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OP Malhotra Class 9 ICSE Maths Solutions Chapter 19 Trigonometrical Ratios
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 19, Trigonometrical Ratios. Here students can easily find step by step solutions of all the problems for Trigonometrical Ratios, Exercise 19a, 19b, 19c and 19d Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 19 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Trigonometrical Ratios Exercise 19(a) Solution :
Question no – (1)
Solution :
Figure – 1 |
Figure – 2 | Figure – 3 | |
| Tan A = tan B = cot A = cot B = |
3/5 4/5 4/3 3/4 |
3/12 12/9 12/9 9/12 |
15/20 20/15 2/15 15/20 |
| Sin A = sin B = cosec A = cosec B = |
3/5 4/5 5/3 3/4 |
9/15 12/15 15/9 15/12 |
15/25 2025 25/15 25/20 |
| Cos A = cos B = sec A = sec B = |
4/5 3/5 3/5 2/3 |
11/15 9/15 15/12 15/9 |
20/25 15/25 25/20 25/15 |
Question no – (2)
Solution :
According to the figure,
sin θ = 5/13
cos θ = 12/13
tan θ= 5/12
∴ sin2 θ = (5/13)2 = 25/169
cos2 θ = (12/15)2 = 144/225
tan2 θ = (5/12)2 = 25/144
Question no – (4)
Solution :
(i) Sin RQM
= Sin RM/QM
(ii) Sin QMP
= Sin QMP = PQ/PM
(iii) Sin OQR
= Sin OQR = OR/OQ
(iv) cos QMP
= Cos QMP = QM/PM
(v) tan RQM
= tan RQM = RM/QR
(vi) cot MOP
= Cot = OM/MP
(vii) sec RQO
= sec ROQ = OQ/OR
Trigonometrical Ratios Exercise 19(b) Solution :
Question no – (1)
Solution :
Required figure,
Tan = 5/12
If the △ABC, ∠3 = 90°
AB = 12, BC = 5
By theorem, AC2 = AB2 + BC2
= (12)2 + (5)2
= 169
= AC = 13
∴ Sin θ = 5/13
= cos θ = 12/13
Question no – (2)
Solution :
Given, Sin θ = 3/5
From, △ ABC
AC = 3, AB = 5
AB2 = BC2 + AC2
= 52 = BC2 + 32
= BC2 – 52 – 32
= 16
= BC = 4
(i) cos θ,
= Cos θ 4/5
(ii) tan θ = 3/4
Question no – (3)
Solution :
(i) AB = 7 cm, AC = 24 cm
Here, ∠A = 90°
By theorem,
BC2 = AB2 + AC2
= BC2 = (7)2 + (24)2
= 49 + 576
= 625
= BC = 25 cm
tan B = 24/7
tan c = 7/24
sin B = 24/25
sin c = 7/28
cosB = 7/25
cose = 24/25
(ii) AB = 12 cm, AC = 9 cm
From, △ ABC,
BC2 = AB2 + AC2
= (12)2 + (9)2
= 144 + 81
= 225
= BC = 15 CM
tan B = 12/9
tan c = 9/12 = 3/4
sin B = 9/15
sin c = 12/15
Cos B = 12/15
Cos C = 9/15
Question no – (4)
Solution :
Required figure,
(i) PQ = 10 cm, QR = 8 cm
From fig, △PQR
∠R = 90°
By theorem,
(PQ)2 = (QR)2 + PR2
= (10)2 = 82 + PR2
= PR2 = 100 – 64
= 36
= PR = 6 CM
∴ tan P = 8/6 = 4/3
tan Q = 6/8
Sin P = 8/10 = 4/5
Sin Q = 6/10 = 3/5
Cos P = /6/10 = 3/5
Cos Q = 8/10 = 4/5
Required figure,
