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Contents
(Click on the Chapter Name to download the solutions for the desired Chapter)
Chapter 1 : Real Numbers
Chapter 2 : Polynomials
Chapter 3 : Pair of Linear Equations in Two Variables
Chapter 4 : Triangles
Chapter 5 : Trigonometric Ratios
Chapter 6 : Trigonometric Identities
Chapter 7 : Statistics
Chapter 8 : Quadratic Equations
Chapter 9 : Arithmetic Progression
Chapter 10 : Circles
Chapter 11 : Constructions
Chapter 12 : Some Applications of Trigonometry
Chapter 13 : Probability
Chapter 14 : Coordinate Geometry
Chapter 15 : Areas related to Circles
Chapter 16 : Surface Areas and Volumes
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Class X Chapter 5 β Trigonometric Ratios Maths
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Exercise 5.1
1. In each of the following one of the six trigonometric ratios is given. Find the values of the
other trigonometric ratios.
Sol:
(i) Sin A = 2
3
We know that Sin π = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π
Let us Consider a right angled βle ABC.
By applying Pythagorean theorem we get π΄πΆ2 = π΄π΅2 + π΅πΆ2
9 = π₯2 + 4
π₯2 = 9 - 4
x = β5
We know that cos = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π and
tanπ = πππππ ππ‘π π πππ
ππππππππ‘ π πππ
So, cosπ = β5
3 ;
sec = 1
πππ π=
3
β5
tanπ = 2
β5 ;
cot = 1
π‘πππ =
β5
2
cosecπ = 1
π πππ =
3
2
(ii)
Cos A = 4
5
We know that cosπ = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π
Let us consider a right angled βle ABC.
Class X Chapter 5 β Trigonometric Ratios Maths
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Let opposite side BC = x.
By applying pythagornβs theorem, we get π΄πΆ2 = π΄π΅2 + π΅πΆ2
25 = x + 16
x = 25 - 16 = 9
x = β9 = 3
We know that cosA = 4
5
sinA = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
3
5
tanA = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
3
4
cosecA = 1
π πππ΄ =
1
3
5 =
5
3
secA = 1
πππ π΄ =
1
4
5 =
5
4
cotA = 1
π‘πππ΄ =
1
3
4 =
4
3
(iii)
tanπ = 11.
We know that tanπ = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
11
1
Consider a right angled βle ABC.
Let hypotenuse AC = x, by applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 112 + 12
π₯2 = 121 + 1
π₯ = β122
We know that sinπ = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
11
β122
cosπ = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π =
1
β122
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cosecπ = 1
π πππ =
1
11
β122 =
β122
11
secπ = 1
πππ π =
1/1
β122 = β122
cotπ = 1
π‘πππ =
1
11 =
1
11
(iv)
Sin π = 11
5
We know Sin π = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
11
15
Consider right angled βle ACB.
Let x = ππππππππ‘ π πππ
By applying Pythagoras π΄π΅2 = π΄πΆ2 + π΅πΆ2
225 = 121+π₯2
π₯2 = 225 -121
π₯2 = 104
π₯ = β104
cos = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π = β
104
15
tan = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
11
β104
cosecπ = 1
π πππ =
15
11
sec = 1
πππ π=
15
β104
cot = 1
π‘πππ =
β104
11
(v)
tanΞ± = 5
12
We know that tanΞ±= πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
5
12
Now consider a right angled βle ABC.
Class X Chapter 5 β Trigonometric Ratios Maths
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Let x = hypotenuse .By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 52 + 122
π₯2 = 25 + 144 = 169
π₯ = 13
sinΞ± = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
5
13
cosΞ± = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π =
12
13
cotΞ± = 1
π‘ππΞ± =
12
15
cosecΞ± = 1
π πππΌ =
1/5
13 =
13
5
secΞ± = 1
πππ πΌ =
1
12
13 =
13
12 .
(vi)
Sin π = β3
2
We know Sin π = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
β3
2
Now consider right angled βle ABC.
Let x = ππππππππ‘ π πππ
By applying Pythagoras
π΄π΅2 = π΄πΆ2 + π΅πΆ2
4 = 3+π₯2
π₯2 = 4 β 3
π₯2 = 1
π₯ = 1
cos = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π =
1
2
tan = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
β3
1 = β3
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cosecπ = 1
π πππ =
1
β3
2
=2
β3
sec = 1
πππ π=
1
1
2= 2
cot = 1
π‘πππ =
1
β3
(vii)
Cosπ = 7
25.
We know that cosπ = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π
Now consider a right angled βle ABC,
Let x be the opposite side.
By applying pythagornβs theorem π΄πΆ2 = π΄π΅2 + π΅πΆ2
(25)2 = π₯2 + 72
625 - 49 = π₯2
576 = β576 = 24
sinπ = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
24
25
tanπ = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
24
7
cosecπ = 1
π πππ =
1
3
5 =
25
24
secπ = 1
πππ π =
1
4
5 =
25
7
cotπ = 1
π‘πππ =
1
3
4 =
7
24
(viii)
tanπ = 8
15
We know that tanπ= πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
8
15
Now consider a right angled βle ABC.
Class X Chapter 5 β Trigonometric Ratios Maths
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By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 82 + 152
π₯2 = 225 + 64 = 289
π₯ = β289 = 17
sinπ = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
8
17
cosπ = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π =
15
17
tanπ = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
8
15
cotπ = 1
π‘πππ =
18
15
= 15
8
cosecπ = 1
π πππ =
1
8
17 =
17
8
secπ = 1
πππ π =
1
15
17 =
17
15
(ix)
cotπ = 12
5
cotΞ±= ππππππππ‘ π πππ
πππππ ππ‘π π πππ =
12
5
Now consider a right angled βle ABC,
By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 25 + 144
π₯2 = 169 = β169
π₯ = 13
tanπ = 1
πππ‘π =
1
12
5 =
5
12
sinπ = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
5
13
cosπ = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π =
12
13
cosecπ = 1
π πππ =
1
5/13=
13
5
secπ = 1
πππ π =
1
12/13 =
13
12
(x)
secπ = 13
5
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secπ = βπ¦πππ‘πππ’π π
ππππππππ‘ π πππ =
13
5
Now consider a right angled βle ABC,
By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
169 = π₯2 + 25
π₯2 = 169 β 25 = 144
π₯ = 12
cosπ = 1
π πππ =
1
13
5=
5
13
tanπ = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
12
5
sinπ = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
12
13
cosecπ = 1
π πππ =
1
12/13=
13
12
secπ = 1
πππ π =
1
5/13 =
13
5
cotπ = 1
π‘πππ=
1
12/5=
5
12
(xi)
cosecπ = β10
cosecπ = βπ¦πππ‘πππ’π π
πππππ ππ‘π π πππ = β10
consider a right angled βle ABC, we get
Let x be the adjacent side.
By applying pythagoraβs theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
(β10)2
= 12 + π₯2
π₯2 = 10 β 1 = 9
π₯ = 3
sinπ = 1
πππ πππ =
1
β10
cosπ = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π =
3
β10
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tanπ = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
1
3
secπ = 1
πππ π =
β10
3
cotπ = 1
π‘πππ=
1
1
3= 3.
(xii)
Cosπ = 12
5
cosπ = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π=
12
15 .
Let x be the opposite side.
By applying pythagornβs theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
225 = π₯2 + 144
225 β 144 = π₯2
π₯2 = 81
π₯ = 9
sinπ = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
9
15
tanπ = πππππ ππ‘π π πππ
ππππππππ‘ π πππ =
9
12
cosecπ = 1
π ππΞΈ =
1
9
15 =
15
9
secπ = 1
πππ ΞΈ =
1
12
15=
15
12
cotπ = 1
π‘ππΞΈ =
1
9
12 =
12
9
2. In a βABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) Sin A, Cos A
(ii) Sin C, cos C
Sol:
βABC is right angled at B
AB = 24cm, BC = 7cm.
Class X Chapter 5 β Trigonometric Ratios Maths
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Let βxβ be the hypotenuse,
By applying Pythagoras
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 242 + 72
π₯2 = 576 + 49
π₯2 = 625
π₯ = 25
a. Sin A, Cos A
At β A, opposite side = 7
adjacent side = 24
hypotenuse = 25
sin A = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π =
7
25
cos A = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π=
24
25
b. Sin C, Cos C
At β C, opposite side = 24
adjacent side = 7
hypotenuse = 25
sin C = 24
25
cos C = 7
25
3. In Fig below, Find tan P and cot R. Is tan P = cot R?
Sol:
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Let x be the adjacent side.
By Pythagoras theorem
ππ 2 = ππ2 + π π2
169 = π₯2 + 144
π₯2 = 25
π₯ = 5
At LP, opposite side = 5
Adjacent side = 12
Hypotenuse = 13
tan P =
1
12
5β
5
12
At LR, opposite side = 12
Adjacent side = 5
Hypotenuse = 13
cot R = 1
tan π =
112
5
=5
12
[β΅ Tan R = πππππ ππ‘π π πππ
ππππππππ‘ π πππ]
β΅ tan P = cot R
4. If sin A = 9
41, πππππ’π‘π cos π΄ πππ tan π΄
Sol:
sin π΄ =9
41
Sin A = πππππ ππ‘π π πππ
ππππππππ‘ π πππ=
9
41
Consider right angled triangle ABC,
Let x be the adjacent side
By applying Pythagorean
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π΄πΆ2 = π΄π΅2 + π΅πΆ2
412 = 122 + 92
π₯2 = 412 β 92
π₯ = 40
cos π΄ =ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π=
40
41
tan π΄ =πππππ ππ‘π π πππ
π»π¦πππ‘πππ’π π π πππ=
9
40
5. Given 15 cot A = 8, find Sin A and sec A.
Sol:
15 cot A = 8, find Sin A and sec A
Cot A = 8
15
Consider right angled triangle ABC,
Let x be the hypotenuse,
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = (8)2 + (15)2
π₯2 = 64 + 225
π₯2 = 289
π₯ = 17
Sin A = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π=
15
17
Sec A = 1
cos π΄
cos π΄ =ππππππππ‘ π πππ
π»π¦πππ‘πππ’π π=
8
17
Sec A = 1
cos π΄=
1
8/17=
17
8
6. In βPQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec
P and sec R.
