Prime Optional Mathematics 10

8) Cos6q + Sin6q = 1 (5 + 3Cos4q)
8
Solution :

L.H.S. = Cos6q + Sin6q

= (Cos2q)3 + (Sin2q)3

= (Cos2q + Sin2q)3 – 3Cos2q.Sin2q(Cos2q + Sin2q)

= 1 – 3Cos2q.Sin2q

=1– 3 (2Sinq.Cosq)2
4

=1– 3 (Sin22q)
4

=1– 3 × 1
4 2 (2Sin22q)

=1– 3 (1 – Cos4q)
8

= 8 – 3 + 3Cos4i
8

= 1 (5 + 3Cos4q)
8

= R.H.S. is proved

9) Cos3q.Cos3q + Sin3q.Sin3q = Cos32q
Solution :
L.H.S. = Cos3q.Cos3q + Sin3q.Sin3q

= 1 [4cos3q.Cos3q + 4Sin3q.Sin3q]
4

= 1 [(3Cosq + Cos3q)Cos3q + (3Sinq – Sin3q)Sin3q]
4

= 1 [3Cos3q.Cosq + Cos23q + 3Sin3qSinq – Sin23q]
4

= 1 [3(Cos3q.Cosq + Sin3q.Sinq) + Cos23q – Sin23q]
4

= 1 [3Cos(3q – q) + Cos2(3q)]
4

= 1 [3Cos2q + Cos3(2q)]
4

= 1 × 4Cos32q
4

= Cos32q
= R.H.S. is proved

10) Sin18° = 5 –1
4
Solution :

Let, q = 18°

or 5q = 90°

or, 2q + 3q = 90°

or, Sin2q = Sin(90° – 3q°)

or, 2SinqCosq = Cos3q

or, 2SinqCosq = 4Cos3q – 3cosq

196 PRIME Opt. Maths Book - X

or, 2SinqCosq = Cosq(4Cos2q – 3) [\ Cosq ≠ 0]

or, 2Sinq = 4(1 – Sin2q) – 3

or, 4Sin2q + 2Sinq = 1

or, (2Sinq)2 + 2.2.Sinq. 1 + ( 1 )2 = ( 1 )2 + 1
2 2 2

5 2
2
or, (2Sinq + 1 )2 = c! m
2

or, 2Sinq + 1 =± 5
2 2

or, 2Sinq = ± 5 – 1
2 2

or, Sinq = !^ 5 – 1h
4

5 –1
\ Sin18° = 4 [q = 18° is an acute angle lies on 1st quadrant]

11) Cos4 r + Cos4 3p + Cos4 5r + Cos4 7r = 3
8 8 8 8 2

Solution : p 3r 5r 7r
8 8 8 8
L.H.S. = Cos4 + Cos4 + Cos4 + Cos4

= Cos4 p + Cos4 3r + Cos4 a 8r – 3r k + Cos4 a 8r – r k
8 8 8 8

= Cos4 p + Cos4 3r + Cos4 ar – 3r k + Cos4 `r – r j
8 8 8 8

= Cos4 p + Cos4 3r + (– Cos 3r )4 + (–Cos r )4
8 8 8 8

= 2Cos4 p + 2Cos4 3r
8 8

= 1 (2Cos2 p )2 + 1 (2Cos2 3r )2
2 8 2 8

= 1 (1 – Cos2. p )2 + 1 (1 + Cos2. 3r )2
2 8 2 8

= 1 (1 + Cos p )2 + 1 (1 + Cos 3r )2
2 4 2 4

= 1 (1 + 1 )2 + 1 (1 – 1 )2
2 2 2 2

= 1 [1 + 2 + 1 +1– 2 + 1 ]
2 2 2 2 2

= 1 (3)
2

3
=2

R.H.S. is proved

PRIME Opt. Maths Book - X 197

12) (2Cosq + 1)(2Cosq – 1)(2Cos2q – 1)(2Cos4q – 1) = 2cos8q + 1
Solution :
L.H.S = (2Cosq + 1)(2Cosq – 1)(2Cos2q – 1)(2Cos4q – 1)
= (4Cos2q – 1)(2Cos2q – 1)(2Cos4q – 1)
= {2 (2Cos2q) – 1)}(2Cos2q – 1)(2Cos4q – 1)
= {2 (1 + Cos2q) – 1}(2Cos2q – 1)(2Cos4q – 1)
= (2 + 2Cos2q – 1)(2Cos2q – 1)(2Cos4q – 1)
= (4Cos22q – 1)(2Cos4q – 1)
= {2(2Cos22q) – 1}(2Cos4q – 1)
= {2(1 + Cos4q) – 1}(2Cos4q – 1)
= (2 + 2Cos4q – 1)(2Cos4q – 1)
= (2Cos4q + 1)(2Cos4q – 1)
= 4Cos24q – 1
= 2(2Cos24q) – 1
= 2(1 + Cos8q) – 1
= 2 + 2Cos8q – 1
= 2Cos8q + 1
= R.H.S. is proved

13) Prove that Tan7 1c = 6– 3+ 2 –2
2
Solution :
1c
L.H.S. = Tan7 2

15°
Sin 2
=
15°
Cos 2

2Sin2 15°
2
=
2Cos2 15°
2

=

= a 1 – Cos15° 2
Sin15°
k

=

=

=

2 2 – 3 –1 3 +1
= 3 –1 × 3 +1

198 PRIME Opt. Maths Book - X

=

=

=

= 6– 3+ 2–2

= R.H.S is proved

14) Prove that Cos4A = 1 (3 + 4Cos2A + Cos4A)
8
Solution :

L.H.S. = Cos4A

= 1 (2Cos2A)2
4

= 1 (1 + Cos2A)2
4

= 1 (1 + 2Cos2A + Cos22A)
4

= 1 × 1 (2 + 4Cos2A + 2Cos22A)
4 2

= 1 (2 + 4Cos2A + 1 + Cos4A)
8

= 1 (3 + 4Cos2A + Cos4A)
8

= R.H.S. is proved

15) Prove that : Cos3q + Cos3(120° – q) + Cos3(120° + q) = 3 Cos3q
4

Solution :

L.H.S. = Cos3q + Cos3(120° – q) + Cos3(120° + q)

= 1 [4Cos3q + 4Cos3(120° – q) + 4Cos3(120° + q)]
4

= 1 [3Cosq + Cos3q + 3Cos(120° – q) + Cos3(120° – q) + 3Cos (120° + q) + Cos3(120° + q)]
4

= 1 [3Cosq + Cos3q + 3{Cos(120° – q) + Cos(120° – q)} + Cos (90° × 4 + 3q) + Cos(90° × 4 + 3q)]
4

= 1 [3Cosq + Cos3q + 3{– 1 Cosq + 3 sin i – 1 Cosq – 3 sin i } + Cos3q + Cos3q]
4 2 2 2 2

= 1 [ 3Cosi + 3Cos3q – 3Cosi ]
4

= 3 Cos3q
4

= R.H.S. is proved

PRIME Opt. Maths Book - X 199

16) If 2 Tana = 3Tanb, prove that Tan(a – b) = Sin2b
Solution : 2Tana = 3Tanb 5 – Cos2b

or, Tana = 3Tanb
2

or, Tana = 3Sinb
Then, 2Cosb

L.H.S.= Tan(a – b)
Tana – Tanb
= 1 + Tana.Tanb

=

= Sinb ×
2Cosb

=

=

= Sin2b
5 – Cos2b

= R.H.S. is proved

Exercise 5.1

1. Answer the following questions.
i) Write down the formula of Sin2A in terms of TanA.
ii) Express Cos2A in terms of CotA.
iii) Write down the formula of Sin3A.
iv) Express Tan3A in terms of TanA.
v) Express Sin2A in terms of Cos2A.

2. i) If SinA = 3 , find the value of Sin2A and Cos2A.
5

ii) If CosA = 12 , find the value of Cos2A and Sin2A.
13

iii) If Tanq = 1 , find the value of sin2q, Cos2q & Tan2q.
3

iv) If Cota = 3 , find the value of Tan2a.
4

v) If CosA = 1 (a + 1 ), prove that Cos2A = 1 (a2 + 1 ).
2 a 2 a2

3. i) If SinA = 4 , find the value of Sin3A and Cos3A.
5

ii) If Cosa = 3 , find the value of Sin3a.
2

200 PRIME Opt. Maths Book - X

iii) If Cosq = 1 (m + 1 ), prove that Cos3q = 1 (m3 + 1 )
2 m 2 m3

iv) If SinA = 1 (p + 1 ), prove that Sin3A = – 1 (p3 + 1 )
2 p 2 p3

v) If TanA = 4 , find the value of Tan3A.
3

4. Prove that the followings:

i) Sin3A = 3SinA – 4Sin3A ii) CosA = ± 1 + Cos2A
2

iii) TanA = ± 1 – Cos2A iv) Tan3A = 3TanA – Tan3 A
1 + Cos2A 1 – 3Tan2 A

v) Cos2A = 1 – Tan2 A
1 + Tan2 A

5. Prove that :

i) 1 – Cos2A = TanA ii) 1 + Cos2A = CotA
Sin2A Sin2A

iii) Sin2A = CotA iv) 1 + Cos2A = Cot2A
1 – Cos2A 1 – Cos2A

v) 1 – Cos2A = Tan2A
1 + Cos2A

6. Prove that : = CotA ii) = Tanq
i) iv) = TanA

iii) = Cotq
v) = Cot2a

7. Prove that :

i) Cos2i = 1 – Tani ii) 1 – Sin2A = CosA – SinA
1 + Sin2i 1 + Tani Cos2A CosA + SinA

iii) 1 – Sin2A = a CotA – 1 2 iv) Cos2A = Tan(45° + A)
1 + Sin2A CotA + 1 1 – Sin2A
k

v) 1 + Sin2A = Tan(4p + A)
Cos2A

8. Prove that :

i) Sin2A + Tan2A = 4TanA ii) CosA.Cos2A = 1 Sin4A.CosecA.
1 – Tan4 A 4

iii) SinA.Cos2A = 1 Sin4A.SecA iv) Sin3 A + cos3 A =1– 1 Sin2A
4 SinA + CosA 2

v) 1 + TanA.Tan2A = Sec2A

9. Prove that :

i) Sin3A – Cos3A =2 ii) Cos5A + Sin5A = 2Cos4A.Cosec2A
SinA CosA SinA CosA

iii) Sin5A – Cos5A = 4Cos2A iv) = Sin2A
SinA CosA

PRIME Opt. Maths Book - X 201

v) Cos2(45° – A) – Sin2(45° – A) = Sin2A

10. Prove that :

i) 2Sin2(45° – A) = 1 – Sin2A ii) = 2CosA
Sin8A = 8SinA.CosA.Cos2A.Cos4A
iii) Cotq – Tanq = 2Cot2q iv)

v) (2Cosq + 1)(2Cosq – 1) = 2Cos2q + 1

11. Prove that :

i) 1 – 1 = Cot4q
Tan3i + Tani Cot3i + Coti

ii) 1 – 1 = Cot2q
Tan3i – Tani Cot3i – Coti

iii) CotA + Cot(60° + A) – Cot(60° – A) = 3Cot3A

iv) (2Cosq + 1)(2Cosq – 1)(2Cos2q – 1)(2Cos4q – 1) = 2Cos8q + 1

v) Sec2A – Tan2A = CosA – SinA
CosA + SinA

12. Prove that : ii) Cosec4A + Cot8A = Cot2A – Cosec8A
i) Cosec2A + Cot4A = CotA – cosec4A iv) 3Cosec20° – Sec20° = 4
iii) Cosec10° – 3Sec10° = 4
v) Sec80° – 3Cosec80° = 4

