tan35º + tan10º = 1 – tan35º. tan10º
\ tan35º + tan10º + tan35º. Tan10º = 1 proved.

9. Prove that : cot57° . cot21° . cot12° = cot57° + cot21° + cot12°

Solution :

We know that : 57º + 21º + 12º = 90º
or, 57º + 21º = 90º – 12º
Now, cot(57° + 21°) = cot(90° – 12°) = tan12°

or, cocso5s75°7°. cot21° – 1 = 1
+ cot21° cot12°

or, cot57° . cot21° . cot12° – cot12° = cot57° + cot21°

\ cot57º. cot21º. cot12º = cot57º + cot21º + cot12º proved.

230 Infinity Optional Mathematics Book - 9

10. Show that : tan80° = 2tan70° + tan10°

Solution:
We know that, 80º = 70º + 10º

Now, tan80º = tan (70º + 10º)

or, tan 80° = tan70° + tan10°
1 – tan70° . tan10°

or, tan80º – tan80º.tan70º.tan10º = tan70º + tan10º

or, tan80º – tan80º. tan70º . tan(90° – 80º) = tan70º + tan10º

or, tan80º – tan80º . tan70º. cot80º = tan70º + tan10º

or, tan80º – tan70º = tan70º + tan10º

\ tan80º = 2tan70º + tan10º proved.

11. If tanA = k 1 and tanB = 1 1 , Show that : (A + B) = 
k+ 2k + 4
Solution :

We have, tanA + tanB
1 – tanA° . tanB°
Tan (A + B) =

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

2k2 + k + k + 1

or, tan(A + B) = (k + 1) (2k + 1)
2k2 + k + 2k + 1 – k
(k + 1) (2k + 1)

= 2k2 + 2k + 1
2k2 + 2k + 1

or, tan(A + B) = 1

or, tan(A + B) = tan45º

∴ A + B =  proved.
4

12. If sinA = 1 and sinB = 1 , then prove that : A + B = 45°
Solution : 10 5

Here, sinA = 1
10

Now, cosA = 1 – sin2A = 1– 12 = 3
10 10
1
. Sin B = 5

Now, cosB = 1 – sin2B = 1– 1 2= 2
55

Infinity Optional Mathematics Book - 9 231

Now, sin (A + B) = sinA . cosB + cosA.sinB

= 1 . 2+ 3 . 1
10 5 10 5

= 2+3
50 50

= 2 = 1
52 2
or, sin (A + B) = sin45º

\ A + B = 45º

13. Prove that :
sin(A + B + C) = sinA . cosB . cosC + cosA . sinB . cosC + cosA .

cosB . sinC – sinA . sinB . sinC

Solution :

L.H.S. = sin(A + B + C)

= sin{(A + B) + C}

= sin(A + B).cosC + cos(A + B). sinC

= (sinA . cosB + cosA . sinB) . cosC + (cosA . cosB – sinA . sinB) . sinC

= sinA . cosB . cosC + cosA . sinB . cosC + cosA . cosB . sinC – sinA . sinB . sinC

= R.H.S. proved.

14. If A + B + C = pc, prove that: tanA2 . tanB2 + tanB2 . tanC2 + tan C . tanA2 = 1
2
Solution :

Here, A + B + C = pc

or, A + B = pc – C
or, A2 B  C
+ 2 = 2 – 2 [Dividing both sides by 2)

Now, tan A + B = tan c – C
2 2 2 2

oorr,, 1ttaa–nnAA22ta+.ntA2taan.nC2tB2a+nBt2an=B2c.ottaC2n=C2t=a1n1C2­– tanA . tanB
2 2

\ tanA . tanB2 + tanB2 . tanC + tan C . tanA = 1 proved.
2 2 2 2

232 Infinity Optional Mathematics Book - 9

Exercise 5.8

Section 'A'
1. Find the value of the followings by using compound angle formulae.