Sol:
βPQR, right angled at Q.
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Let x be the hypotenuse
By applying Pythagoras
ππ 2 = ππ2 + ππ 2
π₯2 = 42 + 32
π₯2 = 16 + 9
β΄ π₯ = β25 = 5
Find sin π , sin π , sec π, sec π
At LP, opposite side = 3 cm
Adjacent side = 4 cm
Hypotenuse = 5
sin π =πππππ ππ‘π π πππ
π»π¦πππ‘πππ’π π=
3
5
sec π =π»π¦πππ‘πππ’π π
ππππππππ‘ π πππ=
5
4
At LK, opposite side = 4 cm
Adjacent side = 3 cm
Hypotenuse = 5 cm
Sin R = 4
5
Sec R = 5
3
7. If cot π = 7
8, ππ£πππ’ππ‘π:
(i) (1+sin π) (1βsin π)
(1+cos π)(1βcos π)
(ii) Cot2π
Sol:
Cot π = 7
8
(i) (1+sin π) (1βsin π)
(1+cos π)(1βcos π)
= 1βsin2 π
1βcos2 π [β΅ (a + b) (a β b) = π2 β π2] a = 1, b = sin π
We know that πππ2π + cos2 π = 1
1 β sin2 π = cos2 π = cos2 π
1 β cos2 π = sin2 π
=cos2 π
sin2 π
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= cot2 π
= (cot π)2 = [7
8]
2
=49
64
(ii) cot2 π
β (cot π)2 = [7
8]
2
= 49
64
8. If 3 cot A = 4, check whether 1βtan2 π΄
1+tan2 π΄= cos2 π΄ β sin2 π΄ or not.
Sol:
3 cot A = 4, check =1βtan2 π΄
1+tan2 π΄= cos2 π΄ β sin2 π΄
Cot A = ππππππππ‘ π πππ
πππππ ππ‘π π πππ=
4
3
Let x be the hypotenuse
By Applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 42 + 32
π₯2 = 25
π₯ = 5
Tan A = 1
cos2 π΄=
3
4
Cos A = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π=
4
5
Sin A = 3
5
LHS = 1βtan2 π΄
1+tan2 π΄=
1β(3
4)
2
1+(3
4)
2 =16β9
1616+9
16
=7
25
RHS cos2 π΄ β sin2 π΄ = (4
5)
2
β (3
5)
2
=16β9
25
= 7
251
9. If tan π = π
π , find the value of
cos π+sin π
cos πβsin π
Sol:
Tan π = π
π find
cos π+sin π
cos πβsin π β¦.(i)
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Divide equation (i) with cos π, we get
β
cos π+sin π
cos πcos πβsin π
cos π
β 1+
sin π
cos π
1βsin π
cos π
β 1+tan π
1βtan π
= 1+
π
π
1βπ
π
= π+π
πβπ
10. If 3 tan π = 4, ππππ π‘βπ π£πππ’π ππ4 cos πβsin π
2 cos π+sin π
Sol:
3 tan π = 4 find 4 cos π βsin π
2 cos π+sin π β¦(i)
Tan π = 4
3
Dividing equation (i) with cos π we get
=4 cos πβsin π
cos π2 cos π+sin π
cos π
=4βtan π
2+tan π [β΅
sin π
cos π= tan π]
= 4βtan π
2+tan π [β΅
sin π
cos π= tan π]
= 4β
4
1
2+4
5
= 12β4
6+4
= 8
10
= 4
5
11. If 3 cot π = 2, find the value of =4 sin πβ3 cos π
2 sin π+6sin π
Sol:
3 cot π = 2 find 4 sin πβ3 cos π
2 sin π+6 cos π β¦ (i)
Cot π = 2
3
=4 sin πβ3 cos π
sin π2 sin π+6 cos π
sin π
=4β3 cot π
2+6 cot π
=4β3Γ
2
3
2+6Γ2
3
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=4+2
2+4=
2
6
=1
3
12. If tan π = π
π, prove that
π sin πβπ cos π
π sin π +π cos π=
π2βπ2
π2+π2
Sol:
Tanπ = π
π. PT
π sin πβπ cos π
π sin π +π cos π=
π2βπ2
π2+π2
Let π sin πβπ cos π
π sin π+π cos π β¦(i)
Divide both Nr and Dr with cos π of (a)
=
π sin πβπ cos π
cos ππ sin π+π cos π
cos π
= π tan πβπ
π tan π+π
= πΓ(
π
π)βπ
πΓ(π
π)+π
= π2βπ2
π2+π2
13. If sec π = 13
5, show that
2 cos πβ3 cos π
4 sin πβ9πππ π = 3
Sol:
πππ π =13
5
Sec π = π»π¦πππ‘πππ’π π
ππππππππ‘ π πππ=
13
5
Now consider right angled triangle ABC
By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
169 = π₯2 + 25
π₯2 = 169 β 25 = 144
π₯ = 12
πππ π =1
sec π=
1
13=
5
3
tan π =πππππ ππ‘π π πππ
ππππππππ‘ π πππ=
12
13
sin π =πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π=
12
13
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Cosec π = 1
sin π=
1
12/13=
13
12
sec π =1
cos π=
1
5/13=
13
5
Cot π = 1
tan π =
1
12/5=
5
12
14. If cos π = 12
13, show that sin π (1 β tan π) =
35
156
Sol:
Cos π = 12
3 S.T Sin π (1 β tan π) =
35
156
Cos π = ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π=
12
13
Let x be the opposite side
By applying Pythagoras
π΄πΆ2 = π΄π΅2 + π΅πΆ2
169 = π₯2 + 144
x = 25
x = 5
sin π =π΄π΅
π΄πΆ=
5
3
Tan π = π΄π΅
π΅πΆ=
5
12
sin π (1 β tan π) =5
13(1 β
5
12)
= 5
13[
7
12] =
35
156
15. If cot π = 1
β3, show that
1βcos2 π
2βsin2 π=
3
5
Sol:
cot π =1
β3 1βcos2 π
2βsin2 π=
3
5
Cot π = ππππππππ‘ π πππ
πππππ ππ‘π π πππ=
1
β3
Let x be the hypotenuse
By applying Pythagoras
π΄πΆ2 = π΄π΅2 + π΅πΆ2
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π₯2 = (β3)2
+ 1
π₯2 = 3 + 1
π₯2 = 3 + 1 β π₯ = 2
Cos π = π΅πΆ
π΄πΆ= β
1
2
Sin π =π΄π΅
π΄πΆ=
β3
2
1βcos2 π
2βsin2 πβ
1β(1
2)
2
2β(β3
2)
2
β1β
1
4
2β3
4
β3
45
4
= 3
5
16. If tan π = 1
β7
πππ ππ2πβsec2 π
πππ ππ2π+sec2 π=
3
4
Sol:
Tanπ = 1
β7
πππ ππ2πβsec2 π
πππ ππ2π+sec2 π=
3
4
Tan π = πππππ ππ‘π π πππ
ππππππππ‘ π πππ
Let βxβ be the hypotenuse
By applying Pythagoras
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 12 + (β7)2
π₯2 = 1 + 7 = 8
π₯ = 2β2
Cosec π = π΄πΆ
π΄π΅= 2β2
Sec π = π΄πΆ
π΅πΆ=
2β2
β7
Substitute, cosec π, sec π in equation
β (2β2)
2β (2β
2
7)
2
(2β2)2
+(2β2
β7)
2
8β4Γ2
7
8+4Γ2
7
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β8β
8
7
8+8
7
=56β8
756+8
7
=48
64
=3
4
πΏ. π». π = π . π». π
17. If Sin π = 12
13 find
sin2 πβcos2 π
2 sin π cos πΓ
1
π‘ππ2π
Sol:
Let x be the adjacent side
By applying Pythagoras
π΄πΆ2 = π΄π΅2 + π΅πΆ2
169 = 144 + π₯
π₯2 = 25
π₯ = 5
Cos ΞΈ =π΅πΆ
π΄πΆ=
5
13
Tan ΞΈ = π΄π΅
π΅πΆ=
12
5
β (
12
13)
2β(
5
13)
πΌ Γ2
13Γ
5
13
Γ1
[12
5]
2
β 144β25
169
24Γ5
169
Γ25
144
β 119
169120
169
Γ25
144=
129
120Γ
25
144=
595
3456
18. If sec π = 5
4 , find the value of
sin πβ2 cos π
tan πβcot π
Sol:
Not given
19. If cos π = 5
13 , find the value of
sin2 πβcos2 π
2π ππ π cos π=
3
5
Sol:
Not given
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20. Tan π = 12
13 Find
2 sin π cos π
cos2 πβsin2 π
Sol:
Tan π = πππππ ππ‘π π πππ
ππππππππ‘ π πππ
Let x be, the hypotenuse
By Pythagoras we get
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 144 + 169
π₯ = β313
Sin π = π΄π΅
π΄πΆ=
12
β313
Cos π = π΅πΆ
π΄πΆ=
13
β313
Substitute, Sin π, cos π in equation we get
2 sin π cos π
cos2 πβsin2 πβ
2Γ12
β313Γ
13
β313169
313β
144
313
=312
31325
313
=312
25
21. If cos π = 3
5, find the value of
sin πβ1
tan π
2 tan π
Sol:
Cos π = 3
5 find value of
sin πβ1
tan π
2 tan π