13. Prove that : = CotA ii) = Tanq
i)

iii) = 2Cosq iv) Sec8i – 1 = Tan8i
v) (1 + Sin2q + Cos2q)2 = 4Cos2q(1 + Sin2q) Sec4i – 1 Tan2i

14. Prove that :

i) Cos2A + Sin2A.Cos2B = Cos2B + Sin2B.Cos2A

ii) Sin2a – Cos2a.Cos2b = Sin2b – Cos2b.Cos2a

iii) 4(Cos310° + Sin320°) = 3(Cos10° + Sin20°)

iv) Tan(A + 45°) – Tan(A – 45°) = 2Sec2q

v) Cot(q + 45°) – Tan(q – 45°) = 2Cos2i
1 + Sin2i

15. Prove that :

i) Cos6q – sin6q = Cos2q(1 – 1 Sin22q)
4

ii) Cos6q + Sin6q = 1 (5 + 3Cos4q)
8

iii) Cos8a – Sin8a = 1 Cos2a(3 + Cos4a)
4

iv) Sin8A + Cos8A = 1 (8 – 8Sin22A + Sin42A)
8

v) Sin4A + Cos4A = 1 (3 + Cos4A)
4

202 PRIME Opt. Maths Book - X

16. PRIME more creative questions:

a. Prove that:

i) Tanq + 2Tan2q + 4Tan4q + 8cot8q = cotq

ii) 2SinA + 2Sin3A + 2Sin9A = Tan27A – Tan A
Cos3A Cos9A Cos27A

iii) Cos3qCos3q + Sin3qSin3q = cos32q

iv) Sin2q + Sin2(120° – q) + Sin2(120° + q) = 3
2
3
v) Sin3ACos3A + Cos3ASin3A = 4 Sin4A

vi) Cos3q + Cos3(120° + q) + Cos3(240° + q) = 3 Cos3q.
4

b. Prove that:

i) Sin4q = 1 (3 – 4cos2q + Cos4q) ii) Cos36° = 5+1
8 4

iii) Cot 82 1 c = 6– 3+ 2–2 iv) Cos144° = – c 5+ 1 m
2 4

v) (Sin2A + Sin2B)2 + (Cos2A + Cos2B)2 = 4Cos2(A – B)

c. Prove that:

i) `1 + Cos r j a1 + Cos 3r k a1 + Cos 5r k a1 + Cos 7r k = 1
8 8 8 8 8

ii) Sin4 p + Sin4 3r + Sin4 5r + Sin4 7r = 3
8 8 8 8 2

iii) Sin3q + Sin3(120° – q) + Sin3(120° + q) = 3 Sin3q
4

iv) Cosec2A + Cosec4A + Cosec8A = CotA – Cot8A

v) (Sin2a – Sin2b)2 + (Cos2a – Cos2b)2 = 4Sin2(a – b)

vi) (Sec2q + 1) (Sec4q + 1) (Sec8q + 1) = Tan8q.Cotq

d. i) If 2TanA = 3TanB, prove that Tan(A – B) = Sin2B
5 – Cos2B

ii) If 2Tana = 3Tanb, prove that Tan(a + b) = 5Sin2b 1
iii) If Tand =
1 , Tang = 1 , prove that Sin4g = 5Cos2b –
7 3 Cos2d

iv) If aTana = b, prove that acos2a + bsin2a = a

v) If 2TanA = TanB, prove that Cos2A = 3 + 5Cos2B
5 + 3Cos2B

Answer

1. Show to your teacher.

2. i) 24 , 7 ii) 119 , 120 iii) 3 , 1 , 3 iv) – 24
25 25 169 169 2 2 7

2. i) 44 , – 117 ii) 1 v) – 44
125 125 117

PRIME Opt. Maths Book - X 203

5.2 Sub - Multiple Angles

In Previous chapter we discussed about the multiple angles 2A, 3A, 4A, etc for an angle A.

Similarly A , A , A , A etc are the sub-multiples of the angle A. Here we have to discuss the
2 3 4 8
A A A A A
terminological ratios for the sub- multiple angles 2 and 3 as Sin 2 , Cos 2 , Sin 3 , Cos

A , Tan A , etc.
3 3

Let, 2a =A 3a = A
\a
= A \a = A
2 3

It means all the multiple angles 2a and 3a can be expressed as a and resultant angles a can

be expressed as A and A as the formulae given on table.
2 3

Let us see the examples,

i) Sin2a = 2SinaCosa

i.e. SinA = 2Sin A Cos A [Replacing a by A ]
2 2 2

ii) Sin3a = 3Sina – 4Sin3 a

\ SinA = 3Sin A – 4Sin3 A [replacing a by A ]
3 3 3

S. No. Multiple angles Sub-multiple angles
1.
Sin2A = 2SinA.CosA SinA = 2Sin A . Cos A
2 2
Sin2A = 2 tan A
1 + tan2 A A
2Tan 2
SinA =
1+ Tan2 A
2

2. Cos2A = Cos2A – Sin2A CosA = Cos2 A – Sin2 A
2 2
Cos2A = 2Cos2A – 1
A
Cos2A = 1– 2Sin2A CosA = 2Cos2 2 –1

Cos2A = 1 – Tan2 A CosA = 1 – 2Sin2 A
1 + Tan2 A 2

CosA = 1 – Tan2 A
1 + Tan2 2
A
2

3. Tan2A = 2TanA A
1 – Tan2 A 2Tan 2
TanA =
Tan2 A
1– 1

4. Cot2A = Cot2 A – 1 Cot2 A –1
2CotA 2
CotA =
A
2Cot 2

204 PRIME Opt. Maths Book - X

5. Sin3A = 3SinA – 4Sin3A SinA = 3Sin A – 4Sin3 A
Cos3A = 4Cos3A – 3CosA 3 3

CosA = 4Cos3 A – 3Cos A
3 3

6. Tan3A = 3TanA – Tan3 A
1 – 3Tan2 A
TanA =

7. Cot3A = Cot3 A – 3CotA
3Cot2 A – 1
CotA =

8. 2Cos2A = 1 + Cos2A 2Cos2 A = 1 + CosA
2Sin2A = 1 – Cos2A 2
4Cos3A = 3CosA + Cos3A
4Sin3A = 3SinA – Sin3A 2Sin2 A = 1 – CosA
2

4Cos3 A = 3Cos A + CosA
3 3

4Sin3 A = 3Sin A – SinA
3 3

Worked out Examples

1. If cos A = 12 , find the value of TanA and SinA.
2 13
Solution:

Cos A = 12
2 13

Now, Then,
SinA
Sin A = 1– cos2 A = 2sin A cos A .
2 2 2 2

= 1– cos2 A =2× 5 × 12
2 13 13

= 1 – a 12 2 = 120
13 169
k

169 – 144 TanA = SinA
= 169 CosA

= 5 =
13

5 12
2 × 13 × 13
=
a 12 2 a 5 2
13 – 13
k k

120
169
= 119
169

= 120 .
119

PRIME Opt. Maths Book - X 205

2. Prove that : 1 + SinA – CosA = Tan A
1 + SinA + CosA 2

Solution:
1 + SinA – CosA
L.H.S. = 1 + SinA + CosA

=

=

=

= Tan A .
2

R.H.S. proved.

3. If Cos330° = 3 , Prove that Cos165° = – e 3 +1 o
Solution: 2 2 2

Let, A = 330°

\ A = 330° = 165°
2 2

We have,

CosA = 2Cos2 A –1
2

or, Cos330° + 1 = 2Cos2 165°
3

or, 2Cos2 165° = 1 + 2

or, Cos2 165° = 2+ 3 × 2
4 2

or, Cos2 165° =

or, Cos2 165° = ^ 3 + 1h2
8

or, Cos165° = – 3 + 1
22

\ Cos 165° = – e 3 + 1 o
22

4. Prove that: Tan a r + i k = Secq + Tanq
4 2

R.H.S. = Secq + Tanq

= 1 + Sini
Cosi Cosi

= 1 + Sini
Cosi

206 PRIME Opt. Maths Book - X

=

=

=

Cos i + Sin i
2 2
=
i i
Cos 2 – Sin 2

Cos i + Sin i
2 2

Cos i q
2 2
= [ Dividing each term by cos ]
i i
Cos 2 – Sin 2

Cos i
2

1 + Tan i
2
=
i
1 – Tan 2

=

= Tan (45° + 2q)

= Tan a r + i k
4 2

R.H.S. proved.