(ii) PQ = 29 mm, PR = 21 mm
Here, If △ PQR
PQ2 = PR2 + QR2
= (29)2 = PR2 + QR2
= (29)2 = (21)2 + QR2
= QR2 = 841 – 441
= 400
= QR = 20 mm
Now, tan P = Q12/PR = 20/21
tan = 21/20
sin P = 20/29 ;
sin Q = 21/29
cos P = 21/29
cos Q = 20/29
Question no – (5)
Solution :
Required figure :
In △ABC,
∠A = θ, ∠B = 90°
∴ sin θ = AB/CA
CA2 = AB2 = BC2
Now, Sin2θ + cos2θ = (BC/CA)2 + (AB/CA)2
= BC2/CA2 + AB2/CA2
= BC2 + AB2/CA2
∴ CA2/CA2
= 1
Therefore, Sin2θ + cos2θ = 1
Question no – (7)
Solution :
Required figure,
In ABC, ∠A = θ , ∠B = 90°
Cos θ = 6/10 = 3/5 = AB/CA
Now, CA2 = AB2 + BC2
= (5)2 = (3)2 + BC2
= BC2 = 25 – 9
= 16
= BC = 4
Sin θ = 4/5, tan θ= 4/3
5 sin – 5an = 5 × 4/3 – 3 × 4/3
= 4 – 4
= 0
Therefore, the value will be 0
Question no – (8)
Solution :
Required figure,
In figure, △ABC
∠B = 90 ° , ∠A acute angle
tan A = 3/4
BC = 3, AB = 4
AC2 = AB2 + BC2
= 42 + 32
= 16 + 9
= 25
= AC = 5
Cos A = 4/5
Therefore, the value of cos A will be 4/5
Question no – (9)
Solution :
Required figure,
In △ABC,
∠B = 90 °, ∠A = θ
Sinθ = 6/10= BC/AC
= AC2 = AB2 + BC2
= (10)2 = AB2 + 36
= AB2 = 100 – 36
= 64
= AB = 8
Cosθ = 8/10 = 4/5
tanθ = 6/8 = 3/4
cosθ + tanθ = 4/5 + 3/4
= 16 + 15/20
= 31/20
Therefore, the value of (cos θ + tan θ) will be 31/20
Question no – (10)
Solution :
(i) In the diagram (Fig. 19.14)
From figure
∴ tan θ = AB/BC
tan θ = BC/AB is false
(a) tan θ BC/AB ;
Sec θ = AC/BC
∴ sec θ= BC/AC is (false)
(b) sec θ BC/AC
AB : BC = Sin A : A COS a
= BC/AC = tan A
∴ AB : BC = Sin A : Cos A (False)
Question no – (11)
Solution :
(i) Find the value of (a) sin ⌀, (b) tan θ
= sin θ = CD/BC = 5/13
DE || BC
BD = BC = 12
ED = 5
∴ AE = AB – EB
= 14 – 5
= 9
∴ tan θ = DE/AE
= 12/9
= 4/3
(ii) Write an expression for AD in terms of θ
= In △AED,
AD2 = AE2 + ED2
= 92 + 122
= 81 + 144
= 225
= AD = 15
∴ sin θ = DE/AD
= AD = DE/sin θ
= 12/sin θ
∴ cos θ = AE/AD
= AD = AE/cos θ
= 9/cos θ
∴ AD = 12/sin θ and 9/cos θ
Question no – (13)
Solution :
Required figure,
(i) Here, sin θ = 12/13
= BC/AC
∴ AC2 = AB2 + BC2
= (13)2 = AB2 + (12)2
= (AB)2 = 169 – 144
= 25
= AB = 5
∴ cos θ = 5/13
tan θ = 12/5
Now, cos θ + tan θ = 5/13 + 12/5
= 25 + 156/65
= 181/65
= 2 51/65
Therefore, the value will be 2 51/65
Question no – (14)
Solution :
Here, tan θ = 5/12
Now, cos θ – sin θ/cos θ + sin θ
= 1 – tan θ/1 + tan θ [divide by cos θ]
= 1 – 5/12/1 + 5/12
= 12 – 5/12/12 + 5/12
= 7/12 × 12/17
= 7/17
Therefore, the value will be 7/17.