(a) sin15º (b) cos75º (c) cos105º (d) sin75º

(e) tan15º (f) cot75º (g) sec75º (h) cosec105º

(i) cot105º (j) sin(– 165º) (k) tan(– 195º) (l) cos(– 255º)

2. Prove the followings :

(a) cos15º ­– sin15º = 1 (b) sin75º – sin15º = 1
2 2

(c) tan15º + cot15º = 4 (d) sin105º + cos105º = 1
2

(e) sin105º + cos75º = 3 (f) cos 75° + sin 75° = 3
2 2

3. (a) If sinA = 3 and cosB = 12 , then find the value of
5 13

(i) sin(A + B) (ii) cos(A + B)

(b) If cosA = 4 and sinB = 5 1 , then find the value of
5 2

(i) cos(A – B) (ii) sin(A – B)

(c) If tanA = 1 and tanB = 13, then find the value of
2

(i) tan(A + B) (ii) tan(A –­ B) (iii) cot(A + B) (iv) cot(A –­ B)
(iv) cot(A ­+ B)
(d) If sinA = cosB = 23, then find the value of
(i) sin(A + B) (ii) cos(A –­ B) (iii) tan(A – B)

4. Prove the followings :

(a) cos(A + 45º) = 1 (cosA – sinA) (b) sin(A – 45º) = 1 (sinA – cosA)
2 2

(c) cos θ –  = 1 (cosq + 3 sinθ) (d) tan(45º – B) = cosB – sinB
3 2 cosB + sinB

(e) cot(q + 45º) = cosq – sinq (f) sin(45º + A) – cos(45° – A) = 0
cosq + sinq

(g) sin(45º + A) + cos(45° + A) = 2 cosA.

(h) sin(45º + A).cos(45° – A) = 1 (1 + 2sinA.cosA)
2

Infinity Optional Mathematics Book - 9 233

(i) 2sin(A + 45°) . sin(A – 45°) = sin2A – cos2A

(j) tan(45° + A) . tan (45° – A) = 1 (k) cos(45° – A) = 1
(l) cos2(45° + A) + cos2(45° – A) = 1 sin(45° + A)

(m) sec2(45° ­– A) – cot2(45° + A) = 1

(n) sin2(45° –­ A) + sin2(45° + A) = 1 (o) cosec2(45°­+ A) – tan2(45° – A) = 1

5. Prove the followings :

(a) sin(A + B) + sin(A – B) = 2sinA.cosB

(b) cos(A – B) – cos(A + B) = 2sinA.sinB

(c) sin(A + B) . sin(A – B) = sin2A – sin2B

(d) cos(A + B) . cos(A – B) = cos2A – sin2B = cos2B – sin2A

(e) cot (A + B). cot (A – B) = cot2A . cot2B – 1
cot2B . cot2A
tan2A – tan2B sin2A – sin2B
(f) tan (A + B). tan (A – B) = 1 – tan2A . tan2B = cos2B – sin2A

(g) sinA.sin(B – C) + sinB.sin(C – A) + sinC.sin(A – B) = 0

(h) cosA + cos(120º + A) + cos(120º – A) = 0

(i) sinA + sin(A + 120º) + sin(A – 120º) = 0

6. Prove the followings :

(a) csoisnA(A. + B) = tanA + tanB (b) cos(A + B) = cotB – tanA
cosB cosA . sinB

(c) csoisnA(A. – B) = cotB – cotA (d) sin(A + B) . sin(A – B) = tan2A – tan2B
sinB cos2A . cos2B

(e) cos(Asi+n2BA) . cos(A – B) = cot2A – tan2B
. cos2B

(f) csoins((AA + B) – sin(A – B) = tanB
+ B) + cos(A – B)

(g) ssiinnA(A. – B) + sin(B – C) + sin(C s–inAC) = 0
sinB sinB . sinC sinA .

(h) sin(A + B + C) = cosA . cosB . cosC (tanA + tanB + tanC – tanA . tanB . tanC)

7. Prove the followings :
(a) sin3A . cosA + cos3A . sinA = sin4A
(b) cos130º . cos40º + sin130º . sin40º = 0

234 Infinity Optional Mathematics Book - 9

(c) sin25º . cos5º + cos25º . sin5º = 1
2

(d) tan25° + tan35° = 3
1 – tan35° . tan35

(e) cos(A – B) . cos(A + B) + sin(A – B) . sin(A + B) = cos2B

(f) sin(60° + q).cos(30° – q) + cos(60° + q) . sin(30° – q) = 1

(g) 1 tan5 – tan2 = tan3 α
+ tan5 . tan2

(h) cot(A – B) . cotB – 1 = cotA
cot(A – B) + cotB

(i) 1 + tan(A+ B) . tanB = cot A
tan(A+B) – tanB

(j) cos5 + sin5 = sin13 α
cosec8 sec8

(k) Cot 2q + tan q = cosec 2q

(l) (cosA + cosB)2 + (sinA – sinB)2 = 2 + 2cos(A + B)