We know that cos π =ππππππππ‘ π πππ
βπ¦πππ‘πππ’π π
Let us consider right angled βle ABC
Let x be the opposite side, By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
25 = π₯2 + 9
π₯2 = 16 β π₯ = 4
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Sin π = π΄π΅
π΄πΆ=
4
5
Tan π = π΄π΅
π΅πΆ=
4
3
Substitute sin π, tan π in equation we get
sin πβ1
tan π
2 tan π=
4
5β
3
4
2Γ4
3
=16β15
208
3
=1
208
3
=1
20Γ
3
8=
3
160
22. If sin π = 3
5, ππ£πππ’ππ‘π
cos πβ1
tan π
2 cot π
Sol:
Not given
23. If sec A = 5
4, verify that
3 sin π΄β4 sin3 π΄
4 cos3 π΄β3 cos π΄=
3 tan π΄βtan3 π΄
1β3 tan2 π΄
Sol:
Not given
24. If sin π = 3
4, prove that β
πππ ππ2πβcot2 π
sec2 πβ1=
β7
3
Sol:
Not given
25. If sec A = 17
8, verify that
3β4 sin2 π΄
4 cos2 π΄β3=
3βtan2 π΄
1β3 tan2 π΄
Sol:
Sec A = 17
8 verify that
3β4 sin2 π΄
4 cos2 π΄β3=
3βtan2 π΄
1β3 tan2 π΄
We know sec A = βπ¦πππ‘πππ’π π
ππππππππ‘ π πππ
Consider right angled triangle ABC
Let x be the adjacent side
By applying Pythagoras we get
π΄πΆ2 = π΄π΅2 + π΅πΆ2
(17)2 = π₯2 + 64
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π₯2 = 289 β 64
π₯2 = 225 β π₯ = 15
Sin A = π΄π΅
π΅πΆ=
15
17
Cos A = π΅πΆ
π΄πΆ=
8
17
Tan A = π΄π΅
π΅πΆ=
15
8
L.H.S = 3β4 sin2 π΄
4 cos2 π΄β3=
3β4Γ(15
17)
2
4Γ(8
17)
2β3
=3β4Γ
225
289
4Γ64
289β3
=867β900
256β867=
β33
β611=
33
611
R.H.S = 3βtan2 π΄
1β3 tan2 π΄=
3β(15
8)
2
1β3Γ(15
8)
2 =3β
225
64
1β3Γ225
64
=β33
64β611
64
=β33
β611=
33
611
β΄ LHS = RHS
26. If cot π = 3
4, prove that β
sec πβπππ ππ π
sec π+πππ ππ π=
1
β7
Sol:
Cot π = 3
4 P.T β
sec πβπππ ππ π
sec π+πππ ππ π=
1
β7
Cot π = ππππππππ‘ π πππ
πππππ ππ‘π π πππ
Let x be the hypotenuse by applying Pythagoras theorem.
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 16 + 9
π₯2 = 25 β π₯ = 5
sec π =π΄πΆ
π΅πΆ=
5
3
Cosec π = π΄πΆ
π΄π΅=
5
4
On substituting in equation we get
βsec πβπππ ππ π
sec π+πππ ππ π= β
5
3β
5
45
3+
5
4
= β20β15
1220+15
12
= β5
35=
1
β7
27. If tan π = 24
7, find that sin π + cos π
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Sol:
Tan π = 24
7ππππ sin π + cos π
Let x β 1 be the hypotenuse By applying Pythagoras theorem we get
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = (24)2 + (7)2
π₯2 = 576 + 49 = 62.5
x = 25
sin π = π΄π΅
π΄πΆ=
24
25
cos π = π΅πΆ
π΄πΆ=
7
25
sin π + cos π = 24
25+
7
25
= 31
25
28. If sin π = π
π, find sec π + tan π in terms of a and b.
Sol:
Sin π = π
π find sec π + tan π
We know sin π = πππππ ππ‘π π πππ
βπ¦πππ‘πππ’π π
Let x be the adjacent side
By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π2 = π2 + π₯2
π₯2 = π2 β π2
π₯ = βπ2 β π2
sec π =π΄πΆ
π΅πΆ=
π
βπ2βπ2
Tan π = π΄π΅
π΅πΆ=
π
βπ2βπ2
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Sec π + tan π = π
βπ2βπ2+
π
βπ2βπ2
= π+π
βπ2βπ2=
π+π
β(π+π)(πβπ)=
π+π
βπ+πβ
1
βπβπ= β
π+π
πβπ
29. If 8 tan A = 15, find sin A β cos A.
Sol:
8 tan A = 15 find. Sin A β cos A
Tan A = 15
8
Tan A = πππππ ππ‘π π πππ
ππππππππ‘ π πππ
Let x be the hypotenuse By applying theorem.
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π₯2 = 152 + 82
π₯2 = 225 + 64
π₯2 = 289 β x = 17
πππ π΄ =π΄π΅
π΄πΆ=
15
17
Sin A β cos A = 15
17β
8
17
=7
17
30. If 3cos π β 4sin π = 2cos π + sinπ Find tan π
Sol:
3 cos π β 2 cos π = 4 sin π + sin π find tan π
3 cos π β 2 cos π = sin π + 4 sin π
Cos π = 5 sin π
Dividing both side by use we get cos π
cos π=
5 sin π
cos π
1 = 5 tan π
β tan π = 1
31. If tan π = 20
21, show that
1βsin π+cos π
1+sin π+cos π=
3
7
Sol:
Tan π = 20
21 S.T
1βsin π+cos π
1+sin π+cos π=
3
7
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Tan π = πππππ ππ‘π π πππ
πππππππππ‘ π πππ=
20
21
Let x be the hypotenuse By applying Pythagoras we get
π΄πΆ2 + π΄π΅2 + π΅πΆ2
π₯2 = (20)2 + (21)2
π₯2 = 400 + 441
π₯2 = 841 β x = 29
πππ π =π΄π΅
π΄πΆ=
20
29
Cos π = π΅πΆ
π΄πΆ=
21
29
Substitute sin π, cos π in equation we get
β 1βsin π+cos π
1+sin π+cos π
β 1β
20
29+
21
29
1+20
29+
21
29
=29β20+21
2929+20+21
29
=30
70=
3
7
32. If Cosec A = 2 find 1
πππ π΄+
sin π΄
1+cos π΄
Sol:
Cosec A = βπ¦πππ‘πππ’π π
πππππ ππ‘π π πππ=
2
1
Let x be the adjacent side
By applying Pythagoras theorem
π΄πΆ2 = π΄π΅2 + π΅πΆ2
4 = 1 + π₯2
π₯2 = 3 β π₯ = β3
Sin A = 1
πππ ππ π΄=
1
2
Tan A = π΄π΅
π΅πΆ=
1
β3
Cos A = π΅πΆ
π΄πΆ=
β3
2
Substitute in equation we get
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1
ππππ΄+
sin π΄
1+cos π΄=
11
β3
+1
2
1+β3
2
= β3 +1
22+β3
2
= β3 +1
2+β3=
2β3+3+1
2+β3=
2β3+4
2+β3=
2(2+β3)
2+β3= 2
33. If β A and β B are acute angles such that cos A = cos B, then show that β A = β B.
Sol:
β A and β B are acute angles.
Cos A = cos B S.T β A = β B
Let us consider right angled triangle ACB.
We have cos A = ππππππππ‘ π πππ
π»π¦πππ‘πππ’π π
= π΄πΆ
π΄π΅
Cos B = π΅πΆ
π΄π΅
πΆππ π΄ = cos π΅ π΄πΆ
π΄π΅=
π΅πΆ
π΄π΅
AC = BC
β A = β B
34. If β A and β P are acute angles such that tan A = tan P, then show that β A = β P.
Sol:
A and P are acute angle tan A = tan P
S. T. β A = β P
Let us consider right angled triangle ACP,
We know tan π = πππππ ππ‘π π πππ
ππππππππ‘ π πππ
Tan A = ππΆ
π΄πΆ
Tan A = π΄πΆ
ππΆ
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Tan A = π΄πΆ
ππΆ
Tan = tan P π·πΆ
π΄πΆ=
π΄πΆ
ππΆ
(ππΆ)2 = (π΄πΆ)2
ππΆ = π΄πΆ [β΅ Angle opposite to equal sides are equal]
β π = β π΄
35. In a βABC, right angled at A, if tan C = β3, find the value of sin B cos C + cos B sin C.
Sol:
In a βle ABC right angled at A tan C = β3
Find sin B cos C + cos B sin C
Tan c = β3
Tan C = πππππ ππ‘π π πππ
ππππππππ‘ π πππ
Let x be the hypotenuse By applying Pythagoras we get
π΅πΆ2 = π΅π΄2 + π΄πΆ2
π₯2 = (β3)2
+ 12
π₯2 = β β π₯ = 2
At β B, sin B = π΄πΆ
π΅πΆ=
1
2
Cos B = β3
2
At β C, sin = β3
2
Cos c = 1
2
On substitution we get
β 1
2Γ
1
2+
β3
2Γ
β3
2
β 1
4+
(β3)
4Γ (β3) =
β3Γβ3+1
4=
3+1
4=
4
4= 1
36. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) Sec A = 12
5 for some value of angle A.