5. Prove that: (1 + cos r ) (1 + cos 3r ) (1 + cos 5r ) (1 + cos 7r ) = 1
8 8 8 8 8
Solution:
L.H.S = (1 + cos8p) (1 + cos 3r 5r 7r
8 ) (1 + cos 8 ) (1 + cos 8 )

= `1 + cos r j a1 + cos 3r k [1 + cos 8r – 3r ] + [1 + cos 8r – r ]
8 8 8 8

= `1 + cos r j a1 + cos 3r k [1 + cos ar – 3r k ] + [1 + cos `r – r j ]
8 8 8 8

= `1 + cos r j a1 + cos 3r k a1 – cos 3r k `1 – cos r j
8 8 8 8

= `1 – cos2 r j a1 – cos2 3r k
8 8

PRIME Opt. Maths Book - X 207

= Sin2 p . Sin2 3r
8 8

= Sin2 p Sin2 a 4r – r k
8 8

= Sin2 p . Sin2 ` r – r j
8 2 8

= Sin2 p . Cos2 p
8 8
p p
= 1 [2Sin 8 .Cos 8 ]2
4

= 1 `sin r 2
4 4
j

= 1 c 12
4 m

2

= 1 .
8

R.H.S. is proved

Exercise 5.2

1. Answer the following questions.

i. Express SinA in terms of Cot A .
2

ii. Express CosA in terms of Tan A
2

iii. Express 1 – CosA in term of Sin A
2

iv. Express 1 – CosA in term of Tan A .
1 + CosA 2

v. Express Sin4A in expanded form.

2. i If Sin A = 2
ii. 2 , Find SinA and CosA.

3

If Cos A = 4 , find SinA.
3 5

iii. If Tan q = 1 , find Tanq
3 3

iv. if Sin q = 3 , find Cosq.
3 5

v. If Cos q = 7 , find Cos q
25 2

3 3 –1
3. i. If Cos30° = 2 , prove that Sin15° = 2 2

3
ii. If Cos30° = 2 ,prove that Tan15° = 2 – 3

iii. If Cos45° = 1 , prove that Tan 22 1 c = 3–2 2
2 2

208 PRIME Opt. Maths Book - X

iv. If Cos330° = 3 1 _ 2+ 3i
2 , prove that Cos 165° = 2

v. If Cos330° = 3 , prove that Sin165° = 3 –1
2 22

4. i. If Sin A = 1 `a + 1 j , find SinA.
3 2 a

ii. If Cos3q = 1 `m + 1 j , prove that Cosq = 1 am3 + 1 k
2 m 2 m3

iii. If Cos A = 1 ap + 1 k prove that CosA = 1 cp2 + 1 m
2 2 p 2 p2

iv. If Sina2 = 1 `x + 1 j , prove that Sina = 1 ax2 + 1 k
2 x 2 x2

v. If Cos A = 3 3.
2 2 , prove that TanA =

5. Prove that the followings.

i. SinA = 3Sin A – 4Sin3 A ii. CosA = 4Cos3 A – 3Cos A .
3 3 3 3

iii. SinA = Cot A iv. 1 + Sini + Cosi = Cot2q.
1 – CosA 2 1 + Sini – Cosi

v. = Cot q
2

6. Prove that the followings:

1 ° 1 ° aSin i i 2
2 2 2 2
i. Cot 22 – Tan 22 = 2. ii. + Cos k = 1 + Sinq

iii. Sin2i × Cosi = Tan q iv. Cot q – 2Cotq = Tan 2q
1 + Cos2i 1 + Cosi 2 2

v. Cosecq – Cotq = Tan q
2

7. Prove the the followings:

i. (SinA + SinB)2 + (CosA + CosB)2 = 4Cos2 A– B .
2

ii. (Sina – Sinb)2 + (Cosa – Cosb)2 = 4Sin2 a–b .
2

iii. = Cot2q.

iv. Cosec A + CotA = Cot A – CosecA.
2 2

AA
Cos 2 – Sin 2
v. SecA – TanA =
A + A
Cos 2 Sin 2

PRIME Opt. Maths Book - X 209

8. Prove that the followings:

i. Tan a r + i k = SecA + TanA.
4 2

ii) Tan a r – A k = 1–SinA
4 2 1 + SinA

iii) Sec a r + i k .Sec a r – i k = 2Secq
4 2 4 2

iv) Tan a r + i k + Tan a r – i k = 2Secq
4 2 4 2

v) Cot a A + r k – Tan a A – r k = 2Cosi
2 4 2 4 1 + Sini

9. PRIME more creative questions
a. Prove that the followings.

i) Cos q .Cos 2r .Cos 3r = 1
7 7 7 8

ii) Sin6°.Cos48°sin66°Sin78° = 1
16

iii) `1 + Sin r j a1 + Sin 3r k a1 – Sin 5r k a1 – Sin 7r k = 1
8 8 8 8 8

iv) Cot 7 1 c 6+ 3 + 2 +2
v) 2 = Cosec A + Cosec
2
CosecA + A = Cot A – CotA
4 8

b. Prove that the followings.

i) 2Cos r =
32

A A1
ii) Cos6 2 + Sin6 2 = 8 (5 + 3Cos2A)

iii) (Cosa + Cosb)2 + (Sina + Sinb)2 = 4Cos2 a – b
2

iv) (Cosa – Cosb)2 + (Sina – Sinb)2 = 4Sin2 a – b
2

v) = Cot q
2

Answer

1. Show to your teacher.

2. i) 2 2 , – 1 ii) 117 iii) 13 iv) – 44 v) ! 4
3 3 125 9 125 5

4. i) – 1 :a3 + 1 D
2 a3

210 PRIME Opt. Maths Book - X

5.3 Transformation formula for trigonometry

The formula of compound angle are transferred to the product from sum or difference and
vice versa in this section

1. Transformation from sum and difference to the product.
i. Sin(A + B) + Sin(A – B)
= SinA.CosB + CosA.SinB + SinA.CosB – CosA.SinB
= 2sinA.CosB.

ii. Sin(A + B) – sin(A – B)
= (sinA.CosB + CosA.SinB) – (SinA.CosB – CosA.SinB)
= 2CosA.SinB

iii. CosA(A + B) + Cos(A – B)
= CosA.CosB – SinA.SinB + CosA.CosB + SinA.SinB
= 2CosA.CosB

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

\ Cos (A – B) – Cos(A + B) = 2SinASinB.

2. Transformation from product to sum or difference.
i. 2SinA.CosB = Sin(A + B) + Sin(A – B)
ii. 2CosA.SinB = Sin(A + B) – Sin(A – B)
iii. 2CosA.CosB = Cos(A +B) + Cos(A – B)
iv. 2SinA.SinB = Cos(A – B) – Cos(A + B)

3. Transformation of A + B & A – B to C and D respectively.

Solution:

Let, A + B = C

A–B=D
C+D
Then, A = 2 &B= C–D
2

i. SinC + SinD = 2Sin C+D Cos C–D
2 2

ii. SinC – SinD = 2Cos C + D Sin C–D
2 2

iii. CosC + CosD = 2Cos C+D Cos C–D
2 2

iv. CosC – CosD = –2Sin C + D Sin C – D
2 2

= 2Sin C +D Sin D – C
2 2

\ CosD – CosC = 2Sin C + D Sin C – D
2 2

PRIME Opt. Maths Book - X 211

4. Conclusion of transformation formula: (Ideas to remember formula)

i. Sum to product: A+B A–B
22
Angles: A B$

Format: S +S = 2S C

S–S= 2 C S

C+C= 2 C C

C – C = –2 S S

ii. Product to sum: A B $ (A + B) (A – B)
Angles:
S S
Format: 2 C C= S + S
C S= S – C
2 S C= C + C
S= C –
2

–2

iii. Always arrange the angles of trigonometric ratio in descending order to use the formula.

Worked out Examples

1. Sin10° + Cos 10° = 2 Cos35°

L.H.S. = Sin10° + Cos10°

= Sin(90° – 80°) + Cos10°

= Cos 80° + Cos10°

= 2Cos 80° + 10° Cos 80° – 10° [\ C + C = 2CC]
2 2

= 2Cos45° Cos35°

= 2 × 1 Cos 35°
2

= 2 Cos35°

2. Prove that: Sin105° + Sin15° = 3
2
Solution:

L.H.S. = Sin105° + Sin15°

= 2Sin 105 + 15 Sin 105 – 15
2 2

= 2sin 60° Sin45°

=2× 3 1
2× 2

= 3 .
2

R.H.S. is proved

212 PRIME Opt. Maths Book - X

3. Prove that: Sin15° Sin75° = 1
4
Solution:

L.H.S. = Sin15° Sin75°

= 1 [2Sin75° Sin15°]
2

= 1 [ Cos (75° – 15°) – Cos(75° + 15°)]
2

= 1 [Cos60° – Cos90°]
2

= 1 [ 1 – 0]
2 2

= 1
4

R.H.S is proved

4. Prove that: Cos40° + Cos80° + Cos160° = 0

Solution:

L.H.S. = Cos40° + Cos80° + Cos160°

= Cos40° + [Cos160° + Cos80°]

= Cos40° + 2Cos 160° + 80° Cos 160 – 80
2 2

= Cos40° + 2cos120° cos40°

= Cos40° + 2 a– 1 k cos40°
2

= Cos40° – Cos40°

= 0.
R.H.S. is proved

5. Prove that: Sin10° Sin30° Sin50° Sin70° = 1 .
16
Solution:

L.H.S. = Sin10° Sin30° Sin50° Sin70°

= Sin30° × Sin10° × 1 [2Sin70° Sin50°]
2

= 1 1 Sin10° [Cos (70 – 20) – Cos (70 + 50)]
2 ×2

= 1 Sin10° Cos20° – 1 Sin10° Cos120°
4 4

= 11 [2cos20° sin10°] – 1 Sin10° a– 1 k
4×2 4 2

= 1 [Sin(20° + 10°) – Sin(20° – 10°)] + 1 Sin10°
8 8

= 1 Sin30° – 1 Sin10° + 1 Sin10°
8 8 8

= 1 × 1
8 2

= 1
16

R.H.S. is proved

PRIME Opt. Maths Book - X 213

6. Cos2A Cos3A – Cos2A Cos7A = Sin7A + Sin3A
Sin4A Sin3A – Sin2A Sin5A SinA

L.H.S. =

=

=
=
=

=

= 2Sin5A Cos2A
SinA

=

= Sin7A + Sin3A
SinA

R.H.S is proved

7. Prove that: = – Tan(A + B)
Solution:
× 2
L.H.S = 2

=

= Cos2A – Cos2B
Sin2A – Sin2B

=

= – Sin^A + Bh
Cos ^A + Bh

= – Tan (A + B)

R.H.S. is Proved.