Section 'B'

8. Prove the followings :

(a) cos20º – sin20º = 2 sin25º (b) cos55º + cos35º = 2 cos10º

(c) 3 cos10º + sin10º = 2sin70º (d) cos40º + 3 sin40º = 2cos160º

(e) ccooss1100°° – sin10° = tan35º (f) cos35° + sin35° = cot10º
+ sin10° cos35° – sin35°

(g) ccooss99°° + sin9° = tan54º (h) sin20° + cos20° = cot155º
– sin9° sin20° – cos20°

9. Prove the followings :
(a) 1 – tan5º . tan40º = tan5º + tan40º
(b) tan22º + tan23º + tan22º . tan23º = 1
(c) tan65º – tan20º – tan65º. Tan20º = 1
(d) cot25º. cot20º – cot25º – cot20º = 1
(e) tan8A – tan5A – tan3A = tan8A . tan5A . tan3A
(f) cot10A . cot6A + cot10A . cot4A – cot6A . cot4A + 1 = 0

Infinity Optional Mathematics Book - 9 235

(g) tan20º + tan72º + tan88º = tan20º . tan72º . tan88º
(h) tan10º + tan70º + tan100º = tan10º . tan70º . tan100º
(i) tan80º + tan65º + tan35º = tan80º . tan65º . tan35º
(j) tan40º . tan35º + tan40º . tan15º + tan35º . tan15º = 1
(k) cot57º . cot21º . cot12º = cot57º + cot21º + cot12º

10. Show that : (b) tan80º = 2tan70º + tan10º
(a) tan50º = tan40º + 2tan10º (d) 2tan50° + tan 20° = cot20°
(c) tan65º = tan25º + 2tan40º

11. If A + B = 4, then show that:
(a) tanA + tanB = 1 – tanA . tanB (b) (1 + tanA) (1 + tanB) = 2

12. If A + B = 225º then show that : (b) (1 – cotA) (1 – cotB) = 2
(a) cotA + cotB = cotA.cotB – 1

Section 'C'

13. (a) If tanA = 5 and tanB = 111, show that A + B = c
6 4
=k1 2
(b) If tanA and tanB = k, show that A + B = 2

(c) If cotA = m+ 1 and cotB = 2m + 1, show that A + B = c
m 4
c
(d) If sinα = 1 and sinβ = 1 , show that α + β = 4
10 5

(e) If cos q = 4 and cosβ = 7 , show that q + β = c
14. Expand: 5 52 4

(a) sin(A + B + C) (b) cos(A + B + C) (c) tan(A + B + C)
15. If A + B + C = pc, prove that:

(a) tan A + tanB + tanC = tanA . tanB . tanC

(b) cotA . cotB + cotB . cotC + cotC . cotA = 1

(c) cot A + cot B + cot C = cot A . cot B . cot C2
2 2 2 2 2

16. (a) An angle b is divided into two parts α and q such that tan α : tan q = a : b

Show that sin(α – q ) = a – b sinβ.
a + b
tanα m+1 = sin(α + β)
(b) If tanb = m, prove that m–1 sin(α – β)

236 Infinity Optional Mathematics Book - 9

UNIT VECTOR

6

6.1 Vector
Review

Discuss the following questions.
(a) What is your height?
(b) What is your school distance from your home?
(c) A man is driving a motorcycle with the speed of 40km per hour.
(i) How far does he reach in 2 hours?
(ii) Where does he reach in 2 hours?

Introduction to vector

Which of the following are measurable ?

Weight, force, distance, knowledge, density, area, volume, sentiment etc.