(iii) Cos A is the abbreviation used for the cosecant of angle A.
(iv) Sin π = 4
3 for some angle π.
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Sol:
(a) Tan A β 1
Value of tan A at 45Β° i.e., tan 45 = 1
As value of A increases to 90Β°
Tan A becomes infinite
So given statement is false.
(b) Sec A = 12
5 for some value of angle of
M-I
Sec A = 2.4
Sec A > 1
So given statement is True
M-II
For sec A = 12
5
For sec A = 12
5 we get adjacent side = 13
We get a right angle βle
Subtending 9i at B.
So, given statement is true
(c) Cos A is the abbreviation used for cosecant of angle A.
The given statement is false. β΄ Cos A is abbreviation used for cos of angle A but not for
cosecant of angle A.
(d) Cot A is the product of cot A and A
Given statement is false
β΅ cot A is co-tangent of angle A and co-tangent of angle A = ππππππππ‘ π πππ
πππππ ππ‘π π πππ
(e) Sin π = 4
3 for some angle π
Given statement is false
Since value of sin π is less than (or) equal to one. Here value of sin π exceeds one, so
given statement is false.
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Exercise 5.2
Evaluate each of the following (1 β 19):
1. sin 45Β° sin 30Β° + cos 45Β° cos 30Β°
Sol:
sin 45Β° sin 30Β° + cos 45Β° cos 30Β° β¦ . (π)
We know that by trigonometric ratios we have,
Sin 45Β° =1
β2 sin 30Β° =
1
2
Cos 45Β° =1
β2 cos 30Β° =
β3
2
Substituting the values in (i) we get
1
β2β
1
2+
1
β2β
β3
2
= 1
β2β
β3
2β2=
β3+1
2β2
2. Sin 60Β° cos 30Β° + cos 60Β° sin 30Β°
Sol:
Sin 60Β° cos 30Β° + cos 60Β° sin 30Β° β¦(i)
By trigonometric ratios we have,
Sin 60Β° = β3
2 sin 30Β° =
1
2
Cos 30Β° = β3
2 cos 60Β° =
1
2
Substituting above values in (i), we get
β3
2β
β3
2+
1
2β
1
2
= 3
4+
1
4=
4
4= 1
3. Cos 60Β° cos 45Β° - sin 60Β° β sin 45Β°
Sol:
Cos 60Β° cos 45Β° - sin 60Β° β sin 45Β° β¦(i)
By trigonometric ratios we know that,
Cos 60Β° = 1
2 cos 45Β° =
1
β2
Sin 60Β° = β3
2 sin 45Β° =
1
β2
By substituting above value in (i), we get
1
2β
1
β2β
β3
2β
1
β2 β
1ββ3
2β2
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4. πππ2 30Β° + sin2 45Β° + sin2 60Β° + sin2 90Β°
Sol:
πππ2 30Β° + sin2 45Β° + sin2 60Β° + sin2 90Β° β¦(i)
π΅π¦ π‘πππππππππ‘πππ πππ‘πππ π€π βππ£π
Sin 30Β° = 1
2 sin 45Β° =
1
β2
Sin 60Β° = β3
2 sin 90Β° = 1
By substituting above values in (i), we get
= [1
2]
2
+ [1
β2]
2
+ [β3
2]
2
+ [1]2
= 1
4+
1
2+
3
4+ 1 β
1+3
4+
1+2
2
β 1 +3
2=
2+3
2=
5
2
5. cos2 30Β° + cos2 45Β° + cos2 60Β° + cos2 90Β°
Sol:
cos2 30Β° + cos2 45Β° + cos2 60Β° + cos2 90Β° β¦(i)
By trigonometric ratios we have
Cos 30Β° = β3
2 cos 45Β° =
1
β2
Cos 60Β° = 1
2 cos 90Β° = 0
By substituting above values in (i), we get
[β3
2]
2
+ [1
β2]
2
+ [1
2]
2
+ [1]2
3
4+
1
2+
1
4= 0 β 1 +
1
2=
3
2
6. tan2 30Β° + tan2 60Β° + tan245Β°
Sol:
tan2 30Β° + tan2 60Β° + tan245Β° β¦(i)
By trigonometric ratios we have
Tan 30Β° = 1
β3 tan 60Β° = β3 tan 45Β° = 1
By substituting above values in (i), we get
[1
β3]
2
+ [β3]2
+ [1]2
β 1
3+ 3 + 1 β
1
3+ 4
β 1+12
3=
13
3
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7. 2 sin2 30Β° β 3 cos2 45Β° + tan2 60Β°
Sol:
2 sin2 30Β° β 3 cos2 45Β° + tan2 60Β° β¦(i)
By trigonometric ratios we have
Sin 30Β° = 1
2 cos 45Β°
1
β2 tan 60Β° = β3
By substituting above values in (i), we get
2 β [1
2]
2
β 3 [1
β2]
2
+ [β3]2
2.1
4β 3.
1
2+ 3
1
2β
3
2+ 3 β
3
2+ 2 = 2
8. sin2 30Β° cos2 45 Β° + 4 tan2 30Β° +1
2sin2 90Β° β 2 πππ 2 90Β° +
1
24cos2 0 Β°
Sol:
sin2 30Β° cos2 45 Β° + 4 tan2 30Β° +1
2sin2 90Β° β 2 πππ 2 90Β° +
1
24cos2 0 Β° β¦(i)
By trigonometric ratios we have
Sin 30Β° = 1
2 cos 45Β° =
1
β2 tan 30Β° =
1
β3 sin 90Β° = 1 cos 90Β° = 0 cos 0Β° = 1
By substituting above values in (i), we get
[1
2]
2
β [1
β2]
2
+ 4 [1
β3]
2
+1
2[1]2 β 2[0]2 +
1
24[1]2
1
4β
1
2+ 4 β
1
3+
1
2β 0 +
1
24
1
8+
4
3+
1
2+
1
24=
48
24= 2
9. 4(sin4 60Β° + cos4 30Β°) β 3(tan2 60Β° β tan2 45Β°) + 5 cos2 45Β°
Sol:
4(sin4 60Β° + cos4 30Β°) β 3(tan2 60Β° β tan2 45Β°) + 5 cos2 45Β° β¦(i)
By trigonometric ratios we have
Sin 60Β° = β3
2 cos 30Β° =
β3
2 tan 60Β° = β3 tan 45Β° = 1 cos 45 Β° =
1
β2
By substituting above values in (i), we get
4 ([β3
2]
4
+ [β3
2]
4
) β 3([3]2 β [1]2) + 5 [1
β2]
2
β 4 [9
16+
9
16] β 3[3 β 1] + 5 [
1
2]
β 4 β18
16β 6 +
5
2
β 1
4β 6 +
5
2
= 9
2+
5
2β 6
= 14
2β 6 = 7 β 6 = 1
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10. (πππ ππ245Β° sec2 30Β°)(sin2 30Β° + 4 cot2 45Β° β sec2 60Β°)
Sol:
(πππ ππ245Β° sec2 30Β°)(sin2 30Β° + 4 cot2 45Β° β sec2 60Β°) β¦(i)
By trigonometric ratios we have
Cosec 45Β° = β2 sec 30Β° = 2
β3 sin 30Β° =
1
2 cot 45Β° = 1 sec 60Β° = 2
By substituting above values in (i), we get
([β2]2
β [2
β3]
2
) ([1
2]
2
+ 4[1]2 β [2]2)
β [2 β4
3] [
1
4+ 4 β 4] β 3 β
4
3β
1
4=
2
3
11. πππ ππ3 30Β° cos 60Β° tan3 45Β° sin2 90Β° sec2 45Β° cot 30Β°
Sol:
πππ ππ3 30Β° cos 60Β° tan3 45Β° sin2 90Β° sec2 45Β° cot 30Β° β¦(i)
By trigonometric ratios we have
Cosec 30Β° = 2, cos 60Β° = 1
2, tan 45Β° = 1 sin 90Β° = 1 sec 45Β° = β2 cot 30Β° = β3
By substituting above values in (i), we get
[2]3 β1
2β (1)3 β (1)2(β2)
2β β3
β 8 β1
2β 1 β 2 β β3 β 8β3
12. cot2 30Β° β 2 cos2 60Β° β3
4sec2 45Β° β 4 sec2 30Β°
Sol:
cot2 30Β° β 2 cos2 60Β° β3
4sec2 45Β° β 4 sec2 30Β° β¦(i)
By trigonometric ratios we have
cot 30Β° = β3 cos 60Β° = 1
2 sec 45Β° = β2 sec 30Β° =
2
β3
By substituting above values in (i), we get
(β3)2
β 2 [1
2]
2
β3
4(β2)
2β 4 [
2
β3]
2
3 β 2 β1
4β
3
4β 2 β 4 β
4
3
3 β1
2β
3
2β
8
3β β
5
3
13. (cos 0Β° + sin 45Β° + sin 30Β°)(sin 90Β° β cos 45Β° + cos 60Β°)
Sol:
(cos 0Β° + sin 45Β° + sin 30Β°)(sin 90Β° β cos 45Β° + cos 60Β°) β¦(i)
By trigonometric ratios we have
Cos 0Β° = 1, sin 45Β° = 1
β2 , sin 30Β° =
1
2 , sin 90Β° = 1, cos 45Β° =
1
β2 cos 60Β° =
1
2
By substituting above values in (i), we get
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(1 +1