214 PRIME Opt. Maths Book - X

8. 2Cos r Cos 3r + Cos 9r + Cos 11r = 0.
13 13 13 13

Solution: r 3r 9r 11r
13 13 13 13
L.H.S. = 2Cos Cos + Cos + Cos

= Cos a 3r + r k + Cos a 3r – r k + Cos 9r + Cos 11r
13 13 13 13 13 13

= Cos 4r + Cos 2r + Cos a 13r – 4r k + Cos a 13r – 2r k
13 13 13 13

= Cos 4r + Cos 2r + Cos ar – 4r k + Cos ar – 2r k
13 13 13 13

= Cos 4r + Cos 2r – Cos 4r – Cos 2r
13 13 13 13

=0

R.H.S. is Proved.

9. If 1 + 1 = 1 + 1 Prove that : TanA.TanB = Cota A + B k
SinA CosA SinB CosB 2

Solution:

Given : 1 + 1 = 1 + 1
SinA CosA SinB CosB

or, 1 – 1 = 1 – 1
SinA SinB CosB CosA

or, SinB – SinA = CosA – CosB
SinA SinB CosA CosB

or, = SinA SinB
CosA CosB

or, Cot A+B = TanA. TanB
2

\ TanA TanB = Cot a A + B k
2

\ L.H.S = R.H.S. Proved

10. Prove that: Sinq.Sin(60° – q) Sin(60° + q) = 1 Sin3q
4
Solution:

L.H.S. = Sinq.Sin(60° – q) Sin(60° + q)

= Sinq × 1 [2Sin(60° + q). Sin(60° – q)]
2

= 1 Sinq[Cos(60° + q – 60° + q) – Cos(60° + q + 60° – q)]
2

= 1 Sinq.Cos2q – 1 SinqCos120°
2 2

= 1 × 1 [2Cos2q.Sinq] – 1 Sinq a– 1 k
2 2 2 2

= 1 [Sin(2q + q) – Sin(2q – q)] + 1 Sinq
4 4

PRIME Opt. Maths Book - X 215

= 1 Sin3q – 1 Sini + 1 Sini
4 4 4

= 1 Sin3q
4

R.H.S is proved

Exercise 5.3

1. Express the following in product form & evaluate.

i) Sin105° + Sin15° ii. Cos15° – Cos75°

iii) Sin10° – Cos10° iv. Cos20° + Sin20°

v) Sin(60° + q) – Sin (60° – q)

2. Express the followings in terms of sum or difference form & evaluate.

i) 2Cos3q Cosq ii. Cos15° Cos105°

iii) Sin15°Sin75° iv. Sin75°Cos105°

v) Cos(45° + A) Sin(45° – A).

3. Prove that the followings:
i) Sin75° – Sin15° = 1
2

ii) Sin20°Sin40° = 1 (Cos20° – 1 )
2 2

iii) Cos80° + Cos40° = – Cos20°
iv) Sin(A + 3B) – Sin(3A + B) = 2Cos (2A + 2B). Sin(B – A)
v) Cos(A + 120°) + Cos(A – 120°) = CosA.

4. Prove that the followings:
i) Cos80° + Cos40° – Cos20° = 0
ii) Sin70° – Sin50° – Sin10° = 0
iii) Cos40° + Cos80° + Cos160° = 0
iv) 3Cos20° – Sin40° – Sin80° = 0
v) Cos(45° + A) + Cos(45° – A) – 2Cos2A = 0

5. Prove that the followings:
Sin2A + Sin2B
i) Sin2A + Cos2B = Tan(A + B). Cot(A – B).

ii) CosA + CosB = Cot A+ B Cot B – A
CosA – CosB 2 2

iii) SinA + Sin(A + 120) + Sin(A – 120°) = 0

iv) = Cot59°

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

216 PRIME Opt. Maths Book - X

6. Prove that the followings:
i) = Tan2A

ii) = TanA

iii) = Tan3q

iv) = Sin7A + Sin3A
SinA

v) = 1.

7. Prove that the followings: = 4Cos2(a – b)
i) ^Sin2a + Sin2bh2 +

ii) (CosA – CosB)2 + (SinA – SinB)2 = 4 Sin2 A–B
2

iii) = Tan(a + b)

iv) = Tan(a – b)

v) Cos2A + Cos2(A +120°) + Cos2(A –120°) = 3
2

8. Prove that the followings:

i) Sinq Sin(60° – q) Sin(60° + q) = 1 Sin3q
4

ii) Cosq Cos(60° – q) Cos (60° + q) = 1 Cos3q
4

iii) 2 Cos r Cos 3r + Cos 9r + Cos 11r =0
13 13 13 13

iv) Cos2A sin3A = 1 (2SinA + Sin3A – Sin5A)
16

v) Cos2A Sin4A = 1 [Cos6A – 2Cos4A – Cos2A + 2]
32

9. Prove that the followings:

i) Cos10° Cos30° Cos50° Cos70° = 3 .
16

ii) Sin20° Sin40° Sin60° Sin80° = 3 .
16

iii) Cos80 Cos160° Cos320° = – 1 .
8

iv) Sin50° Sin100 Sin200° = 1 .
8

v) Sin10° Cos20° Sin30° Cos40° = 1 .
16

11. PRIME more creative questions:
r 2r 4r 8r
a. i) Cos 15 Cos 15 Cos 15 Cos 15 = – 1
16

ii) Sin6° Sin42° Sin66° Sin78° = 1
16

PRIME Opt. Maths Book - X 217

iii) Sin2A + Sin2B + 2SinA SinB Cos(A + B) = Sin2(A + B)

iv) Sin3q + Sin3(120° + q) + Sin3(240 + q) = – 3 Sin3q
4

v) Sin r Sin 3r Sin 5r = 1
14 14 14 8

b. i) If Cosa + Cosb = 1 and Sina + Sinb = 1 , prove that a+b = 3
3 4 Tan 2 4

ii) If SinA + SinB = 1 and CosA + CosB = 1 , prove that Cos(A + B) = 3 .
4 2 5

iii) If xCosq + ySinq = xCosa + ysina, prove that Cot a i + a k = x
2 y

iv) If Sin(a – b) = kSin(a + b), prove that Tanb = k –1
Tana k +1

v) If SinA = kSinB, prove that: Cot A+ B = k –1 Cot A+B
2 k +1 2

1. i) 3 ii) 1 Answer
2 2
iii) – 2 Sin35°
iv) 2Cos25° v) Sinq

3. i) Cos4q + Cos2q ii) – 1 iii) 1
4 4

iv) – 1 v) 1 (1 – Sin2A)
4 2

218 PRIME Opt. Maths Book - X

5.4 Conditional trigonometrical ratios:

The given condition helps to prove the trigonometrical identities in this section where all the

formulae and ideas used in previous chapters of trigonomery should be applied here in this

section. p
2
The conditions may be A + B + C = r, A + B + C = and any more, where
A+B+C= p

A+B= p–C

Sin (A + B) = Sin(p – C) = SinC

Cos (A + B) = Cos(p – C) = – CosC

Tan (A + B) = Tan(p – C) = – TanC

A+B+C=p
r
A + B = 2 – C
2 2 2

Sin a A + B k = Sin a r – C k = Cos C
2 2 2 2 2

Cos a A + B k = Cos a r – C k = Sin C
2 2 2 2 2

Tan a A + B k = Tan a r – C k = Cot C
2 2 2 2 2

Worked out Examples

1. If A + B + C = p, Prove that : CotA.CotB + CotB.CotC + CotC.CotA = 1

Solution:

A+B+C=p

or, A + B = p – C

or, Cot(A + B) = Cot(p – C)

or, CotACotB – 1 = – CotC
CotB + CotA

or, CotA.CotB – 1 = –CotB.CotC – CotC.CotA
or, CotA.CotB + CotB.CotC + CotC.CotA = 1
\ L.H.S = R.H.S is proved

2. If A + B + C = p, Prove that: SinA + SinB + SinC = 4Cos A Cos B Cot C
2 2 2
Solution:

Here, A + B + C = p

or, A + B = p – C
r
or, A + B = 2 – C
2 2 2

or, Sin a A + B k = Sin a r – C k = Cos C
2 2 2 2 2

or, Cos a A + B k = Cos a r – C k = Sin C
2 2 2 2 2
Then,

L.H.S = SinA + SinB + SinC

= (SinA + SinB) + SinC
+B
= 2Sin A 2 Cos A– B + SinC
2

PRIME Opt. Maths Book - X 219

= 2Cos C Cos a A – B k + 2Sin C Cos C
2 2 2 2 2

= 2Cos C [Cos a A – B k + Sin C ]
2 2 2 2

= 2Cos C [Cos a A – B k + Cos a A + B k ]
2 2 2 2 2

= 2Cos C [Cos A Cos B + Sin A .Sin B + Cos A .Cos B – Sin A .Sin B ]
2 2 2 2 2 2 2 2 2

= 4Cos A Cos B Cos C
2 2 2

3. Cos2A + Cos2B – Cos2C = 1 – 4 SinA.SinB.CosC

Solution:

Here, A + B + C = p

or, A + B = p – C

or, Sin(A + B) = Sin (p – C) = SinC

or, Cos(A + B) = Cos (p – C) = – CosC.

Then,

L.H.S. = Cos2A + Cos2B – Cos2C
2A + 2B
= 2Cos 2 Cos 2A – 2B – Cos2C
2

= 2Cos (A + B) Cos (A – B) – Cos2C
= –2CosC Cos(A – B) – 2Cos2C + 1
= 1 – 2CosC [Cos (A – B) + CosC]
= 1 – 2CosC [Cos(A – B) – Cos(A + B)]
= 1 – 2CosC [ CosA.CosB + SinA.SinB – CosA.CosB + SinA.SinB]
= 1 – 4SinA.SinB.CosC.
= R.H.S.

4. Sin2A + Sin2B + Sin2C = 2 + 2CosA CosB CosC

Solution:

A+B+C=p

A+B=p–C

Sin(A + B) = Sin(p – C) = SinC

Cos(A + B) = Cos(p – C) = – CosC.