Question no – (15)
Solution :
5 sin θ = 4
= sin θ = 4/5
Now, 1 + sin θ/1 – sin θ
= 1 + 4/5/1 – 4/5
= 5 + 4/5/5 – 4/5
= 9/5 × 5/1 = 9
Therefore, the value will be 9
Question no – (16)
Solution :
b tan θ = a …(given)
= tan θ = a/b
Now, cos θ + sin θ/cos θ – sin θ”
= 1 + tan θ/1 – tan θ
= 1 + a/b/ 1 – a/b
= b + a/b/b – a/b
= b + a/b × b/b – a
= b + a/b – a
Therefore, the value will be b + a/b – a
Question no – (17)
Solution :
In the given question,
5 tan θ = 4
= tan θ = 4/5
∴ 5 sin θ – 3 cos θ/5 sin θ + 2 cos θ
= 5 tan θ – 3/5 tan θ + 2 [divide by cos θ]
= 5 × 4/3 – 3/5 × 4/5 + 2
= 4 – 3/4 + 2
= 1/6
Therefore, the value will be 1/6
Question no – (18)
Solution :
Required figure,
Given in the question,
13 sin A = 5
= sin A = 5/13
= BC/AC
Now, AC2 = AB2 + BC2
= (13)2 = AB2 + 52
= AB2 = 169 – 25
= 144
= AB = 12
∴ cos A = 12/13
tan A = 5/12
∴ 5 sin A – 2 cos A/tan A
= 5 × 5/13 – 2 × 12/13/5/12
= 25/13 – 24/13/5/12
= 1/13 × 12/5
= 12/65
Therefore, the value will be 12/65
Question no – (19)
Solution :
As per the question we get,
5 cos A – 12 sin A = 0
= 5 cos A = 12 sin A
= sin A/cos A = 5/12
∴ sin A + cos A/2 cos A – sin A
= tan A + 1/2 – tan A
= 5/12 + 1/2 – 5/12
= 5 + 12/12/24 – 5/12
= 17/12 × 12/19
= 17/19
Therefore, the value will be 17/19
Trigonometrical Ratios Exercise 19(c) Solution :
Question no – (1)
Solution :
(i) Here, L.H.S, sin60° = √3/2
R.H.S 2 sin 30 cos30
= 2 1/2 √ √3/2
= √3/2
∴ L.H.S = R.H.S …(Proved)
(ii) 3 sin2 45 + 2 cos60°
Now, 3 sin2 45 + 2 cos60°
= 3 × (1√2)2 + 2 × (1/2)2
= 3/2 + 2 × 1/4
= 3/2 + 1/2
= 4/2
= 2
Therefore, the value will be 2
Question no – (2)
Solution :
Here, sin x = cos x
= sin/cos = 1
= tan x = 1
= tan x = tan 45°
= x = 45°
Therefore, the numerical value of x will be 45°
Question no – (3)
Solution :
(i) Given, sin2 60° + cos2 45°
∴ sin2 60° + cos2 45°
= (√3/2)2 + (1/√2)2
= 3/4 + 1/2
= 3 + 2/4
= 5/4
Therefore, the value will be 5/4
(ii) Given, 3 cos2 30° + tan2 60°
∴ 3 cos2 30° + tan2 60°
= 3 × (√2/2)2 + (√3)2
= 3 × 3/4 + 3
= 9/4 + 3
= 21/4
= 5 1/4
Thus, the value will be 5 1/4
(iii) Given, 4 sin2 60° + tan2 30° – 8 sin 45° cos 45°
∴ 4 sin2 60° + tan2 30° – 8 sin 45° cos 45°
= 4 × (√3/2)2 + 3. (1/√3)2 – 8 × 1/√2 × 1/√2
= 4 × 3/4 + 3 1/3 – 8 × 1/2
= 3 + 1 – 4
= 0
Hence, the value will be 0.