β2+
1
2) (1 β
1
β2+
1
2)
[3
2+
1
β2] [
3
2β
1
β2] β [
3
2]
2
β [1
β2] =
9
4β
1
2=
7
4
14. sin 30Β°βsin 90Β°+2 cos 0Β°
tan 30Β° tan 60Β°
Sol: sin 30Β°βsin 90Β°+2 cos 0Β°
tan 30Β° tan 60Β° β¦(i)
By trigonometric ratios we have
Sin 30Β° = 1
2 sin 90Β° = 1 cos 0Β° = 1 tan 30Β° =
1
β3 tan 60Β° = β3
By substituting above values in (i), we get 1
2β1+2
β3β1
β3
=3
2+1
1=
3
2
15. 4
cot2 30Β°+
1
sin2 60Β°β cos2 45Β°
Sol: 4
cot2 30Β°+
1
sin2 60Β°β cos2 45Β° β¦(i)
By trigonometric ratios we have
Cot 30Β° = β3 sin 60Β° = β3
2 cos 45Β° =
1
β2
By substituting above values in (i), we get
4
(β3)2 +
1
(β3
2)
2 β (1
β2)
2
4
3+
4
3β
1
2=
13
6
16. 4(sin4 30Β° + cos2 60Β°) β 3(cos2 45Β° β sin2 90Β°) β sin2 60Β°
Sol:
4(sin4 30Β° + cos2 60Β°) β 3(cos2 45Β° β sin2 90Β°) β sin2 60Β° β¦(i)
By trigonometric ratios we have
Sin 30Β° = 1
2 cos 60Β° =
1
2 cos 45Β° =
1
β2 Sin 90Β° = 1 sin 60Β° =
β3
2
By substituting above values in (i), we get
4 [(1
2)
4
+ (1
2)
2
] β 3 [[1
β2]
2
β 1] β [β3
2]
2
4 [1
16+
1
4] β 3 [
1β[β2]
(β2)2 ] β
3
4
1
4+ 1 β 3 [
1β[β2]
[β2]]
2
β3
4
= 1
4+ 1 β
3
4+
3
2= 2
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17. tan2 60Β°+4 cos2 45Β°+3 sec2 30Β°+5 cos2 90Β°
πππ ππ 30Β°+sec 60Β°βcot2 30Β°
Sol:
tan2 60Β°+4 cos2 45Β°+3 sec2 30Β°+5 cos2 90Β°
πππ ππ 30Β°+sec 60Β°βcot2 30Β° β¦(i)
By trigonometric ratios we have
Tan 60Β° = β3 cos 45Β° = 1
β2 sec 30Β° =
2
β3
cos 90Β° = 0 cosec 30Β° = 2 sec 60Β° = 2 cot 30Β° = β3
By substituting above values in (i), we get
(β3)2
+4β(1
β3)
2+2+[
2
β3]
2+5(0)2
2+2β2(+β3)2
=3+4β
1
2+3β
4
3
4β3=
3+2+4
1= 9
18. sin 30Β°
sin 45Β°+
tan 45Β°
sec 60Β°β
sin 60Β°
cot 45Β°β
cos 30Β°
sin 90Β°
Sol: sin 30Β°
sin 45Β°+
tan 45Β°
sec 60Β°β
sin 60Β°
cot 45Β°β
cos 30Β°
sin 90Β° β¦(i)
By trigonometric ratios we have
Sin 30Β° = 1
2 sin 45Β° =
1
β2 tan 45Β° = 1 sec 60Β° = 2 sin 60Β° =
β3
2
cot 45Β° = 1 cos 30Β° = β3
2 sin 90Β° = 1
By substituting above values in (i), we get
1
2β β2 +
1
2β
β3
2β 1 β
β3
2. 1
=2+1β
2
3
2
19. πππ 45Β°
πππ ππ 30Β°+
sec 60Β°
cot 45Β°β
5 sin 90Β°
2 cos 0Β°
Sol: πππ 45Β°
πππ ππ 30Β°+
sec 60Β°
cot 45Β°β
5 sin 90Β°
2 cos 0Β° β¦(i)
By trigonometric ratios we have
Tan 45Β° = 1 cosec 30Β° = 2 sec 60Β° = 2 cot 45Β° = 1 sin 90Β° = 1 cos 0Β° = 1
By substituting above values in (i), we get 1
2+
2
1β 5 β
1
2
β4
2+ 2 = β2 + 2 = 0
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20. 2sin 3x = β3 s = ?
Sol:
Sin 3x = β3
2
Sin 3x = sin 60Β°
Equating angles we get,
3x = 60Β°
x = 20Β°
21. 2 sin π₯
2= 1 x = ?
Sol:
Sin π₯
2=
1
2
Sin π₯
2= sin 30Β°
π₯
2= 30Β°
π₯ = 60Β°
22. β3 sin π₯ = cos π₯
Sol:
β3 tan x = 1
tan π₯ =1
β3
β΄ Tan x = Tan 30Β°
x = 30Β°
23. Tan x = sin 45Β° cos 45 Β° + sin 30Β°
Sol:
Tan x = 1
β2β
1
β2+
1
2 [β΅ sin 45Β° =
1
β2 cos 45Β° =
1
β2 sin 30Β° =
1
2]
Tan x = 1
2+
1
2
Tan x = 1
Tan x = tan 45Β°
x = 45Β°
24. β3 tan 2π₯ = cos 60Β° + sin 45Β° cos 45Β°
Sol:
β3 tan 2π₯ =1
2+
1
β2β
1
β2 [β΅ cos 60Β° =
1
2sin 45Β° = cos 45Β° =
1
β2]
β3 tan 2π₯ =1
β3β tan 2π₯ = tan 30Β°
2x = 30Β°
x = 15Β°
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25. Cos 2x = cos 60Β° cos 30Β° + sin 60Β° sin 30Β°
Sol:
Cos 2x = 1
2β
β3
2+
β3
2β
1
2 [β΅ cos 60Β° = sin 30Β° =
1
2sin 60Β° = cos 30Β° =
β3
2]
Cos 2x = 2 ββ3
4
β cos 2x = β3
2
Cos 2x = cos 30Β°
2x = 30Β°
x = 15Β°
26. If π = 30Β° verify
(i) Tan 2π = 2 tan π
1βtan2 π
Sol:
Tan 2π = 2 tan π
1βtan2 π β¦(i)
Substitute π = 30Β° in (i)
LHS = Tan 60Β° = β3
RHS = 2 tan 30Β°
1βtan2 30Β°=
2β1
β3
1β(1
β3)
2
=
2
β3
1β1
3
= β3
β΄ LHS = RHS
(ii) Sin π = 2 tan π
1βtan2 π
Substitute π = 30Β°
Sin 60Β° = 2 tan 30Β°
1+(tan 30Β°)2
= β3
2=
2 .1
β3
1+(1
β3)
2
= β3
2=
2
β3β
3
4β
β3
2=
β3
2
β΄ LHS = RHS
(iii) Cos 2π = 1βtan2 π
1+tan2 π
Substitute π = 30Β°
LHS = cosec π RHS = 1βtan2 π
1+tan2 π
= cos 2(30Β°) = 1βtan2 30Β°
1+tan2 30Β°
Cos 60Β° = 1
2 =
1β(1
β3)
2
1+(1
β3)
2 =1β
1
3
1+1
3
=2
34
3
=1
2
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β΄ LHS = RHS
(iv) Cos 30π = 4 cos3π β 3 cos π
LHS = Cos 30Β° RHS 4 cos3 π β 3 cos π
Substitute π = 30Β° 4cos3 30Β° - 3 cos 30Β°
Cos 3 (30Β°) = cos 90Β° 4β [β3
2]
3
β 3 ββ3
2
= 0 β 3β3
2β
3β2
2= 0
27. If A = B = 60Β°. Verify
(i) Cos (A β B) = Cos A cos B + sin A sin B
Sol:
Cos (A β B) = Cos A cos B + sin A sin B β¦(i)
Substitute A & B in (i)
β cos (60 - 60Β°)= cos 60Β° cos 60Β° + sin 60Β° sin 60Β°
Cos 0Β° = (1
2)
2
+ (β3
2)
2
1 = 1
4+
3
4= 1 = 1 LHS = RHS
(ii) Substitute A & B in (i)
Sin (60Β° - 60Β°) = Sin 60Β° Cos 60Β° β cos 60Β° sin 60Β°
= sin 0Β° = 0 = 0
LHS = RHS
(iii) Tan (A β B) = πππ π΄βtan π΅
1+tan π΄ tan π΅
A = 60Β° B = 60Β° we get
Tan (60Β° - 60Β°) = tan 60Β°βtan 60Β°
1βtan 60 tan 60Β°
Tan 0Β° = 0
0 = 0
LHS = RHS
28. If A = 30Β° B = 60Β° verify
(i) Sin (A + B) = Sin A Cos B + cos A sin B
Sol:
A = 30Β°, B = 60Β° we get
Sin (30Β° + 60Β°) = Sin 30Β° cos 60Β° + cos 30Β° sin 60Β°
Sin 90Β° = 1
2β
1
2+
β3
2β
β3
2
Sin 90Β° = 1 β 1 = 1
LHS = RHS
(ii) Cos (A + B) = cos A cos B β Sin A Sin B
A = 30Β° B = 60Β°
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Cos (90Β°) = Cos 30Β° cos 60Β° β sin 30Β° sin 60Β°
= cos 90Β° = 1
2β
β3
2β
β3
2β
1
2
0 = 0
LHS = RHS
29. Sin (A β B) = Sin A Cos B β cos A sin B
Cos (A β B) = cos A Cos B β sin A sin B
Find sin 15Β° cos 15Β°
Sol:
Sin (A β B) = Sin A Cos B β cos A sin B β¦(i)
Cos (A β B) = cos A Cos B β sin A sin B β¦(ii)
Let A = 45Β° B = 30Β° we get on substituting in (i)
β Sin(45Β° β 30Β°) = Sin 45Β° cos 30Β°
Sin 15Β° = 1
β2β
β3
2β
1
β2β
1
2
β΄ Sin 15Β° = β3β1
2β2
(ii) A = 45Β° B = 30Β° in equation (ii) we get
Cos (45Β° β 30Β°) cos 45Β° cos 30Β° + sin 45Β° sin 30Β°
Cos 15Β° - 1
β2β
β3
2+
1
β2β
1
2
Cos 15Β° β β3+1
2β2
30. In right angled triangle ABC. β C = 90Β°, β B = 60Β°. AB = 15units. Find remaining angles
and sides.