Now,

L.H.S. = Sin2A + Sin2B + Sin2C

= 1 [2sin2A + 2sin2B] + Sin2C
2

= 1 [1 – Cos2A + 1 – Cos2B] + Sin2C
2

= 1 [2 – (Cos2A + Cos2B)] + Sin2C
2

= 1 [2 – 2 Cos 2A + 2B Cos 2A – 2B ] + Sin2C
2 2 2

= 1 ×2 [1 – Cos (A + B) Cos(A – B)] + Sin2C
2

= 1 + CosC Cos (A – B) + 1 – Cos2C
= 2 + CosC [Cos (A – B) – CosC]

220 PRIME Opt. Maths Book - X

= 2 + CosC [Cos (A – B) + Cos (A + B)]
= 2 + CosC [CosA.CosB + SinA SinB + CosA.CosB – SinA SinB ]
= 2 + 2 CosA CosB CosC. = R.H.S

5. Cos2 A + Cos2 B – Cos2 C = 2Cos A Cos B Sin C
2 2 2 2 2 2
Solution:

A+B+C=p

or, A + B = p – C r
2
or, Sin a A + B k = Sin a – C k = Cos C
2 2 2 2

or, Cos a A + B k = Cos a r – C k = Sin C
2 2 2 2 2
Now,

L.H.S. = Cos2 A + Cos2 B – Cos2 C
2 2 2

= 1 [2Cos2 A + 2Cos2 B ] – Cos2 C
2 2 2 2

= 1 [1 + CosA + 1 + CosB] – Cos2 C
2 2

= 1 [2 + (CosA + CosB)] – Cos2 C
2 2

= 1 [2 + 2Cos A+ B .Cos A+ B ] – Cos2 C
2 2 2 2

= 1 × 2 [ 1 + Sin C .Cos a A – B k ] – Cos2 C
2 2 2 2 2

=1 + Sin C .Cos a A – B k –1 + Sin2 C
2 2 2 2

= Sin C [Cos a A – B k + Sin C ]
2 2 2 2

= Sin C [Cos a A + B k + Cos a A + B k ]
2 2 2 2 2

= Sin C [Cos A .Cos B + Sin A B + Cos A .Cos B – Sin A B ]
2 2 2 2 .Sin 2 2 2 2 .Sin 2

= Sin C [2Cos A .Cos B ]
2 2 2

= 2Cos A .Cos B .Sin C
2 2 2

= R.H.S. prove

6. If A + B + C = p, prove that: Cos(B + C – A) + Cos(C + A – B) + Cos(A + B – C) = 1 + 4 CosA.CosB.
CosC
Solution :
Given, A + B + C = p.
or, A + B = p – C
or, Sin(A + B) = Sin(p – C) = SinC
or, Cos(A + B) = Cos(p – C) = – CosC
Now,
L.H.S. = Cos(B + C – A) + Cos(C + A – B) + Cos(A + B – C)

= Cos(p – A – A) + Cos(p – B – B) + Cos(p – C – C)

PRIME Opt. Maths Book - X 221

= Cos(p – 2A) + Cos(p – 2B) + Cos(p – 2C)

= – Cos2A – Cos2B – Cos2C

= –2Cos a 2A + 2B k .Cos a 2A – 2B k – Cos2C
2 2

= –2Cos(A + B)Cos(A – B) – Cos2C
= 2CosC.Cos(A – B) – 2Cos2C + 1
= 1 + 2CosC[Cos(A – B) – CosC]
= 1 + 2CosC[Cos(A – B) + Cos(A + B)]

A+B+A–B A +B– A +B
= 1 + 2CosC[2Cos 2 .Cos 2 ]

= 1 + 2CosC(2CosA.CosB)
= 1 + 4CosA.CosB.CosC
R.H.S. proved.

7. If A + B + C = p, prove that : Sin5A + sin5B + Sin5C = 4Cos 5A .Cos 5B .Cos 5C
2 2 2
Solution:

Given A + B + C = p

or, 5A + 5B = 5r – 5C
2 2 2 2

or, Sin( 5A + 5B ) = Sin( 5r – 5C ) = Cos 5C
2 2 2 2 2

or, Cos( 5A + 5B ) = Cos( 5r – 5C ) = Sin 5C
2 2 2 2 2
Now,

L.H.S. = Sin5A + Sin5B + Sin5C

= 2Sin 5A + 5B .Cos 5A – 5B + Sin5C
2 2

= 2Sin 5A + 5B .Cos 5A – 5B + 2Sin 5C .Cos 5C
2 2 2 2

= 2Cos 5C [Cos( 5A – 5B ) + Sin 5C ]
2 2 2 2

= 2Cos 5C [Cos( 5A – 5B ) + Cos( 5A + 5B )]
2 2 2 2 2

= 2Cos 5C [2Cos .Cos ]
2

= 2Cos 5C [2Cos 5A Cos 5B ]
2 2 2

= 4Cos 5A Cos 5B Cos 5C
2 2 2

R.H.S. is proved.

8. If A + B + C = p, prove that : Sin A + Sin B + Sin C = 1 + 4Sin r – A .Sin r– B .Sin r– C
Solution: 2 2 2 4 4 4

Given A + B + C = p

or, A + B = p – C

or, B + C = p – A

or, C + A = p – B

222 PRIME Opt. Maths Book - X

Then, r r–B r–
4 4
R.H.S. = 1 + 4Sin – A .Sin .Sin C
4

= 1 + 4Sin B+ C .Sin C+A .Sin A +B
4 4 4

= 1 + 2Sin B+C [Cos( C +A – A +B ) – Cos( C+A + A +B )]
4 4 4 4 4

= 1 + 2Sin B+ C .Cos C–B –2Cos 2A +B+ C .Sin B +C
4 4 4 4

= 1 + [Sin( B + C + C –B ) + Sin( B+C – C–B )] – [Sin( 2A +B+ C + B+ C ) –
4 4 4 4 4 4

[Sin( 2A +B + C + B+C )
4 4

= 1 + [Sin 2C + Sin 2B – Sin 2r + Sin 2A ]
4 4 4 4

= Sin A + Sin B + Sin C
2 2 2

L.H.S. is proved

9. If A + B + C = p, prove that : SinA.CosB.CosC + CosA.SinB.CosC + CosA.CosB.SinC = SinA.SinB.
SinC.
Solution:
Given A + B + C = p
or, A + B = p – C
or, Sin(A + B) = Sin(p – C) = SinC
or, Cos(A + B) = Cos(p – C) = – CosC
Then,
L.H.S. = SinA.CosB.CosC + CosA.SinB.CosC + CosA.CosB.SinC
= CosC[SinA.CosB + CosA.SinB] + CosA.CosB.SinC
= CosC.Sin(A + B) + CosA.CosB.SinC
= SinC[CosC + CosA.CosB]
= SinC[–Cos(A + B) + CosA.CosB]
= SinC[– CosA.CosB + SinA.SinB + CosA.CosB ]
= SinA.SinB.SinC
R.H.S. is proved

PRIME Opt. Maths Book - X 223

Exercise 5.4

1. If a + b + g = 180°, prove that the followings:

i. Tana + Tanb + Tang = Tana. Tanb. Tang

ii. CotaCotb + CotbCotg + CotgCota = 1

iii. Tan2a + Tan2b + Tan2g = Tan2a . Tan2b. Tan2g
a Cot 2b Cot 2g a b Cot 2g
iv. Cot 2 + + = Cot 2 . Cot 2 .

v. Tan 3a . Tan 3b + Tan 3b . Tan 3c + Tan 3c Tan 3a =1
2 2 2 2 2 2

2. If A + B + C = p, prove that the followings:

i. SinA + SinB + SinC = 4Cos A Cos B Cos C
2 2 2

ii. CosA + CosB + CosC = 1 + 4 Sin A Sin B Sin C
2 2 2

iii. SinA – SinB + SinC = 4Sin A Cos B Sin C
2 2 2

iv. CosA + CosB – CosC = 4Cos A Cos B Sin C –1
2 2 2

v. CosA – CosB – CosC = 1 – 4Sin A Cos B Cos C
2 2 2

3. If A, B and C are the angles of a triangle prove that the followings.

i. Sin2A + Sin2B – Sin2C = 4CosA CosB SinC

ii. Cos2A + Cos2B + Cos2C = – 1 – 4 CosA.CosB.CosC

iii. Sin3A + Sin3B + Sin3C = –4Cos 3A Cos 3B Cos 3C
2 2 2

iv. Cos5A + Cos5B + Cos5C = 1 + 4 sin 5A Sin 5B Sin 5C
2 2 2

v. Sin6A + Sin6B + Sin6C = 4Sin3A Sin3B Sin3C.

4. If A + B + C = p, prove that the followings:
i. Sin2A + Sin2B + Sin2C = 2 + 2CosA CosB CosC.
ii. Cos2A + Cos2B + Cos2C = 1 – 2CosA CosB CosC.
iii. Sin2A + Sin2B – Sin2C = 2SinA SinB CosC.
iv. Cos2A + Cos2B – Cos2C = 1 – 2SinA SinB CosC.
v. Sin2A – Sin2B – Sin2C = – 2CosA SinB SinC

5. If A, B and C are the angles of a triangle, prove that the followings:

i. Sin2 A + Sin2 B + Sin2 C = 1 – 2Sin A Sin B Sin C .
2 2 2 2 2 2

ii. Cos2 A + Cos2 B – Cos2 C = 2Cos A Cos B Sin C .
2 2 2 2 2 2

iii. Cos22A + Cos22B + Cos22C = 1 + 2Cos2A cos2B Cos2C.
iv. Sin(B + C – A) + Sin(C + A – B) + Sin(A + B – C) = 4SinA SinB SinC.
v. Cos(B + C – A) + Cos(C + A – B) + Cos(A + B – C) = 1 + 4CosA cosB CosC.