(iv) Given, 2 sin2 30° – 3 cos2 45° + tan2 60°
∴ 2 sin2 30° – 3 cos2 45° + tan2 60°
= 2 × (1/2)2 – 3 (1/√2)2 + (√3)2
= 2 × 1/4 – 3 1/2 + 3
= 1/2 – 3/2 + 3
= 1 – 3 + 6/2
= 4/2
= 2
Therefore, the value will be 2
(v) Given, 4 sin2 60° + tan2 30° – 8 sin 45° cos 45°
∴ 4 sin2 60° + tan2 30° – 8 sin 45° cos 45°
= 4 × (√3/2)2 + 3. (1/√3)2 – 8 × 1/√2 × 1/√2
= 4 × 3/4 + 3 1/3 – 8 × 1/2
= 3 + 1 – 4
= 0
Therefore, the value will be 0.
(vi) cos 90° + cos2 45° sin 30° tan 45°
∴ Cos 90° + cos2 45° sin 30° tan 45°
= 0 + (1/√2)2 × 1/2 × 1
= 1/2 × 1/2
= 1/4
Therefore, the value will be 1/4
(vii) Given, cos2 45° + sin2 60° + sin2 30°
∴ cos2 45° + sin2 60° + sin2 30°
= (1/√2)2 + (√3/2)2 + (1/2)2
= 1/2 + 3/4 + 1/4
= 2 + 3 + 1/4
= 6/4
= 3/2
Therefore, the value will be 3/2
(viii) Given, sin2 30° + cos2 60°
∴ sin2 30° + cos2 60°
= (1/2)2 + (1/2)
= 1/4 + 1/4
= 2/8
= 1/4
Therefore, the value will be 1/4
(ix) sin2 45° + cos2 45°/tan2 60°
∴ sin2 45° + cos2 45°/tan2 60°
= (1/√2)2 + (1/√2)2/(√3)2
= 1/2 + 1/2/3
= 1/3
Therefore, the value will be 1/3
(x) 5 sin2 30° + cos2 45° – 4 tan2 30°/2 sin 30° cos 30° + tan 45°
∴ 5 sin2 30° + cos2 45° – 4 tan2 30°/2 sin 30° cos 30° + tan 45°
= 5 (1/2)2 + (1/√2)2 – 4 (1/√5)2/2 × 1/2. 3/2 + 1
= 5/4 + 1/52 – 4/√3/2 + 1
= 15 + 6 – 16/12/√3 + 2/2
= 5/12 × 2/√3+ 2
= 5/6(√3 + 2)
Therefore, the value will be 5/6(√3 + 2)
(xi) 2√2 cos 45°. Cos 60° + 2√3 sin 30°. Tan 60° – cos 0°
∴ 2√2 cos 45°. Cos 60° + 2√3 sin 30°. Tan 60° – cos 0°
= 2√2 × 1/√2 × 1/√2 + 2√3 × 1/2 × √3 – 1
= 1 + 3 – 1
= 4 – 1
= 3
Therefore, the value will be 3
Question no – (4)
Solution :
(i) Here, In △ABC
= AB = AC = x
Now, Draw AD ⊥ BC
∴ ∠B = 45° then ∠C = 45°
A = 90°
∴ ∠BAD = CAD = 45°
Now, if ∠A = 90°
BC2 = AB2 + AC2
= x2 + x2
= 2x2
BC = √2x
∴ BD = DC = AD = √2X/2
So, Sin45° = √2x/2/2/x
= 2/√2
= 1/√2
Thus, the value of each of sin 45° will be 1/√2.
