In this chapter, we provide RD Sharma Class 10 Ex 5.3 Solutions Chapter 5 Trigonometric Ratios for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 5.3 Solutions Chapter 5 Trigonometric Ratios pdf, Now you will get step by step solution to each question.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 5 |
| Chapter Name | Trigonometric Ratios |
| Exercise | 5.3 |
| Category | RD Sharma Solutions |
RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Ex 5.3 Download PDF
Chapter 5: Trigonometric Ratios Exercise – 5.3
Question: 1
Evaluate the following:
Solution:
(i) Given that,
Since sin (90 – θ) = cos θ
Therefore,
(ii) Given that,
Since cos (90 – θ) = sin θ
Therefore
(iii) Given that,
Since (90 – θ) = cos θ
(iv) We are given that,
Since tan (90 – θ) = cot θ
Therefore
(v) Given that,
Since sec (90 – θ) = cosec θ
Therefore
Question: 2
Evaluate the following:
(ii) cos 48°- sin 42°
(vii) cosec 31° – sec 59°.
(viii) (sin 72° + cos 18°) (sin 72° – cos 18°)
(ix) sin 35° sin 55° – cos 35° cos 55°
(x) tan 48° tan 23° tan 42° tan 67°
(xi) sec 50° sin 40° + cos 40° cosec 50°
Solution:
(i) We have to find:
sin(90° – θ) = cos θ and cos(90°- θ) = sin θ
= 1 + 1
= 2
(ii) We have to find: cos 48° – sin 42°
Since cos (90° − θ) = sin θ.
So cos 48° – sin 42° = cos (90° − 42°) – sin 42° = sin 42° – sin 42°= 0
So value of cos 48° – sin 42° is 0
(iii) We have to find:
Since cot (90°- θ) = tan θ and cos (90°- θ) = sin θ
(iv) We have to find:
Since sin(90°- θ) = cos θ and cos (90 – θ)
= 1 – 1
= 0
(v) We have to find:
Since tan (90°- θ) = cot θ and cot (90°- θ) = tan θ
= 1
(vi) We have to find:
= 1 + 1 = 2
(vii) We have to find:
cosec 31° − sec 59°
Since cosec (90° − θ) = sec θ.
So = cosec 31° − sec 59°
= cosec (90° − 59°) – sec 59° = sec 59° − sec 59° = 0
So value of cosec 31° − sec 59° is 0
(viii) We have to find:
(sin 72° + cos 18°)(sin 72° − cos 18°)
Since sin(90° − θ) = cos θ
So (sin 72° + cos 18°) (sin 72° − cos 18°)
= (sin 72°)2 – (cos 18°)2 = [sin(90° − 18°)]2 − (cos18°)2
= (cos18°)2 – (cos18°)2
= cos218°−cos218° = 0
So value of (sin 72° + cos 18°) (sin 72° − cos 18°) is 0.
(ix) We find:
sin 35°sin 55° – cos 35°cos 55°
Since sin(90° − θ) = cos θ and cos(90° − θ) = sin θ sin 35°sin 55°- cos 35°cos 55°
= sin (90° − 55°) sin 55° – cos(90° − 55°) cos 55°
= 1 – 1
= 0
So value of sin 35° sin 55° – cos 35° cos 55° is 0
(x) We have to find:
tan 48° tan 23° tan 42° tan 67°
Since tan(90°− θ) = cot θ. So tan 48° tan 23° tan 42° tan 67°
= tan (90° − 42°) tan (90° − 67°) tan 42° tan67°
= cot 42°cot 67° tan 42° tan 67°
= (tan 67°cot 67°) (tan 42°cot 42°)
= 1 x 1 = 1
So value of tan 48° tan 23° tan 42° tan 67° is 1
(xi) We have to find:
sec 50° sin 40° + cos 40°cosec 50°
Since cos (90° − θ) = sin θ, sec (90°- θ) = cosec θ and sin θ. cosec θ = 1.