We often go to the market to buy necessary things we get things by measurement.
For example, we get rice in kilogram (kg.), cloth in metres, oil in litres and so on.
Similarly, when we walk from home to school, we may walk a certain distance such
as 1km to the east and so on. Thus, measurement is an essential aspect of physical
science and our daily life as well.

The quantities which we can measure are called physical quantities. We can
classify the physical quantities into two classes which are as follows.

(i) Scalar quantities or scalar

(ii) Vector quantities or vector
(i) Scalar :
From the above question (a) what is your height? If you say, that my height

is 5.5 inches or 5.6 inches it gives the complete, answer about height. In
general, what is the boiling point of water? Discuss on it.
The boiling point of water is 100°c. A number 100 and its unit (°c) completely
describes about the temperature. The physical quantities which have such
values are known as scalar. Length, mass, density, height, area, volume
temperature etc. are examples of scalar.

Infinity Optional Mathematics Book - 9 237

The physical quantity which have only magnitude but no direction is called
scalar quantity or simply scalar.

(ii) Vector

From above question (c) (i) the distance cover a man in 2 hours is (40km × 2)

= 80km. But we can't say about the position of a man in 2 hours. We are not

known about the direction of him. So, to know his position (displacement)

we should know the direction. The physical quantities which are depends

on direction are known as vector. Velocity, acceleration, displacement, force,

weight, pressure etc. are examples of vector.


The physical quantity which have both magnitude as well as direction is
called vector quantity or simply vector.


Difference between vectors and scalars


Vectors Scalars

1. Vectors have both magnitude 1. Scalars have magnitude only.
and direction.

2. They are represented by directed 2. No arrows are required in

line segment and with symbols symbols representing scalars like

with an arrow at the top like AB. AB.

3. They cannot be added or 3. They can be added or subtracted
subtracted by simple algebraic by simple algebraic methods.
methods i.e. they are operated
through vector geometry.

Notation of a vector:
A vector is generally denoted by bold faced type of letters. For example: a, PQ.
But due to inconveniency of writing it is not so used. So for a vector a letter or a

combination of two letters with an arrow over it is used. For example : a or PQ.

For a scalar, a letter or a combination of two letters without an arrow over it is
used. For example a or PQ are the scalar.

Representation of a vector: B
A vector is generally represented by a directed line segment. The
length of the line segment is chosen according to some scale to
represent the magnitude and the arrow to represent the direction

A

238 Infinity Optional Mathematics Book - 9

of the vector. A vector from point A to point B is denoted by AB. We call the point
A, the initial (starting) point or tail and B, the terminal (end) point or head of the

vector AB. The directed arrow (→) was first introduced by English scientist Sir
Isaac Newton (1642 – 1727A.D).

Representation of vector in terms of components:

Case I When initial point is at origin and terminal Y
point is at A(x, y)

Let, A(x, y) be a point in a Cartesian plane. Join the A(x, y)
origin O to A. Now, OA is directed from O to A. So,

OA is a vector. From A, draw AM perpendicular to O y
OX. Then, OM is called X-component or horizontal x MX
component of OA and AM is called Y-component or X'

vertical component of OA. Y'

Here, OM = x and MA = y.

To displace from O to M and again from M to A is same as to displace from O

to A.

Therefore, OA in terms of components is written as

OA = (OM, MA) = (x, y) or x
y

∴ OA = (x, y) or x
y


Here, OA is said to be position vector of A. x and y are called the x-component

and y-component of OA respectively.

Case II When initial point is at (x1, y1) and terminal point is D(x2, y2)

Let C(x1, x1) and D(x2, y2) be two points Y

in the plane and join C to D. Then CD is a E , y ) D(x2, y2)
vector. To find CD, we need its x-component F P
and y-component. For this, draw CM and 1 1

C(x

DN perpendicular to x-axis.

Similarly, DE and CF are drawn X' OM NX
perpendicular on y-axis. Y'

Here,

OM = x1 ON = x2
CM = y1 DN = y2

Infinity Optional Mathematics Book - 9 239

x- component MN = ON – OM = x2 – x1
and y- component EF = DN – PN

= y2 – CM [ PN = DM]

= y2 – y1

Now, CD = (CP, DP) = (x2 – x1, y2 – y1) or x2 – x1
y2 – y1

Hence, CD = x2 – x1 or (x2 – x1, y2 – y1)
y2 – y1

A vector whose x-component and y-component are arranged horizontally as

(x2 – x1, y2 – y1) is known as row vector. But a vector whose x and y components
x2 – x1
are arranged vertically as y2 – y1 is known as a column vector.