(ii) cos 45°
= √2x/2/x
= √2x/2 × 1/x
= √2/2
= 1/√2
Hence, the value of cos 45° is 1/√2
(iii) tan 45°
= √2x/2/√2x/2
= 1
Thus, the value of tan 45° is 1
Required figure :
Question no – (5)
Solution :
Given, sin 30° – sin 90° + 2 cos 0°/tan 30° × tan 60°
∴ sin 30° – sin 90° + 2 cos 0°/tan 30° × tan 60°
= 1/2 – 1 + 2 × 1/1/3 × 3
= 1/2 – 1 + 2/1
= 1 1/2
Therefore, the value will be 1 1/2
Question no – (6)
Solution :
(i) Given, sin 60° = 2 tan 30°/1 + tan2 30° = √3/2
Here, L.H.S , Sin 60° = 3/2
∴ R.H.S 2 tan 30°/1 + tan2 30°
= 2 × 1/3/1 + (1/3)2
= 2/3/1 + 1/3
= 2/3 × 3/4
= 3/2
Therefore, L.H.S = R.H.S …(Proved)
(ii) cos 60° = 1 – tan2 30°/1 + tan2 30° = 1/2
Here, R.H.S, cos 60° = 1/2
∴ L.H.S, 1 – tan2 30°/1 + tan2 30°
= 1 – (1/3)2/1 + (1/3)2
= – 1 – 1/3/1 + 1/3
= 2/3 × 3/4
= 1/2
Thus, L.H.S = R.H.S ….(Proved)
(iii) cos 60° = cos2 30° – sin2 30°
R.H.S cos2 30° – sin30°
= (3/2)2 – (1/2)2
= 3/4 – 1/4
= 3 – 1/4
= 2/4
= 1/2 …(Proved)
(iv) cos 60° = 1 – 2 sin2 30°= 2 cos2 30 – 1
Here, L.H.S 60° = 1/2
∴ R.H.S = 1 – 2sin2 30°
= 1 – 2 × (1/2)
= 1 – 2 × 1/4
= 1/2
and, 2 cos2 30° – 1
= 2 × (3/2)2 – 1
= 2 × – 1
= 3/2 – 1
= 1/2 …(Proved)
Question no – (7)
Solution :
Given. cos2 30° + sin2 30° + tan2 45° = 2
∴ L.H.S, cos2 30° + sin2 30° + tan2 45°
= (3/2)2 + (1/2)2 + 12
= 3/4 + 1/4 + 1
= 1 + 1
= 2
Therefore, L.H.S = R.H.S …(Proved)
Question no – (8)
Solution :
Required figure,
Here, In rectangle △ABC,
AC = x ∠C = 90°
AB = 15 m
Let, height of wall = x m, and distance any foot from wall = y m
(i) ∴ High up the wall will the pole reach,
= Sin60° = x/15 = 1
= 3/2 × 15 = x
= x = 12.99 m
Hence, the wall will reach 12.99 m.
(ii) Distance of foot of the pole from the wall,
= Cos 60° = y/15
= 1/2 = y/15
= y 15/2 = 7.5 m
Therefore, the distance from foot of wall will be 7.5 m.