224 PRIME Opt. Maths Book - X

6. If A + B + C = 180°, prove that the followings.

i. CosA + CosB + CosC = 2.
SinB SinC SinC SinA SinA SinB

ii. SinA + SinB + SinC = 2TanA TanB TanC.
CosB CosC CosC CosA CosA CosB

iii. SinA CosB CosC + CosA SinB CosC + CosA CosB SinC = SinA SinB SinC.
iv. CosA SinB SinC + SinASinBCosC – CosA CosB CosC = 1 – SinA CosB SinC.

v. = 8Sin A Sin B Sin C .
2 2 2

7. PRIME more creative questions:

a. If A + B + c = p, prove that the followings.
r r r–
i. Cos A + Cos B + Cos C = 4Cos – A Cos – B Cos 4 C
2 2 2 4 4

ii. Sin A + Sin B + Sin C = 1 + Sin A+B Sin B + C Sin C + A
2 2 2 4 4 4

iii. Sin(A + 2B) + Sin(B + 2c) + Sin(c + 2A) = 4sin A – B Sin B – C Sin C – A
2 2 2

iv. Cos A Cos B – C + Cos B Cos C – A + Cos C Cos A– B = SinA + SinB + SinC
2 2 2 2 2 2

v. If A + B + C = 2S, prove that:

Sin( S – A) + Sin(S – B) + Sin(S – C) – SinS = 4Sin A Sin B Sin C .
2 2 2

b. If A +B+C= p , prove that the following.
2
i. Sin2A + Sin2B + Sin2C = 1 – 2SinASinBSinC

ii. Cos2A + Cos2B – Cos2C = 2CosACosBSinC

iii. Cos22A + Cos22B + Cos22C = 1 – 2Cos2ACos2BCos2C

iv. Sin22A + sin22B – Sin22C = 2Sin2ASinBCos2C

v. TanATanB + TanBTanC + TanCTanA = 1

PRIME Opt. Maths Book - X 225

5.5 Trigonometrical Equations

Let, us consider the algebraic equations
ax + b = 0 (Linear equation)
ax2 + bx + c = 0 (Quadratic equation)
ax3 + bx2 + cx + d = 0 (Cubic equation)

Such equations can be expressed by using the trigonometric functions as the variable ‘q'.
aSinq + b = 0
aSin2q + bSinq + c = 0
aSin3q + bsin2q + cSinq + d = 0

They are called the trigonometrical equations. By solving such equations, roots of the
variables Sinq, Cosq, Secq, Tanq, Cotq, Cosecq can be obtained according to the process of
solving the algebraic equations to find the roots of variable ‘x'. Value of the angles q for each
trigonometrical ratios so obtained have to be calculated between 0° to 360° in grade ‘X',
according to the values in quadrants (CAST rule).

An equation involving trigonometric functions with unknown
angles in different forms like linear, quadratic, cubic, exponential is
called trigonometrical equation.
The value of the unknown angle have to be found out by solving
them, called roots of the trigonometrical equations.

Ways of solving the trigonometrical equations
Step - 1: Reduce the given equation into the trigonometrical functions like Sinq, Cosq

etc. by using any suitable methods or operations.

4Sin2q – 1 = 0

or, Sin2q = 1
4

\ Sinq = ± 1
2

Step - 2: Determination of the quadrants where the value of trigonometrical ratios lie
1
by using CAST rule like Sinq = 2
1
Sinq = 2 ü ü

Sinq = Sin30°; Sin(180° – 30°)

\ q = 30°, 150°

Sinq = – 1
2

Sinq = Sin(180° + 30°); Sin(360° – 30°)

\ q = 210; 330 üü

226 PRIME Opt. Maths Book - X

Step - 3: Solve for the unknown angle ‘q' by using the above informations according
to respective standard angles.

Note: Taking an example Sinq = ‘K' = SinA
i) If Sinq = K is positive, the standard angles are A or 180° – A.
ii) If Sinq = K is negative, the standard angle = 180° + A or 360° – A.
iii) If Sinq = K is not the value of standard angle of 0°, 30°, 45°, 60° & 90°, A = Sin–l(k)
iv) If Sinq = K is not the value of standard angle as well as is not in the range –1 ≤
K ≤ 1, angle doesn't exit.
v) The table of standard angles.

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

Write down = 01234

Dividing by 4 = 0123 4
4444 4
Taking 4
Square root = 0123 4
4444 1

Result = 011 3 90°
1
2 22 0
∞
Tabulation of the above values respectively fro different ratios. 1
∞
Angles 0° 30° 45° 60° 0
Ratios

Sin 0 1 13
2 22

Cos 1 31 1

2 22

Tan 0 1 1 3
3

Cosec ∞ 2 2 2
3

2
Sec 1 3 2 2

Cot ∞ 3 1 1
3

PRIME Opt. Maths Book - X 227

Worked out Examples

1. Solve the equations, [0 ≤ q ≤ 360°]

i) 2Sinq – 1 = 0

Solution:

The given equation is,

2Sinq – 1 = 0

or, 2Sinq = 1 ü ü

or, Sinq = 1 ü ü
2 ü ü
ü
It is positive and lies in first as well as second quadrant. ü
ü
or, Sinq = Sin 30° or
ü
Sin (180° – 30°)

\q = 30° or 150°

ii. 2Cosq + 1 = 0

or, 2 Cos = –1.

1

or, Cosq = –

2

It is negative and lies in second and third quadrants.

or, Cosq = Cos (180° – 45°) or

Cos (180° + 45°)

\q = 135° or 225°.

iii. Tan2q – 3 = 0
Solution:
The given equation is,
Tan2q – 3 = 0
or, Tan2q = 3
or, Tanq = ± 3

It is positive as well as negative and it lies in all the quadrants.

or, Tanq = Tan60° or

Tan (180° – 60°)or

Tan (180° + 60°) or

Tan (360° – 60°)

\ q= 60°, 120°, 240° or 300°.

2. Solve the equation sin3q = cos2q (0 ≤ q ≤ 180°)
Solution:
The given equation is, sin3q = cos2q
It is positive and lies in first as well as second quadrants.
or, Sin3q = Sin(90° – 2q) or
Sin (90° + 2q) or
Sin (450° – 2q)
or, 3q = 90° – 2q, 90° + 2q or 450° – 2q

228 PRIME Opt. Maths Book - X

Here, 5q = 90°, q = 90° or 5q = 450°

` The angles, q = 18° or 90°.

3. Solve the equation 2Sin2q + 3Sinq = 0 [0 # i # 360° ] üü
üü
Solution:
The given equation is, üü

2Sin2q + 3Sinq = 0 229
or, Sinq(2Sinq + 3) = 0
Either,
Sinq = 0
or, Sinq = Sin0° or

Sin(180° – 0°) or
Sin (360° + 0°)
\ q = 0° or 180° or 360°

OR,
2Sinq + 3 = 0

or, 2Sinq = – 3

or, Sinq = – 3
2

or, Sinq = Sin(180° + 60°) or
Sin(360° – 60°)

\ q = 240° or 300°
\ The possible angles are 0°, 180°, 240°, 300° or 360°.

4. Solve the equation 2cos2q + Sinq – 1 = 0 ^0 # i # 360°h .
Solution:
The given equation is,
2Cos2q + Sinq – 1 = 0
or, 2(1 – Sin2q) + Sinq– 1= 0
or, 2 – 2Sin2q + Sinq – 1 = 0
or, 2Sin2q – Sinq – 1 = 0
or, 2sin2q – 2sinq + Sinq – 1 = 0
or, 2Sinq (Sinq – 1) + 1 (Sinq – 1) = 0
or, (Sinq – 1) (2Sinq – 1) = 0

Either,
Sinq – 1 = 0

or, Sinq = 1
It is positive and lies in first and second quadrant.
or, Sinq = Sin 90° or

Sin (180° – 90)
\ q = 90°

PRIME Opt. Maths Book - X

OR,

2Sinq + 1 = 0

or, Sinq = – 1
2

It is negative and lies in third and fourth quadrants. üü
or, Sinq = Sin(180 + 30°) or [0° # i # 360°]

Sin (360° – 30°) ü
\ q = 210° or 330° ü
\ The possible angles are, 90°, 210° or 330°.
ü
5. Solve the equation Cot2q + c 3 + 1 m Cotq = – 1 ü
3
Solution:

The given equation is,

Cot2q + c 3+ 1 m Cotq = – 1
3

or, Cot2q + 3Cotq + 1 Cotq + 1 = 0
3

or, Cotq ^Coti + 3h + 1 ^Coti + 3h = 0
3

or, (Cotq + 3) + 1 ^Coti + 3h = 0
3

or, ^Coti + 3h cCoti + 1 m = 0
3

Either,
Cotq + 3 = 0

or, Cotq = – 3

It is negative and lies in second and fourth quadrants.
or, Cotq = Cot(180° – 30°) or

Cot(360° – 30°)
\ q = 150° or 330°

OR,
Cotq + 1 = 0
3

or, Cotq = – 1
3

or, Cotq= Cot (180° – 60 °) or,
Cot (360° – 60°)

\ q = 120° or 300°
\ The possible angles are:

120°, 150°, 300° or 330°

230 PRIME Opt. Maths Book - X

6. Solve the equation 3Cosq + Sinq = 1 [0° # i # 360°]
Solution:
The given equation is,
or, 3Cosq + Sinq = 1
or, 3Cosq = 1 – Sinq

Squaring on both sides,
or, ( 3Cosq)2 = (1 – Sinq)2
or, 3Cos2q = 1 –2 Sinq + Sin2q
or, 3 – 3 Sin2q = 1 –2 Sinq + Sin2q
or, 4Sin2q – 2Sinq – 2 = 0
or, 2Sin2q – Sinq – 1 = 0
or, 2Sin2q – 2Sinq + Sinq – 1 = 0
or, 2Sinq(Sinq – 1) + 1(Sinq – 1) = 0

Either, üü
Sinq – 1 = 0

or, Sinq = 1
or, Sinq = Sin90° or

Sin(180° – 90°)
\ q = 90°

OR,

2Sinq + 1 = 0

or, Sinq = – 1
2

or, Sinq = Sin(180° + 30°) 0r üü
Sin(360° – 30°)

\ q = 210°, 330°

If squaring on both sides is done, the angles so obtained have to be checked.

For q = 90°, For q = 210°,

3Cosq + Sinq = 1 3Cosq + Sinq = 1

or, 3Cos90° + Sin90° = 1 or, 3Cos210° + Sin210° = 1

or, 3 × 0 + 1 = 1 or, 3 c– 3 m + a– 1 k =1
or, 1 = 1 (True) 2 2

or, –2 ! 1 (false)

For q= 330°,

3Cosq + Sinq = 1

or, 3Cos330° + Sin330° = 1

or, 3c 3 m +a– 1 k =1
2 2

or, 1 = 1 (True)
Here, 90° and 330° only satisfied the equation.

Hence, The possible angles are 90° and 330°.