So sec 50° sin 40° + cos 40° cosec 50°
= sec (90° − 40°) sin 40° + cos (90° − 50°) cosec 50°
=1 + 1
= 2
So value of sec 50°sin 40° + cos 40°cosec 50° is 2.
Question: 3
Express cos 75°+ cot 75° in terms of angle between 0° and 30°.
Solution:
Given that: cos 75° + cot 75°
= cos 75° + cot 75° = cos (90° − 15°) + cot (90° − 15°)
= sin 15° + tan 15°
Hence the correct answer is sin 15° + tan 15°
Question: 4
If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.
Solution:
We are given 3A is an acute angle. We have: sin 3A = cos (A – 26°)
⇒ sin 3A = sin(90° – (A – 26°))
⇒ sin3A = sin(116° – A)
⇒ 3A =116° – A
⇒ 4A = 116°
⇒ A = 29°
Hence the correct answer is 29°
Question: 5
If A, B, C are the interior angles of a triangle ABC, prove that,
Solution:
(i) We have to prove:
Since we know that in triangle ABC
A + B + C = 180
⇒ C + A = 180° – B
Hence proved
(ii) We have to prove:
Since we know that in triangle ABC
A + B + C = 180
⇒ B + C = 180° – A
Hence proved
Question: 6
Prove that:
(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1
(ii) sin 48°. Sec 48° + cos 48°.cosec 42° = 2
Solution:
(i) Therefore tan 20° tan 35° tan 45° tan 55° tan 70°
= tan (90° − 70°) tan (90° − 55°) tan 45°tan 55° tan70°
= cot 70°cot 55° tan 45° tan 55° tan 70°
= (tan 70°cot 70°)(tan 55°cot 55°) tan 45°
= 1 × 1 × 1 = 1
Hence proved
(ii) We will simplify the left hand side
Sin 48°. Sec 48° + cos 48°. Cosec 42°
= sin 48°. sec (90° − 48°) cos 48°. cosec (90° − 48°)
= sin 48°.cos 48° + cos 48°.sin 48°
= 1 + 1 = 2
Hence proved
(iii) We have,
So we will calculate left hand side
= 1 + 1 – 2
= 2 – 2
= 0
Hence proved
(iv) We have
We will simplify the left hand side
= 1 + 1
= 2
Hence proved
Question: 7
If A, B, C are the interior of triangle ABC, show that
Solution:
(i) A + B + C = 180°
B + C = 180° – A/2
LHS = RHS
Question: 8
If 2θ + 45° and 30 − θ are acute angles, find the degree measure of θ satisfying sin (20 + 45°) = cos (30° + θ)
Solution:
Here 2θ + 45° = sin (60° + θ)
We know that, ((90° − θ) = cos (θ)
= sin (2θ + 45°) = sin (90° − (30° – θ))
= sin (2θ + 45°) = sin (90° − 30° – θ)
= sin (2θ + 45°)
= sin (60° + θ)
On equating sin of angle of we get,
= 2θ + 45° = 60° + θ
= θ = 60° – 45°
= θ = 15°
Question: 9
If θ is appositive acute angle such that sec θ = cosec 60°, find 2cos2θ − 1
Solution:
We know that,
sec (90° − θ) = cosec2θ = sec θ
= sec (90° − 60°)
= θ = 30°
= 2cos2 θ − 1
= 2cos2 30 − 1
Question: 10
If sin 3θ = cos (θ − 6°) where 3θ and θ – 6° acute angles, find the value of θ.
Solution:
We have, sin 3θ = cos (θ – 6°) cos (90° + 3θ)
= cos (θ − 6°)
= 90° − 3θ = θ − 6°
− 3θ – θ = – 6° − 90°
− 4θ = – 96°
θ = – 96°/- 4
θ = 24°
Question: 11
If sec 2A = cosec (A − 42°) where 2A is acute angle, find the value of A.
Solution:
We know that sec (90 – 3θ) = cosec θ
sec 2A = sec (90 − (A − 42))
sec 2A = sec (90 − A + 42))
sec 2A = sec (132 − A)
Now equating both the angles we get
2A = 132 – A
3A = 132
A = 132/3
A = 44
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