Question no – (9)
Solution :
Required figure
Here, △ABC
= ∠B = 90° ∠C = θ°
AC = 40cm AB = 20 cm
sinθ = AB/AC = 20/40 = 1/2
Sinθ = 1/2
sinθ = sin 30 °
∴ θ = 30°
Question no – (10)
Solution :
(i) A = 30° and c = 40
= ∠A = 30° ∠C = 90°
∴ sin = a/c
= sin 30° = a/c
sin 30 a/c
= 1/2 = a/c
= 1/2 a= a/40
= a = 20
∴ C2 = a2 + b2
= (40)2 (20)2 + b2
= b2 = (40)2 – (20)2
= (40 + 20) (40 – 20)
= 60 × 20
= b = √1200
= 20√3
= 20 × 1.732
= 34.64
(ii) B = 60° and c = 15;
= √B = 60°, C = 15
∴ sinθ = AC/AB
sin 60° = ac
= √3/2 = a/c
= a = 15 √3/2
= 15 × 1.732/2
= 12.99
∴ c2 = a2 + b2
= (15)2 = (12.99)2 + b2
= b2 = (15)2 – (12.99)2
= (225 – 168.74)
= 56.25
= b = 7.5
Now, A = 90° – B
= 90 – 60
= 30°
(iii) A = 45° and a = 7
= A = 45° , a = = 7
∴ sin = a/c
= sin 45° = a/c
= 1/√2 = 7/c
= c = 7√2
= 7 × 1.414
= 9.9
Now, C2 = a2 + b2
= (7√2)2 = (7)2 + b2
= b2 = 98 – 49
= 49
= b = 7
∴ ∠B = 90° – ∠45°
= 45°
Question no – (11)
Solution :
Given in the question,
4 sin2 θ – 1 = 0
= 4sin2 θ = 1
= sin2 θ = 1/4 = (1/2)2
= sin θ = 1/2
= sin θ = sin 30°
= θ = 30°
∴ Value of θ is = 30°
Now, cos2 θ + tan2 θ
= (3/2)2 + (1/3)2
= 3/4 + 1/3
= 9 + 4/12
= 13/12
Question no – (12)
Solution :
Here, sin θ = cos θ
= sinθ /cosθ = 1
= tan = 1
tan θ = tanθ 45
= θ = 45
Now, 2tan2 θ + sin2 θ – 1
= 2tan2 45° + sin2 45 ° = -1
= 2 × (1)2 + (1/3)2 – 1
= 1 1/2
= 1.5
= 3/2
= 1.5
Hence, the value will be 1.5
Question no – (13)
Solution :
Required figure,
Here, Angle of elevation = 60°
AB = x, AC = 75
∠B = 60°
∴ sin θ = 75/x
= sin 60° = 75/x
= √3/2 = 75/x
= x = 75 × 2/√3
x = 50√3
= 50 × 1.732
= 87 m
Thus, the length of the string will be 87 m.
Question no – (14)
Solution :
(i) Given in the question, 2 sin θ – 1 = 0
Now, 2 sin θ – 1 = 0
= 2 sin θ= 1
= sin θ = 1/2
= sin θ = sin30°
= θ = 30°
Thus, the value of θ will be 30°
(ii) In the question we get, cos² θ +tan² θ
∴ cos² 30° + tan² 30°
= √3/2)² + (1/√3)²
= 3/4 + 1/3
= 9 + 4/12
= 13/12
= 1 1/12
Question no – (16)
Solution :
Here, in △ABC,
∠B = 60°, ∠C = 90°
AB = 15 unit
∴ ∠A = 180° – (90 + 60) °
= 180° – 150°
= 30°
∴ sin θ = AC/AB
= sin 30° = AC/15
= √3/2 = AC/15
= AC = 15√3/2
= 15 × 1.7 32/2
= 12.99
and, cos 60° = BC/AB
= 1/2 = BC/15
= BC = 15/2
= 7.5 unit
Question no – (17)
Solution :
In rectangle, ABCD,
AB = 20 cm, ∠BAC = 60°
AC and BD are diagonals
∴ tan 60° = BC/AB
= √3 = BC/20
= BC = 20√3
Now, cos 60° = 20/AC
= 1/2 = 20/AC
= AC = 40 cm
If BD = AC
∴ BD = 40 cm
Question no – (19)
Solution :
From question figure,
∠A = 45°,
BC = 10 m
Let, AC = x
∴ sin A = BC/AC
= sin 45° = 10/x
= 1/√2 = 10/x
= x = 10√2
= 10 × 1.414
= 14.14 m
Thus, the length of wire will be 14.1 m.
Question no – (21)
Solution :
Here, AB = 150 m
∠B = 60°
∠C = 90°
Let AC = x
∴ sin θ = AC/AB
= sin 60° = x/150
= √3/2 = x/150
= x = 150 × √3/2
= 75√3
= 75 × 1.73
= 129.75 m
Hence, the height of the kite from the ground will be 129.75 m.