A vector can be express either in a row vector or in a column vector.

Note:

(i) If P(x, y) be a point then the position vector of P is OP = (x, y) or x
y

(ii) If C(x1, y1) and D(x2, y2) be two points, the position vector of C and D are

OC = x1 and OD = x2 .
y1 y2

WORKED OUT EXAMPLES

1. A point M(–2, 3). Find the position vector of M.
Solution: Here,

M(–2, 3) is a point in a plane. Joined O and M then OM is called position

vector.

x-component of OM = OQ = –2

y-component of OM = QM = 3

∴ OP = ( –2, 3) or –2
3

240 Infinity Optional Mathematics Book - 9

2. Find UV if U(5, –3) and V(3, 4)

Solution: Here,

Let, U(5, –3) = (x1, y1) and V(3, 4) = (x2, y2)

By formula UV = x2 – x1 = 3–5 = –2 = –2
y2 – y1 4 – (–3) 4+3 7

∴ UV = –2
7
3. Represent the following vectors by the directed line segment in a

square grid.

(i) AB = 2 (ii) CD = –4
3 6

(iii) ––74 = MN (iv) –84 = PQ

Solution: Here,

(i) AB = 2 B
3

x-component of AB = 2 and

y-component of AB = 3 A
Starting from A, 2 units right and 3 units

above will give the point B. Then directed line

segment AB will give 2
3

(ii) CD = –4 D
6

x-component of CD = – 4 and

y - component of CD = 6

Starting from C, 4 units left and then 6 units C

above will give the point D. Then directed line

segment CD will give –4
6

Infinity Optional Mathematics Book - 9 241

(iii) Here, M

MN = –4
–7

x-component of MN = –4

y-component of MN = –7

Starting from M, 4 units left and 7 units below N
P
will give the point N.

Then, MN will give –4 .
–7

(iv) Here,

PQ = 6
–8

x-component of PQ = 6

y-component of PQ = –8

Starting from point P, 6 units right and 8 units

below will give the point Q. Then the directed Q

line segment PQ will give 6 .
–8

4. From the given square grid, find x and y-component of the directed

line segment ST and EF . Also, find the vectors represented by them.

Solution:

From given graph, for ST E
Starting from S, x-component (right) = 4 and
T

y-component (above) = 6

∴ ST = (4, 6) or 4
6

For EF,

Starting from E, x -component (right) = 6 and S F

y-component (below) = –7

∴ EF = (6, –7) or 6
–7

242 Infinity Optional Mathematics Book - 9

Exercise 6.1

Section 'A'

1. (a) Define vector and scalar with examples.

(b) What are the directed line segments? Define initial point and terminal

point of directed line segment.

2. (a) What is the meaning of the word vector introduced in mathematics and

who used it arrow (→) as symbol for the first time?

(b) How can vectors be represented?

3. (a) Which of the following quantities are vectors and scalars? Write with

reason.

(i) velocity (ii) force (iii) temperature

(iv) time (v) area (vi) volume

(vii) mass (viii) weight (ix) acceleration

(x) speed (xi) displacement (xii) distance

(xiii) density (xiv) pressure

b. Write down the initial point and the terminal point of the following vectors.

(i) Y (ii) Y

SZ

X' O X X' O X
Y' H (iv)
(iii) Y'
G J

I

4. (a) If 'O' be the origin and the coordinates of H(4, 7), write OH in terms of
components.

(b) If 'O' be the origin and the coordinates of L(–4, –3), write OL in terms of

components.

5. (a) Find the vectors represented by the directed line segment joining the

following points.

(i) A(3, 5) and B(4, 7) (ii) C(2,3 ) and D(–3, 5)

(iii) E(–4, –3) and F(4, 9) (iv) G(–1, 2) and H(–3, 5)

(b) A vector displaces the point P(4, 3) to the point Q(3, 0). Express PQ in
row and column vector.