PRIME Opt. Maths Book - X 231

Alternative method:
3Cosq + Sinq = 1

Dividing both sides by
= ^ 3h2 + ^1h2 = 2

or, 3 Cosi + Sini = 1 esqoulvaitniognf.or
2 2

c 3 m 1 n 1 of
or, 2 Cosq + d Sinq = 2 of
2 Btehsits way
1 kind
or, Sin60°Cosq + Cos60°Sinq = 2

or, Sin(60° + q) = 1
2

or, Sin(60° + q) = Sin 30° or,

Sin(180° – 30°),

Sin(360° + 30°) üü

or, 60° + q = 30°, 150° or 390°

q = – 30°, 90° or 330°

(– 30 is not in range)

\ The possible angles are 90° and 330°

7. Solve the equation Cosq – 3Sinq = 3 [0° # i # 360°]
Solution:
The given equation is,
Cosq – 3Sinq = 3

Dividing both sides by

= 12 + ^– 3h2

=2

or, Cosi – 3 Sini = 3
2 2

or, a 1 k Cosq – c 3 m Sinq = 3
2 2 2

3
or, CosqCos60° – SinqSin60° = 2

3
or, Cos(q + 60°) = 2

or, Cos(q + 60°) = Cos60° or

Cos(360° – 60°) or ü
ü
Cos (360° + 60°) [0° # i # 360°]

or, q + 60° = 60° or 300° or 420° PRIME Opt. Maths Book - X

\q = 0° or 240° or 360°

\ The possible angles are 0°, 240° or 360°.

8. Solve the equation Sinq + Sin2q + Sin3q = 0

Solution:

The given equation is,

Sinq + Sin2q + Sin3q = 0

or, (Sin3q + Sinq) + Sin2q = 0

or, 2Sin 3i + i Cos 3i – i + Sin2q = 0
2 2

232

or, 2Sin2qCosq + Sin2q = 0
or, Sin2q(2Cosq + 1) = 0

Either, ü ü
Sin2q = 0
üü
or, Sin2q = Sin0° or, 60° + 2q = 540° – 4q
Sin(180° – 0°) 6q = 480°
Sin(360° + 0°) or, q = 80°
Sin(540° – 0) or,
Sin(720 + 0°)

or, 2q = 0°, 180°, 360°, 540° or 720°
\ q = 0°, 90°, 180°, 270° or 360°

OR,

2Cosq + 1 = 0

or, Cosq = – 1
2

or, Cosq = Cos (180° – 60°) or ü
Cos(180° + 60°) ü

or, q = 120°, 240°
\ The possible angles are:

0°, 90°, 120°, 180°, 240°, 270° or 360°.

9. Solve the equation 3Cosec2q + Sec2q = 4. ( 0° # i # 90°).

Solution:

The given equation is,

3Cosec2q + Sec2q = 4

or, 3. 1 + 1 =4
Sin2i Cos2i

or, = 4

or, 3Cos2q + Sin2q = 2(2Sin2q.Cos2q)

or, 3 Cos2q + 1 Sin2q = Sin4q
2 2

or, Sin60°Cos2q + Cos60°Sin2q = Sin4q
or, Sin(60° + 2q) = Sin4q

It is positive and lies in first and second quadrants.

or, Sin(60° + 2q) = Sin4q or

Sin (180 – 4q) or

Sin(540° – 4q)

or, 60° + 2q = 4q, 180° – 4q or 540° – 4q

Here, 60° + 2q = 4q 60° + 2q = 180 – 4q

or, 2q = 60° 6q = 120°

\ q = 30° q = 20°

\ The possible angles are 20°, 30° or 80°..........

PRIME Opt. Maths Book - X 233

Exercise 5.5

1. Solve the following trigonometrical equations. [0 # i # 90°)

i) 2Sinq – 1 = 0 ii) 2Cosq – 3 = 0

ii) Tanq – 1 = 0 iv) 3Cotq – 1 = 0

v) Sinq – Cosq = 0

2. Solve the following trigonometrical equations ^0 # a # 180°h

i) 2Cosa + 2 = 0 ii) Tana + 3 = 0

iii) 4Sin2a – 3 = 0 iv) Cot2a – 1 = 0

v) Tana + Cota = 2

3. Solve the following trigonometrical equations [0 ≤ q ≤ 360°]

2 ii) 2 3Cosq.Tanq – 3 = 0
i) Cosi + 2 = 0 iv) 3Cot2q – 1 = 0
iii) 4Cos2q – 1 = 3

v) 3Tanq + 3 = 0

4. Solve the following trigonometrical ratios. ^0 # i # 180°h

i) Sin2q = Sinq ii) Cos3q = Cos3q

iii) Tan2q – Tanq = 0 iv) Sin5q + Sin4q = 0

v) Cot3q + Cot2q = 0

5. Solve the following trigonometrical equations: ^0 # i # 180°h

i) Cos3q = Sin2q ii) Cos2q + Sinq = 0

iii) Cos3q – Sinq = 0 iv) Cot2q = Tan3q

v) Cosec2q – Secq = 0

6. Solve the following trigonometrical equations. ^0 # a # 360°h

i) 2Cos2a – Cosa = 0 ii) 2Sin2a + Sina = 0

ii) 2Cos2a = 3Cosa iv) 2Cos2a = 3Sina

v) 3Tan2a = 3Tana

7. Solve the following trigonometrical equations. ^0 # A # 2rh

i) 2Sin2A + CosA – 1 = 0 ii) Sin2A – CosA = 1
iii) 3Sin2A + 4CosA = 4 4

iv) 2SinA = 5 – 2 CosecA

v) Tan2A – SecA – 1 = 0

8. Solve the following equations. ^0 # x # 2rh
i) 4Sec2x – 3 = 7Tan2x
ii) Tanx + Cotx = 2Cosecx
iii) 2Sinx = Cosecx – Cotx iv) Tan2x – c 3 + 1 m Tanx = –1

3

234 PRIME Opt. Maths Book - X

v) Cot2x – (1 – 3)Cotx = – 3

9. Solve: ^0 # i # 360°h
Secq.Tanq = 2
i) 2 3Cos2q = Sinq ii) 2 3 = 2 3Cos2q + Cosq

iii) Cosecq = 2Tanq iv) ^0 # i # 360°h
Sinq – Cosq = 1
v) Cosec2q + 3Cotq = c2 + 1 Cotim
3 2
3Sinq – Cosq = 1
10. Solve: ii)
i) Sinq + Cosq = 1

iii) 2Sinq + 2Cosq = 1 iv)
v) 3Cosq + Sinq = 3

11. PRIME more creative questions:

a. Solve: ^0 # x # 360°h
3Cosx – Sinx = 3
i) Cosx – 3Sinx = 1 ii) Tanx + Tan2x + Tan3x = Tanx.Tan2x.Tan3x

iii) Cos2x + 3Sin2x = 1 iv)

v) Tan3x – Tanx – 3Tan3x.Tanx = 3

b. Solve: ^0 # A # 360°h
i) CosA + Cos2A + Cos3A = 0 ii) SinA – Sin2A + Sin3A = 0
iii) CosA – Cos2A + Cos3A = 1 iv) SinA + CosA + Sin3A = 0
v) Sin2A + Sin4A = CosA + Cos3A

c. Solve the following: (0 ≤ q ≤ 180°)

i) 3 – 1 =4 ii) Cosec2q + 3Sec2q = 4
Sini Cosi

iii) If 2SinqSina = 3 and Cot q + Cot a = 2, find q + a.
2

iv) Tanq + Tan2q – Tan3q = 0

d. If 2Cosx.Siny = 1 and Tanx + Coty = –2, find x – y.
2

Steps of solving quadratic type:
• Convert the trigonometric ratios to the like ration for each

terms.
• Factorise the equation as algebraic equations.
• Find the value of trigonometric ratios.
• Use quadrant rule to find the possible angles between the

given range.

PRIME Opt. Maths Book - X 235

Answer

1. i) 30° ii) 30° iii) 45° iv) 60° v) 45°

2. i) 135° ii) 120° iii) 60° and 120°

iv) 45° and 135° v) 45°

3. i) 135° and 225° ii) 60° and 120° iii) 0, 180° and 360°

iv) 60°, 120 °, 240° and 360° v) 120° and 300°

4. i) 0°, 60°, and 180° ii) 0°, 90° and 180° iii) 0° and 180°

iv) 40°, 80°, 120° and 180

v) 36°, 72°, 108°, 144° and 180°

5. i) 18°, 90°, 162° ii) 90°

iii) 22.5°, 112.5°, 202.5° iv) 18°, 54°, 90°, 126°, 162°

v) 30°, 90°, 150°

6. i) 60°, 90°, 270°, 300° ii) 0°, 180°, 210°, 330°

iii) 30°, 90°, 270°, 330° iv) 30°, 150°, No solution

v) 0°, 30°, 180°, 210°

7. i) 0°, 120°, 240°, 360 ii) 60°, 300°, No solution

iii) 0°, 360°, cos – l( 1 ) iv) 30°, 150°, No solution
3
v) 180°, No solution

8. i) 30°, 150°, 210°, 330° ii) 0°, 60°, 180°, 300°, 360°

iii) 0°, 60°, 180°, 300°, 360° iv) 60°, 150°, 240°, 330°

v) 30°, 45°, 210°, 225°

9. i) 60°, 120°, No solution ii) 45°, 135°, No solution

iii) 45°, 315°, No solution iv) 30°, 330°, No solution

v) 60°, 150°, 240°, 330°

10. i) 0°, 90°, 360° ii) 75°, 195°

iii) 105°, 245° iv) 60°, 180°

v) 0°, 60°, 360°

11.

a. i) 0°, 240°, 360°, ii) 0°, 300°, 360°

iii) 0°, 60°, 180°, 240°, 360° iv) 0°, 60°, 120°, 180°, 240°, 300°, 360°

v) 30°, 120°, 210°, 300°

b. i) 45°, 120°, 135°, ii) 0°, 60°, 90°, 180°, 270°, 300°, 360°

iii) 0°, 90°, 120°, 240°, 270°, 360°

iv) 90°, 105°, 165°, 270°, 285°, 345°

v) 18°, 90°, 162°, 234°, 270°,

c. i) 20°, 120° ii) 15°, 25° & 85° iii) 60°, 120°

iv) 30°, 60°, 90°, 120°, 150°, 180°

d. 135°

236 PRIME Opt. Maths Book - X

5.6 Height and Distance

As we discussed in the topic ‘measurement of angles' in grade ‘IX', trigonometry is useful
for various purposes at present. Besides of them trigonometry is very useful to measure the
distance between any two points as well as height of pole, tower, houses, tree, columns, height
of mountain, etc. Here in grade ‘X', we are going to study about the measurement of heights
and distance by using trigonometrical ratios in a right angled triangle formed or assuming
upon the given objects. The right angled triangle can be supposed according to the study of
two types of angles, ie. angle of elevation and angle of depression.

i) Angle of elevation:

The angle made by the line of sight with the B
horizontal line at the eye of an observer where
Line of sight
the observing point is above from the observer

is called angle of elevation. Angle of elevation

Here, O A
O is the position of the observer. Observation Horizontal line
OA is the horizontal line,

OB is the line of sight while watching upward.