Sol:
In a βle sum of all angles = 180Β°
β A + β B + β C = 180Β°
β 90Β° + 60Β° + β A = 180Β°
β A = 180Β° β 150Β°
β΄ β A = 30Β°
From above figure
Cos B = π΅πΆ
π΄π΅
Cos 60Β° = π΅πΆ
15
1
2=
π΅πΆ
15
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BC = 15
2
Sin B = π΄πΆ
15
Sin 60Β° = π΄πΆ
15
β3
2=
π΄π
15= π΄πΆ =
15β3
2
31. In βABC is a right triangle such that β C = 90Β° β A = 45Β°, BC = 7 units find β B, AB and
AC
Sol:
Sum of angles in βle = 180Β°
β A + β B + β C = 180Β°
45Β° + β B + 90Β° = 180Β°
β B = 180Β° β 135Β°
β B = 45Β°
From figure cos B = π΅πΆ
π΄π΅
Cos 45Β° = 7
π΄π΅
1
β2β
7
π΄π΅
AB = 7β2 units
From figure sin B = π΄πΆ
π΄π΅
Sin 45Β° = π΄πΆ
7β2
1
β2=
π΄πΆ
7β2 β΄ π΄πΆ = 7 π’πππ‘π
32. In rectangle ABCD AB = 20cm β BAC = 60Β° BC, calculate side BC and diagonals AC and
BD.
Sol:
Consider βle ABC we get
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Cos A = π΄π΅
π΄πΆ Sin A =
π΅πΆ
π΄πΆ
β΄ cos 60Β° = 20
π΄πΆ Sin 60Β° =
π΅πΆ
π΄πΆ
1
π=
20
π΄πΆ β΄ AC = 40 cm
β3
2=
π΅πΆ
40
β΄ AC = 40 cm β΄ BC = 20β3 cm
Consider βle ACD we know β CAD = 30Β°
β΄ Tan 30Β° = πΆπ·
π΄π·=
1
β3=
20
π΄πΆ= π΄π· = 20β3
In rectangle diagonals are equal in magnitude
β΄ BD = AC = 40 cm
33. If Sin (A + B) = 1 and cos (A β B) = 1, 0Β° < A + B β€ 90Β° A β₯ B. Fin A & B
Sol:
Sin(A + B) = 1
β΄ Sin (A + B) = Sin 90Β°
A + B = 90Β° β¦(i)
Cos (A β B) = 1
Cos (A β B) = cos 0Β°
A β B = 0Β° β¦(ii)
Adding (i) & (ii) we get
A + B = 90Β°
A β B = 0Β°
A = 90Β° A = 45Β°
A β B = 0
A = B β B = 45Β°
34. If Tan (A β B) = 1
β3 and Tan (A + B) = β3 , 0Β° < π΄ + π΅ β€ 90Β°, A β₯ B, Find A & B
Sol:
Tan (A β B) = πππ 30Β° Tan (A + B) = Tan 60Β°
β΄ A β B = 30Β° β¦(i) A + B = 60Β° β¦(ii)
Add (i) & (ii)
A β B = 30Β°
A + B = 60Β°
2A = 90Β° A = 40Β°
A β B = 30Β° 45Β° - B = 30Β°
B = 45Β° β 30Β° = 15Β°
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35. If Sin (A β B) = 1
2 and Cos (A + B) =
1
2, 0Β° < A + B β€ 90Β°, A > B, Find A & B
Sol:
Sin (A β B) = sin 30Β° Cos (A + B) = cos 60Β°
A β B = 30Β° β¦(i)
A + B = 60Β° β¦(ii)
Add (i) & (ii) we get
2A = 90Β°, A = 45Β°.
A β B = 30Β°
45 β B = 30Β° B = 45 β 30Β°
B = 15Β°
36. In right angled triangle βABC at B, β A = β C. Find the values of
(i) Sin A cos C + Cos A Sin C
Sol:
In βle ABC β A + β B + β C = 180Β°
β A + 90Β° + β A = 180Β°
2β A = 90Β°
β A = 45Β°
β΄ β A = 45Β°
(ii) Sin 45Β° cos 45Β° + cos 45Β° sin 45Β° 1
β2β
1
β2+
1
β2β
1
β2=
1
2β
1
2= 1
(ππ)πππ π΄ πππ π΅ + cos π΄ cos π΅
β A = 45Β° sin 90Β° + cos 45Β° cos 90Β°
= 1
β2β 1 + 0
= 1
β2
37. Find acute angles A & B, if sin (A + 2B) = β3
2 Cos (A + 4B) = 0, A > B.
Sol:
Sin (A + 2B) = Sin 60Β°
Cos (A + 4B) = cos 90Β°
A + 2B = 60Β° β¦(i)
A + 4B = 90Β° β¦(ii)
Subtracting (ii) from (i)
A + 4B = 90Β°
βA β 2B = β60
2B = 30Β° β΄ B = 15Β°
A + 4B = 90Β°
4B = 4(15Β°) = 4B = 60Β°
β΄ A + 60Β° = 90Β° β΄ A = 30Β°
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38. If A and B are acute angles such that Tan A = 1
2 Tan B =
1
3 πππ Tan (A + B) =
tan π΄+πππ π΅
1βtan π΄ πππ π΅ A + B = ?
Sol:
Tan A = 1
2 Tan B =
1
3
Tan (A + B) =
1
2+
1
3
1β1
2β1
3
=5
6
1β1
6
= 1
Tan (A + B) = Tan 45Β°
β΄ A β B = 45Β°
39. In βPQR, right angled at Q, PQ = 3cm PR = 6cm. Determine β P = ? β R = ?
Sol:
From above figure
Sin R = ππ
ππ
Sin R = 3
6=
1
2
β΄ Sin R = Sin 30Β°
R = 30Β°
We know in βle β P + β Q + β R = 180Β°
β P + 90Β° + 30Β° = 180Β°
β P = 60Β°
Exercise 5.3
Evaluate the following:
1. πππ 20Β°
cos 70Β°
Sol:
(i)
β πππ (90Β°β70Β°)
cos 70Β° β
cos 70Β°
cos 70Β° [β΅ πππ (90Β° β π) = cos π]
β cos 70Β°
cos 70Β°= 1
(ii) cos 19Β°
sin 71Β°
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β cos(90Β°β71Β°)
sin 71Β°β
sin 71Β°
sin 71Β° [β΅ cos(90Β° β π) = sin π]
= 1
(iii) sin 21Β°
cos 69Β° β
sin(cos 69Β°)
cos 69Β°=
cos 69Β°
cos 69Β° [β΅ sin(90Β° β π) = πππ π]
= 1
(iv) πππ 10Β°
πΆππ‘ 80Β°β
tan(90Β°β80Β°)
cot 80Β°=
cot 80Β°
cot 80Β° [β΅ tan(90 β π) = cot π]
= 1
(π£) sec 11Β°
πππ ππ 79Β° β
sec(90Β°β79Β°)
πππ ππ 79Β°=
πππ ππ 79Β°
πππ ππ 79Β° [β΅ sec(90 β π) β πππ ππ π ]
= 1
Evaluate the following:
2. (i) [sin 49Β°
cos 45]
2
+ [cos 41Β°
sin 49Β°]
Sol:
We know that sin(49Β°) = sin(90Β° β 41Β°) = cos 41Β° similarly cos 41Β° = sin 49Β°
β [cos 41Β°
cos 41Β°]
2
+ [sin 49Β°
sin 49Β°]
2
= 12 + 12 = 2
(ii)
Cos 48Β° - sin 42Β°
Sol:
Cos 48Β° = cos (90Β° β 42Β°) sin 42Β°
β΄ sin 42Β° β sin42Β° = 0
(iii) cot 40Β°
cos 35Β°β
1
2[
cos 35Β°
sin 55Β°]
Sol:
Cot 40Β° β cot (90Β° - 50Β°) = tan 50Β°
Cos 35Β° = cos (90Β° - 55Β°) = sin 55Β°
β tan 50Β°
tan 50Β°β
1
2[
sin 55Β°
sin 55Β°]
= 1 β1
2[1]
= 1
2
(iv)
[sin 27Β°
cos 63Β°] β [
cos 63Β°
sin 27Β°]
2
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Sol:
Sin 27Β° = sin (90Β° - 63Β°) = cos 63Β° [β΅ sin (90Β° - π) = cos π]
β sin 27Β° = cos 63Β°
[sin 27Β°
sin 27Β°]
2
β [cos 63Β°
cos 63Β°]
2
= 1 β 1 = 0
(v) tan 35Β°
cot 55Β°+
cot 63Β°
cos 63Β°β 1
Sol:
Tan 35Β° = tan (90Β° - 55Β°) = cos 55Β°
Cot 78Β° = cot (90Β° - 12Β°) = tan 12Β°
β cot 55Β°
cot 55Β°+
tan 12Β°
tan 12Β°β 1
= tan 1 β 1 = 1
(vi) sec 70Β°
πππ ππ 20Β°+
sin 59Β°
cos 31Β°
Sol:
Sec 70Β° = sec (90Β° - 20Β°) = cosec 20Β° [β΅ sec (90 β π) = cosec π]
Sin 59Β° = sin (90Β° - 31Β°) = cos 31Β° [β΅ sin (90 - π) = cos π]
β πππ ππ 20
πππ ππ 20+
cos 31Β°
cos 31Β°= 1 + 1 = 2
(vii)
Sec 50Β° Sin 40Β° + Cos 40Β° cosec 50Β°
Sol:
Sec 50Β° = sec (90Β° - 40Β°) = cosec 40Β°
Cos 40Β° = cos (90Β° - 50Β°) = sin 50Β°
β΄ Sin π cosec π = 1
β cosec 40Β° sin 40Β° + sin 50Β° cosec 50Β°
1 + 1 = 2
3. Express each one of the following in terms of trigonometric ratios of angles lying between
0Β° and 45Β°
(i) Sin 59Β° + cos 56Β°
Sol:
Sin 59Β° = sin (90Β° - 59Β°) = cos 31Β°
Cos 56Β° = cos (65Β° - 34Β°) = Sin 34Β°
β cos 31Β° + sin 34Β°
(ii)
Tan 65Β° + cot 49Β°
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Sol:
Tan 65Β° = tan (90Β° - 25Β°) = cot 25Β°
Cot 49Β° = cot (90Β° - 41Β°) = tan (41Β°)
β cot 25Β° + tan 41Β°
(iii)
Sec 76Β° + cosec 52Β°
Sol:
Sec 76Β° = sec (90Β° - 14Β°) = cosec 14Β°
Cosec 52Β° = cosec (90Β° - 88Β°) = sec 38Β°
β Cosec 14Β° + sec 38Β°
(iv)
Cos 78Β° + sec 78Β°
Sol:
Cos 78Β° = cos (90Β° - 12Β°) = sin 12Β°
Sec 78Β° = sec (90Β° - 12Β°) = cosec 12Β°
β sin 12Β° + cosec 12Β°
(v)
Cosec 54Β° + sin 72Β°
Sol:
Cosec 54Β° = cosec (90Β° - 36Β°) = sec 36Β°
Sin 72Β° = sin (90Β° - 18Β°) = cos 18Β°
β sec 36Β° + cos 18Β°
(vi)
Cot 85Β° + cos 75Β°
Sol:
Cot 85Β° = cot (90Β° - 5Β°) = tan 5Β°
Cos 75Β° = cos (90Β° - 15Β°) = sin 15Β°
= tan 5Β° + sin 15Β°
(vii)
Sin 67Β° + cos 75Β°
Sol:
Sin 67Β° = Sin (90Β° β 23Β°) = cos 23Β°
Cos 75Β° = cos (90Β° β15Β° ) = sin 15Β°
= cos 23Β° + sin 15Β°
4. Express Cos 75Β° + cot 75Β° in terms of angles between 0Β° and 30Β°.
Sol:
Cot 75Β° = cos (90Β° - 15Β°) = sin 15Β°
Cot 75Β° = cot (90Β° - 15Β°) = tan 15Β°
= sin 15Β° + tan 15Β°
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5. If Sin 3A = cos (A β 26Β°), where 3A is an acute angle, find the value of A = ?
Sol:
Cos π = sin (90Β° - π)
β Cos (A β 26) = sin (90Β° β(Aβ 26Β°))
β Sin 3A = sin (90Β° β (A β 26))
Equating angles on both sides
3A = 90Β° β A + 26Β°
4A = 116Β° A = 116
4= 29Β°
β΄ A = 29Β°
6. If A, B, C are interior angles of a triangle ABC, prove that (i) tan (πΆ+π΄
2) = cot
π΅
2
Sol:
(i) Tan [π+π΄
2] = cot
π΅
2
Sol:
Given A + B + C = 180Β°
C + A = 180Β° β B
β Tan [180βπ΅
2] β πππ [90Β° β
π΅
2]
β cot π΅
2 [β΅ tan(90Β° β ΞΈ) = cot π]
β΄ LHS = RHS
(ii) Sin [π΅+πΆ
2] = cos
π΄
2
Sol:
A + B + C = 180Β°
B + C = 180Β° - A
LHS = sin [180Β°βπ΄
2] β sin [90Β° β
A
2]
Cos π΄
2 [β΅ πππ (90Β° β π) β πππ π]
β΄ LHS =RHS
7. Prove that
(i)
Tan20Β° Tan 35Β° tan 45Β° tan 55Β° Tan 70Β° = 1
Sol:
Tan 20Β° = tan (90Β° - 70Β°) = cot 70Β°
Tan 35Β° = tan (90Β° - 70Β°) = cot 55Β°
Tan 45Β° = 1
β cot 70Β° tan 70Β° x cot 55Β° tan 55Β° x tan 45Β° β cot π = tan π = 1
β 1 Γ 1 Γ 1 = 1 Hence proved.
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(ii)
Sin 48Β° sec 42Β° + cosec 42Β° = 2
Sol:
Sin 48Β° = sin (90Β° - 42Β°) = cos 42Β°
Cos (45Β°) = cos (90Β° - 42Β°) = sin 42Β°
Sec π β cos π = 1 β sin π cosec π = 1
β cos 42Β° sec 42Β° + sin 42Β° cosec 42Β°
β 1 + 1 = 2
β΄ LHS = RHS
(iii) sin 70Β°
cos 20Β°+
πππ ππ 20Β°
sec 70Β°β 2 cos 70Β° πππ ππ 20Β° = 0
Sol:
Sin (70Β°) = sin (90Β° - 20Β°) = cos 20Β°
Cosec 20Β° = cosec (90Β° - 70Β°) = sec 70Β°
Cos 70Β° = cos (90Β° - 20Β°) = sin 20Β°
β cos 20Β°
cos 20Β°+
sec 70Β°
sec 70Β°β 2 sin 20 πππ ππ 20Β°
1 + 1 β 2(1) = 0
β΄ LHS = RHS Hence proved
(iv) cos 80Β°
sin 10Β°+ cos 59Β° πππ ππ 31Β° = 2
Sol:
Cos 80Β° = cos (90Β° - 10Β°) = sin 10Β°
Cos 59Β° = cos (90Β° - 31Β°) = sin 31Β°
β sin 10Β°
sin 10Β°+ sin 31Β° πππ ππ 31Β°
= 1 + 1 = 2 [β΅ Sin π cosec π = 1]
Hence proved
8. Prove the following:
(i) Sin π sin (90 - π) β cos π cos (90 - π) = 0
Sol:
Sin (90 β π) = cos π
Cos (90 β π) β cos π sin π
= 0
β΄ LHS = RHS
Hence proved
(ii) cos(90Β°βπ) sec(90Β°βπ) tan π
πππ ππ (90Β°βΞΈ) sin(90Β°βΞΈ) cot(90Β°βΞΈ)+
tan(90Β°βπ)
cot π= 2
Sol:
Cos (90Β° - π) = sin A cosec (90 - π) = sec π
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Sec (90Β° - π) = cosec π sin (90 - π) = cos π
Cot (90 - π) = tan π
β sin π πππ ππ π πππ π
sec π.cos π.tan π=
sin π πππ ππ π
sec π cos π [β΅ sin π πππ ππ π = 1]
= 1 [sec π cos π = 1] tan(90Β°β π)
cot π=
cot π
cot π= 1
β 1 + 1 = 2
β΄ LHS = RHS
Hence proved
(iii) tan(90βπ΄) cot π΄
πππ ππ2π΄β cos2 π΄ = 0
Sol:
Tan (90 β A) = cot A
β cot π΄ .cot π΄
πππ ππ2π΄β cos2 π΄
β cot2 π΄
πππ ππ2π΄β cos2 π΄
= cos2 π΄
sin2 π΄β cos2 π΄ β cos2 π΄ cos2 π΄ = 0
Hence proved
(iv) cos(90Β°βπ΄) sin(90Β°βπ΄)
tan(90Β°βπ΄)β sin2 π΄ = 0
Sol:
Cos (90Β° - A) = sin A Tan (90Β° - A) = cot A
Sin (95Β° A) = cos A sin π΄ cos π΄
cot π΄β sin2 π΄ = 0
sin π΄.cos π΄.