Infinity Optional Mathematics Book - 9 243

(c) A vector RS displaces a point R(–1, –6) to the point S(8, 11) then find
RS in row and column vector.

Section 'B'

6. Draw the directed line segment from origin to the points given below. Express

their position vectors both in row and column form.

(i) Z(3, 5) (ii) W(5, 0) (iii) V(0, 6)

(iv) M(–4, –3) (v) N(5, –8) (vi) (–3, –3)

7. Represent the following vectors by the directed line segment in the square

grid.

(i) AB = 0 (ii) CD = –5 (iii) MN = –4
–6 –6 6

(iv) PQ = 5 (v) UV = (5, 0) (vi) IJ = 5
–3 2

8. In the adjoining figure, some arrow diagrams corresponding to some points
are given. Find the x-component and y- component corresponding to each
directed line segment joining the origin and the given points and find the
position vectors corresponding to the given points.
Y

Z
A

X' O QX
R

M
N

Y'
9. Find the x-component and y-components of each directed line segment and

find the vectors represented by them.

244 Infinity Optional Mathematics Book - 9

Y

A B
Q E
S
F
H U

T
G

V X
P A

10. The points A, B, C and D are the vertices of a D B
parallelogram ABCD, Find the directed line C
segments represented by the points and also write
the equivalent pairs.

11. Divide all the students of class 9 in different groups. According to group, make
a list of physical quantities which are available in our surroundings (home,
society and school). Present the list of physical quantities from each group in
the class and discuss which are vectors and scalars quantities with reason.

Infinity Optional Mathematics Book - 9 245

Magnitude of a vector

Case I When initial point is at origin and terminal point is at point P(x, y)

Let, P(x, y) be any point in the plane. Join the Y P(x, y)
origin O to P. Now OP is directed from O to P.

Then OP is a vector whose magnitude means the y
xM X
length of OP is denoted by |OP | X' O
Draw PM perpendicular on OX.

We have, Y'

OM = x and PM = y

By using pythagoras theorem, in right angled DOPM

OP2 = OM2 + MP2­= x2 + y2

or, OP = x2 + y2

Hence, the length of OP = |OP | = x2 + y2

i.e. |OP | = (x-component)2 + y-component)2

Case II: When initial point is at p(x1, y1) and terminal point is at point
Q (x2, y2)

If P(x1, y1) and Q(x2, y2) be the two points. By joining P and Q and directed

from P to Q so that PQ is a vector. Y Q(x2, y2)
Now, R
P(x 1, y )
x-component of PQ = MN = PR = ON – OM
1

= x2 – x1 X' O M NX
Y'
and y -component of PQ = QR = BN – NR

= y2 – PM = y2 – y1

∴ The magnitude of a vector PQ = | OP |

= (x-component)2 + (y-component)2

or, |PQ | = (x2 – x1)2 + (y2 – y1)2

246 Infinity Optional Mathematics Book - 9

Direction of a vector

Case I When initial point is at origin and terminal point is at point P(x, y)

The direction of a vector in two dimensions is Y
determined by the angle it makes with the positive
x-axis. P(x, y)

Let, θ be the angle made by the vector OP with the

positive x-axis, i.e. ∠XOP = θ. Draw PM perpendicular y
with OX. Then,
q
x-component of OP = OM = x X' O x MX

y -component of OP = MP = y Y'

In right angled ∆POM

tanθ = PM p = y
OM b x

or, tanθ y-component of OP
=
x-component of OP

or, θ = tan-1 y
x

Case II When initial point is at p(x1, y1) and terminal point is at point

(x2, y2) Y

If P(x1, y1) be the initial point and Q(x2, y2) be y 1) Q(x2, y2)
q
the terminal point of the vector PQ then the P(x 1, yS
NX
direction of PQ = θ = tan-1 y = tan-1 y2 – y1
x x2 – x1
y2 – y1 X' OM
θ = tan-1 x2 – x1 Y'
∴

Note:
(i) If the given vector lies in 2nd, 3rd and 4th quadrant, first find the least

positive angle (θ).