Then, \ BOA = Angle of elevation.

ii) Angle of depression:

The angle made by the line of sight with the

horizontal line at the eye of an observer where A q O
C
the observing point is below from the observer Angle of depression

is called angle of depression.

Here, Bq
O is the position of the observer.

AO is the horizontal line.

OB is the line of sight while watching downward.

Then, \ AOB = Angle of depression

Also, \ OBC = \ AOB (Being alternate angle)

iii) Right angled triangle:

In right angle DABC respectively by the objects taken during observation.

Where,

\ B = 90° (right angled) A
\ C = q (refrence angle)

Then, p
h
Sinq = = AB Hypertensions
AC
Perpendicular

Cosq = b = BC
h AC

Tanq = p AB C Base B
= BC

b

PRIME Opt. Maths Book - X 237

iv) Value of standard angles between 0° and 90° which are necessary in this chapter.

Ratios 30° 45° 60°
Angles

Sin 11 3

2 22

Cos 31 1
22 2

Tan 1 3
31

Where, 3 = 1.732... = 1.732
2 = 1.414... = 1.414

Activity:

Clinometer : A hand made device which is used to find
the height of the objects by using the concept of right

angled triangle is called clinometer.

• Let us take a piece of plywood and prepare an isosceles right angled triangle as
shown in diagram.

A

BC

• Connect a delivery tube on the side AC (hypoteneous) of the triangle.
• Hang a weight with a thread by making a hole near the point A as shown in

diagram.
A

BC
• Watch the top of the object whose height have to be measured by taking the

thread conciding with the side AB by moving far or near to the object as shown
in diagram.

238 PRIME Opt. Maths Book - X

P
A

T B 45° C

Q SR
• Height of the tower can be calculated by measuring the height of our body and

distance of tower from our foot as.

PQ = CR + QR
By using the concept of isosceles right angled triangle.

Worked out Examples

1. The angle of elevation of the top of an electric pole A
?
from a point on the ground is 30°. On walking 40m
B
towards the pole, elevation changes to 45°. Find the

height of the pole.

Solution:

Let, AB be the height of the electric pole.

C be the 1st position of the observer.

D be the second position of the observer.

Given: 45° 30°
D 40m
CD = 40m C

\ ADB = 45°

\ ACB = 30°

AB = ?

Now,

In right angled DABD.

Tan45° = AB
BD

or, 1 = AB
BD

\ BD = AB ..........(i)

Again, In right angled triangle ABC,

Tan30° = AB
BC

or, 1 = AB
3 BD + DC

PRIME Opt. Maths Book - X 239

or, 3AB = AB + DC [ a From equation (i) ]

or, ( 3 – 1)AB = DC

or, AB = 40
3 –1

or, AB = 40
0.732

\ AB = 54.64m

\ Height of the pole is 54.64m.

2. A 10m long ladder is taken against a wall A
Q
where top of the ladder is 10m below from
B
the top of the wall. If angle of elevation of the

top of the wall from the foot of the ladder is 10m
?
60°, find the height of the wall.

Solution:

Let, AB be the height of a wall.

PQ be the length of the ladder.

P is the position of the observer.

Given: 60° 10m
AQ = PQ = 10m

\ APB = 60° 30°

AB = ? P
Now, In right angled DABP,

\ APB = 60°

\ \ PAB = 90° – 60° = 30°

Then,

\ QAP = \ APQ = 30° [ a AQ = PQ = 10m ]

\ \ QPB = 60° – 30° = 30°

Again, In right angled DQBP,
QB

Sin30° = PQ

or, 1 QB
2 = 10m

` QB = 5m

` Height of the wall = 10m + 5m = 15m

3. The angle of depression of a boat coming towards A E
the sea - shore from the top of a light house situated 30°
at the sea - shore is found to be 30° and after 10 60°
seconds depression changes to 60°. Find the time
taken to reach the boat at sea - shore. 60° 30° C
Solution: B ?D
Let, AB be the height of a light house.
C be the 1st position of a boat.
D be the 2nd position of the boat.
Given:

240 PRIME Opt. Maths Book - X

\ EAC = \ ACB = 30°
\ EAD = \ ADB = 60°
Time taken to travel CD = 10 seconds
Time taken to travel BD = ?
Now, In right angled DABD,

Tan60° = AB
BD

or, 3 = AB
BD

or, AB = 3 BD ........................(i)

Again, In right angled DABC,

Tan30° = AB
BC

or, 1 = AB
3 BD + DC

or, 1 3 BD
3 = BD + DC [ From equation (i) ]

or, 3BD = BD + DC

\ BD = DC
2

Here, = 10 seconds
Time taken to travel CD
= 10 # CD
Time taken to travel BD CD 2

= 5 seconds

4. The angle of elevation of the foot and top of a D 15° C
flagstaff which is taken at the top of a pole are A 30°
30° and 15° respectively from a point on the 60m
ground which is 60m far from the foot of the B
pole. Find the height of the flagstaff.
Solution:
Let,
AB be the height of a pole.
AD be the height of a flagstaff.
C be the position of the observer on the ground.
Given:
BC = 60m
\ ACB = 30°
\ DCA = 15°
\ DCB = 30° + 15° = 45°

Now, In right angled triangle ABC,

Tan30° = AB
BC

or, 1 = AB
3 60

PRIME Opt. Maths Book - X 241

or, AB = 60 = 34.64 .............................(i)
3

Again, In right angled triangle DBC,

Tan45° = BD
BC

or, 1 = BD
60
` BD
= 60m

Then, Height of flagstaff,
AD = 60m – 34.64m

= 25.36m

5. A bird is flying horizontally from a

height of 200m from the ground where ? C
angle of elevation is changed from 45° A
to 30° within 10 seconds, find the speed

of the bird.

Solution: 200 m

Let, AB (CD) be the height of bird from

the ground.

P be the observation point on the 45°
ground. 30°
AC be the distance travelled by bird in
10 seconds. P B D
Given:

AB = CD = 200m

\ APB = 45°

\ CPD = 30°

Speed of the bird = ?

Now,

In right angled DABP,

Tan45° = AB
PB

1 = 200
PB

\ PB = 200m .........(i)

Again,

In right angled DCPD,

Tan 30° = CP
PD

or, 1 = 200
3 PD

\ PD = 346.20 m ..........(ii)

\ Distance travelled by the bird in 20 sec

AC = BD = 346.20 – 200
= 146.20m

242 PRIME Opt. Maths Book - X

\ Speed of the bird =

= 146.20
10

= 14.62 m/s.

6. The angle of elevation and depression of the top and A

bottom of a tower from the top of 24m high house are

found to be 30° and 60° respectively. Find the height of

the tower.

Solution: 30° C
Let, AB be the height of a tower. E D

CD be the height of a house. 60°
BD be the distance between them.

C be the position of the observer.

Given: 120cm
24m
CD = EB = 24m

\ ACE = 30°

\ ECB = \ CBD = 60°

CE = BD 60°
AB = ? B

Now, In right angled DABC,

Tan30° = AE
EC

or, 1 = AE
3 EC

or, EC = 3AE ..........(i)

Again, In right angled DCDB,

Tan60° = CD
BD

or, 3= 24
EC

or, 3 ^ 3 AEh = 24 [ ` From equation (i) ]

or, 3(AB – EB) = 24

or, 3AB = 96m

` AB = 32m
` Height of tower is 32m.

7. The angle of depression of the top and bottom of a pole A 30° F
from the top of a tower of height 120 m are found to be E 60° C
30° and 45° respectively. Find the height of the pole. B
Solution : 30°
Let, AB be the height of a tower.
CD be the height of a pole. 45° D
A be the position of the observer.
BD be the distance between them.
Given,

PRIME Opt. Maths Book - X 243

AB = 120 m

\ FAC = \ ACE = 30°

\ FAD = \ ADB = 45°

CD = EB = ?

CE = BD.

Now, In right angled DABD,

Tan 45° = AB
BD

\ BD = 120 m

Again, In right angled DAEC,

Tan 30° = AE
EC

or, 1 = AB – EB
3 BD

or, 3(120 – CD) = 120

or, 120( 3 – 1) = 3CD

or, CD = 120 × 0.732
1.732

\ CD = 50.72m
\ Height of the pole is 50.72 m.

8. The angled of elevations of the top of the A

poles where one is double then the other from

mid-way between their bases are found to be

complementary. Find the height of the taller

pole where they are 120m apart. C
x
Solution : 2x
Let, AB and CD be the height of two poles. D

BD be the distance between them.

P be the position of the observer. 90 – q q

Given : B P
AB = 2 CD
120m
BD = 120m

BP = PD = 60m

\ CPD = q (say)

\ APB = 90° – q

Now, In right angled DCDP,

Tanq = CD
PD

or, Tanq = CD ..... (i)
60

Again, In right angled DABP,

Tan(90° – q) = AB
BP

or, Cotq = AB
60

244 PRIME Opt. Maths Book - X

or, 1 = AB
Tani 60

or, 60 = AB
AB 60

2

or, 60 = AB
CD 60

or, AB2 = 3600
2

or, AB = 7200

\ AB = 84.85 m

\ Height of taller pole is 84.85 m.

9. The angle of elevations of the top of a tower 80m C

high from the bottom and top of a house 60m

high are found to be complementary. Find the

distance between their bases.

Solution : 60m
80m
Let, AB be the height of a house. A 90 – q E
CD be the height of a tower. D
q=?
BD be the distance between them. ?

Given :

AB = 60 m B
CD = 80 m

CE = (80 – 60)m = 20 m

\ CBD = q (say) then \ CAE = 90° – q

BD = AE = ?

Now, In right angled DCBD,

Tanq = CD
BD

or, Tanq = 80 .......... (i)
BD

Again, In right angled DCEA

Tan (90° – q) = CE
AE

or, Cotq = 20
BD

or, BD = 20 [from equation (i)]
80 BD

or, BD2 = 1600
\ BD = 40 m
\ Distance between them is 40 m.

PRIME Opt. Maths Book - X 245