Trigonometrical Ratios Exercise 19(d) Solution :
Question no – (1)
Solution :
(i) Given, sin 16°/cos 74°
Now, sin 16°/cos 74°
= sin 16°/cos (90 – 16)°
= sin 16°/sin 16°
= 1
(ii) cos 25°/sin 65°
Now, cos 25°/sin 65°
= cos 25°/sin(90 – 25)°
= cos 25°/cos 25°
= 1
(iii) tan 38°/cot 52°
Now, tan 38°/cot 52°
= tan 38°/cot (90 – 38)°
= tan 38°/tan 38°
= 1
(iv) sec 62°/cosec 28°
Now, sec 62°/cosec 28°
= sec 62°/cosec (90 – 62)°
= sec 62°/sec 62°
= 1
Question no – (2)
Solution :
(i) Given, sin2 67° + sin2 23°
∴ sin2 67° + sin2 23°
= sin2 67° + sin2 (90 – 67) °
= sin2 67° + cos2 67°
= 1
(ii) Given, (sin 49°/cos 41°)2 + (cos 41°/sin 49°)2
Now, (sin 49°/cos 41°)2 + (cos 41°/sin 49°)2
= (sin 49°/cos (90 – 49) °2 + (cos 41°/sin (90 – 41) °2
= (sin 49°/sin 49°)2 + (cos 41°/cos 41°)2
= 1 + 1
= 2
(iii) Given, cos2 20° + cos2 70°/sin2 59° + sin2 31°
Now, cos2 20° + cos2 70°/sin2 59° + sin2 31°
= cos2 (90 – 70) ° + cos2 70°/sin2 59° + sin2 (90 – 59) °2
= sin2 70° + cos2 70°/sin2 59° + cos2 59°
= 1/1
= 1
(iv) Given, cos 70°/sin 20° + cos 59°/sin 31° – 8 sin2 30°
Now, cos 70°/sin 20° + cos 59°/sin 31° – 8 sin2 30°
= cos (90 – 20) °/sin20° + cos (90 – 31) °/sin 31° – 8(sin 30°)2
= sin 20°/sin20° + cos31°/cos31° – 8(1/2)2
= 1 + 1 – 8 × 1/4
= 2 – 2
= 0
(v) tan53° /cot37° – cot80°/tan10°
∴ tan 53°/cot 37° – cot 80°/tan 10°
= 2 tam 53°/cot 37° – cot 80°/tan10°
= 2 tan (90° – 37°)/cot 37° – cot 80/tan (90° – 80°)
= 2 × cot 37°/cot37° – cot 80°/cot 80°
= 2 – 1
= 1
(vi) sec 50° sin40° + cos40° cosec50°
∴ sin40°/cos 50° + cos 40°/sin50°
= sin 40°/cos (90° – 40°) + cos40°/sin (90° – 40°)
= sin 40°/sin 40° + cos40/cos 40°
= 1 + 1
= 2
(vii) cos 80°/sin 10° + cos 59° cosec 31°
Now, cos 80°/sin 10° + cos 59° cosec 31°
= cos 80°/sin 10° + cos 59°/sin 31°
= cos 80°/sin (90° – 80°) + cos 59°/sin (90° – 59°)
= cos 80°/cos 80° + cos 59°/cos 59°
= 1 + 1
= 2
(viii) 2 sin 43°/cos 47° – cot 30°/tan 60° – √2 sin 45°
Now, 2 sin 43°/cos 47° – cot 30°/tan 60° – √2 sin 45°
= 2 sin 43°/cos (90° – 43°) – cot (90° – 60°)/tan60° – √2 × 1/√2
= 2 sin 43°/sin 43° – tan60°/tan60° – 1
= 2 × 1 – 1 – 1
= 0
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