(ii) S.N. x-component y-component Direction

1 Positive (+ve) Positive (+v) Acute angle (θ)

2. Negative (–ve) Positive (+ve) 180° – θ

3. Negative (–ve) Negative (–ve) 180° + θ

4. Positive (+ve) Negative (–ve) 360° – θ

Infinity Optional Mathematics Book - 9 247

Relation between vector components and direction (θ)

x-component y-component Direction

Positive (+ve) Zero (0) θ = 0°

Negative (–ve) Zero (0) θ = 180°

Zero (0) Positive (+ve) θ = 90°

Zero (0) Negative (–ve) θ = 270°

Types of vectors

(i) Row vector: A vector whose x-component and y-component are written
horizontally and enclosed within the round brackets is known as row vector.

For example: AB = (x2 – x1, y2 – y1), →a = (2, – 3), →b = (5, – 2).
(ii) Column vector
Can we write →a = (2, – 3) by another way?

We can write →a = (2, – 3) by →a = 2
–3

∴ A vector whose x-component and y-component are written vertically and
enclosed within the round brackets is known as a column vector.

For example:

(i) AB = x2 – x1 (ii) →a = 5 , →b = –7 etc
y2 – y1 –3 8

Note: A vector can be expressed either as a row vector or as a column vector.
(iii) Position vector:

A vector whose initial point is taken as origin is called Y
position vector of the terminal point. If origin (O) be A
the initial point, then the position vector of any point B

P referred to the origin is defined to the vector OP . O X
X' Y' D

In the figure, OA , OB , OC and OD are the C
position vectors of the terminal points A, B, C and D
respectively.

In position vector, the components of the vector is the

coordinates of the terminal point of the vector. If A(–7, 8) are the coordinate of

the terminal point A then the position vector of A i.e. OA = (–7, 8) or –7
8

248 Infinity Optional Mathematics Book - 9

(iv) Unit vector: Y
What are the magnitudes of →a = (–1, 0) and →b = (0, 1)?

Do their magnitude are 1 ? B(0, 1)

A vector whose length or magnitude is 1 unit i.e. unity X' A X
(1, 0)
is called a unit vector. So, if |CD| = 1 unit then the O

vector CD is called a unit vector. Y'

The unit vector along x-axis is written with the symbol →i = (1, 0) and the unit

vector along y-axis is written as →j = (0, 1).

From figure,

OA = →i = (1, 0) or 1 and OB = →j = (0, 1) or 0
0 1

Again,

Each non-zero vector has unit vector along its own direction. Thus, the unit
vector of a non-zero vector →a in the direction of →a is denoted by a^ (read as 'a'

cap). →a
|→a |
∴ The vector a^ is given by a^ =

(v) Zero of Null vector:

A vector whose length or magnitude is zero is called zero vector or null vector.

So, if |MN| = 0 then the vector MN is called the zero vector. Generally, a zero

vector is denoted by O→, where O→ = 0
0

The initial and the terminal point of a zero vector are coincides i.e.

AA = BB = CC = 0
(vi) Negative vector: 0

4 4 A B
5 5 B
If AB = then can we write BA also ? Discuss

on it. A

The vector AB displaces A to B and BA displaces B to A.
The length of AB and BA are equal but they are opposite in direction.

Infinity Optional Mathematics Book - 9 249

Therefore, negative vector of AB is BA
∴ BA = – AB

or, AB = – BA

For example: If →a = 6 and →b = –6 then →a and →b are said to be negative
2 –2

vectors to each other.

(vii) Like vectors:

Two vectors are said to be like if they have same B
D
direction whatever may be their magnitudes. →a

From figure, two vectors →a and →b are said to be A →

b

like vectors if →a = k→b where k > 0. C

If AB = 2CD, then AB and CD are like vectors.

(viii) Unlike vectors:

Two vectors are said to be unlike if they have opposite directions whatever

may be their magnitudes. M →a N

From figure, two vectors →a and →b are said to be →

b

unlike vectors if →b = k →a where k < 0 P Q

If PQ = – 2MN, then PQ and MN are unlike vectors.

(ix) Equal vectors:

Two vectors are said to be equal if they have equal magnitude and same

direction. B
D
Let, two vectors AB and CD are in same direction

and |AB | = |CD|, then AB and CD are equal A
vectors and written as AB = CD.

C
Similarly, if PQ and QR are in same direction and

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