cos π΄sin π΄ β sin2 π΄
sin2 π΄ β sin2 π΄ = 0
LHS = RHS
Hence Proved
(v) Sin (50Β° + π) β cos (40Β° β π) + tan 1Β° tan 10Β° tan 20Β° tan 70Β° tan 80Β° tan 89Β° = 1
Sol:
Sin (50 + π) = cos (90 β (50 + π)) = cos (40 β π)
Tan 1 = tan (90Β° β 89Β°) β cot 89Β°
Tan 10Β° = tan (90Β° - 80Β°) = cot 80Β°
Tan 20Β° = tan (90Β° - 70Β°) = cot 70Β°
β cos (40Β° - π) β cos (40 - π) = cot 89Β° tan 89Β° . cot 80Β° . cot 70Β° tan 70Β°
Cot . tan π = 1
= 1 β 1 β 1 = 1
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LHS = RHS
Hence proved
9. Evaluate:
(i) 2
3 (cos4 30Β° β sin4 45 Β°) β 3(sin2 60Β° β sec2 45Β°) +
1
4cot2 30Β°
Sol:
Cos 30Β° =β3
2 sin 60Β° =
β3
2 cot 30Β° = β3 sin 45Β° =
1
β2 sec 45Β° =
1
β2
Substituting above values in (i)
2
3 [(
β3
2)
4
β (1
β2)
4
] β 3 [(β3
2)
2
β [1
β2]
2
] +1
4(β3)
2
2
3 [
9
16β
1
4] β 3 [
3
4β
1
2]
1β3
4
2
3[
9β4
16] β 3 [
3β2
4] β
3
4
β 2
3β
5
16β
3
4+
3
4β
5
24
(ii) 4 (sin2 30 + cos4 60Β°) β2
3 3 [(β
3
2)
2
β [1
β2]
2
] +1
4(β3)
2
Sol:
πππ 30Β° =1
2cos 60 =
1
2 sin 60Β° =
β3
2 cos 45Β° =
1
β2 tan 60Β° = β3
β 4 [[1
2]
4
+ [1
2]
4
] β2
3[(
β3
2)
2
β (1
β2)
2
] +1
2(β3)
2
4 [2.1
16] β
2
3[
3
4β
1
2] +
3
2
=1
2β
2
3 β
1
4+
3
2=
11
6
(iii) sin 50Β°
cos 40Β°+
ππ sec 40Β°
sec 50Β°β 4 cos 50Β° πππ ππ 40Β°
Sol:
Sin 50Β° = sin (90Β° - 40Β°) = cos 40Β°
Cosec 40Β° = cosec (90Β° - 50Β°) = sec 50Β°
Cos 50Β° = cos (90Β° - 40Β°) = sin 40Β°
β cos 40Β°
cos 40Β°+
sec 50Β°
π ππ 50Β°β 4 sin 40Β° πππ ππ 40Β°
1 + 1 β 4 = β 2 [β΅ Sin 40Β° cosec 40Β° = 1]
(iv) Tan 35Β° tan 40Β° tan 50Β° tan 55Β°
Sol:
Tan 35Β° = tan (90Β° - 55Β°) = cot 55Β°
Tan 40Β° = tan (90Β° - 50Β°) = cot 55Β°
Tan 65Β° = 1
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Cot 55 tan 55β cot 50 tan 50 β tan 45
1 β 1 β 1 = 1
(v) Cosec (65 + π) β sec (25 β π) β tan (55 β π) + cot (35 + π)
Sol:
Cosec (65 + π) = sec (90 β (65 + π)) = sec (25 β π)
Tan (55 β π) = cot (90 β (55 β π) = cot (35 + π)
β sec (25 β π) β sec (25 β π) tan (55 β π) + tan (55 β π) = 0
(vi) Tan 7Β° tan 23Β° tan 60Β° tan 67Β° tan 83Β°
Sol:
Tan 7Β° tan 23Β° tan 60Β° tan (90Β° - 23) tan (90Β° - 7Β°)
β tan 7Β° tan 23Β° tan 60Β° cot 23Β° tan 60Β°
1 β 1 β β3 = β3
(vii) 2 sin 68
cos 22β
2 cot 15Β°
5 tan 75Β°β
8 tan 45Β° tan 20Β° tan 40Β° tan 50Β° tan 70Β°
5
Sol:
Sin 68Β° = sin (90 - 22) = cos 22
Cot 15Β° = tan (90 - 75) = tan 75
2 βcos 22
cos 22β
2 tan 75Β°
5 tan 75Β°β
3 tan 45Β° tan20Β° tan 40Β° cot 40Β° cot 20Β°
5
= 2 β2
5β
3
5= 2 β 1 = 1
(viii) 3 cos 55Β°
7 sin 35Β°β
4(cos 70 πππ ππ 20Β°)
7 (tan 5Β° tan 25Β° tan 45Β° tan 65Β° tan 85Β°)
Sol:
Cos 55Β° = cos (90Β° - 35Β°) = sin 35Β°
Cos 70Β° = cos (90 β 20) = sin 20Β°
Tan 5 = cot 85Β° tan 25Β° = cot 65Β°
β 3 sin 35Β°
7 sin 35Β°β
4 (sin 20Β° πππ ππ 20Β°)
7(cot 85Β° tan 85Β° cot 65Β° tan 65Β° tan 45Β°)
= 3
7β
4
7= β
1
7
(ix) sin 18Β°
cos 72Β°+ β3 [tan 10Β° tan 30Β° tan 40Β° tan 50Β° tan 80Β°]
Sol:
Sin 18Β° = sin (90Β° - 72) = cos 72Β°
Tan 10Β° = cot 80Β° tan 50Β° = cot 40Β°
β π ππ 18Β°
sin 18Β°+ β3 [tan 80 cos 30 . tan 40 cot 40 .
1
β3]
= 1 + β3 β1
β3= 2
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(x) cos 58Β°
sin 32Β°+
sin 22Β°
cos 68Β°β
cos 38Β° cosec 52Β°
tan 18Β° tan 35Β° tan 60Β° tan 72Β° tan 65Β°
Sol:
Cos 58Β° = cos (90Β° - 32Β°) = sin 32Β°
Sin 22Β° = sin (90Β° - 68Β°) = cos 68Β°
Cos 38Β° = cos (90 β 52) = sin 52Β°
Tan 18Β° = cot 72 tan 35Β° = cot 55Β°
β sin 32Β°
sin 32Β°+
cos 68Β°
cos 68Β°β
sin 52 πππ ππ 52
tan 72 .cot 72 tan 55 cot 55 .tan 60
= 1 + 1 β1
β3 =
2β3β1
β3Γ
β3
β3=
6ββ3
3
10. If Sin π = cos (π β 45Β°), where π β 45Β° are acute angles, find the degree measure of π.
Sol:
Sin π = cos (π β 45Β°)
Cos π = cos (90 β π)
Cos (π β 45Β°) = sin (90Β° - (π β 45Β°)) = sin (90 β π + 45Β°)
Sin π = sin (135 β π)
π = 135 β π
2π = 135
β΄ π = 135Β°/2
11. If A, B, C are the interior angles of a βABC, show that:
(i) Sin (π΅+πΆ
2) = cos
π΄
2 (ii) cos [
π΅+πΆ
2] = πππ
π΄
2
Sol:
A + B + C = 180
B β C = 180 β π΄
2
(i) Sin [90 βπ΄
2] = cos
π΄
2
β΄ LHS = RHS
(ii) Cos [90 βπ΄
2] = sin
π΄
2
β΄ LHS = RHS
12. If 2π + 45Β° and 30Β° β π are acute angles, find the degree measure of π satisfying Sin
(20 + 45Β°) = cos (30 - πΒ°)
Sol:
Here 20 + 45Β° and 30 β πΒ° are acute angles:
We know that (90 β π) = cos π
Sin (2π + 45Β°) = sin (90 β (30 β π))
Sin (2π + 45Β°) = sin (90 β 30 + π)
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Sin (20 + 45Β°) = sin (60 + π)
On equating sin of angle of we get
2π + 45 = 60 + π
2π β π = 60 β 45
π = 15Β°
13. If π is a positive acute angle such that sec π = cosec 60Β°, find 2 cos2 π β 1
Sol:
We know that sec (90 β π) = cosec2 π
Sec π = sec (90 β 60Β°)
On equating we get
Sec π = sec 30Β°
π = 30Β°
Find 2cos2 π β 1
β 2 Γ cos2 30Β° β 1 [cos 30 =β3
2]
β 2 Γ (β3
2)
2
β 1
β 2 Γ3
4β 1
β 3
2β 1
= 1
2
14. If cos 2π = sin 4π where 2π, 4π are acute angles, find the value of π.
Sol:
We know that sin (90 β π) = cos π
Sin 20 = cos 2π
Sin 4π = sin (90 - 2π)
4π = 90 β 20
6π = 90
π = 90
6
π = 15Β°
15. If Sin 3π = cos (π β 6Β°) where 3 π and π β 6Β° are acute angles, find the value of π.
Sol:
30, π β 6 are acute angle
We know that sin (90 β π) = cos π
Sin 3π = sin (90 β (π - 6Β°))
Sin 3π = sin(90 β π + 6Β°)
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Sin 3π = sin (96Β° - π)
3π = 96Β° β π
4π = 96Β°
π = 96Β°
4
π = 24Β°
16. If Sec 4A = cosec (A β 20Β°) where 4A is acute angle, find the value of A.
Sol:
Sec 4A = sec [90 β π΄ β 20] [β΅ sec(90 β π) = πππ ππ π]
Sec 4A = sec (90 β A + 20)
Sec 4A = sec (110 β A)
4A = 110 β A
5A = 110
A = 110
5β π΄ = 22
17. If Sec 2A = cosec (A β 42Β°) where 2A is acute angle. Find the value of A.
Sol:
We know that (sec (90 β π)) = cosec π
Sec 2A = sec (90 β (A β 42))
Sec 2A = sec (90 β A + 42)
Sec 2A = sec (132 β A)
Now equating both the angles we get
2A = 132 β A
3A = 132
3
A = 44