RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3: The following RD Sharma Solutions Class 10 exercise 5.3 contains a number of problems to help students better understand this concept. The solutions are written in simple Language to meet the needs of all students. For a full explanation of the questions in this exercise, students can download RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios Exercise 5.3 PDF, which is provided below.
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RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3
Access answers to RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3- Important Question with Answers
Question 1.
Evaluate the following :
Solution:
Question 2.
Evaluate the following :
Solution:
Question 3.
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin 59° + cos 56°
(ii) tan 65° + cot 49“
(iii) sec 76° + cosec 52°
(iv) cos 78° + sec 78°
(v) cosec 54° + sin 72°
(vi) cot 85″ + cos 75°
(vii) sin 67° + cos 75°
Solution:
(i) sin 59° + cos 56°
= sin (90° – 31°) + cos (90° – 34°)
= cos 31° +sin 34°
(ii) tan 65° + cot 49°
= tan (90° – 25°) + cot (90° – 41°)
= cot 25° + tan 41°
(iii) sec 76° + cosec 52°
= sec (90° – 14°) + cosec (90 0 – 38°)
= cosec 14° + sec 38°
(iv) cos 78° + sec 78°
= cos (90° – 12°) + sec (90°- 12°)
= sin 12° + cosec 12°
(v) cosec 54° + sin 72°
= cosec (90° – 36°) + sin (90°-18°)
= sec 36° + cos 18°
(vi) cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
(vii) sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
Question 4.
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
Solution:
cos 75° + cot 75° = cos (90° – 15°) + cot (90°-15°)
= sin 15° + tan 15°
Question 5.
If sin 3A = cos (A – 26°), where 3A is an acute angle, And the value of A.
Solution:
sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)
Comparing,
90° – 3A = A – 26°
⇒ 90° + 26° = A + 3A ⇒ 4A = 116°
Question 6.
If A, B, C are the interior angles of a triangle ABC, prove
Solution:
Question 7.
Prove that :
Solution:
Question 8.
Prove the following :
Solution:
Question 9.
Evaluate :
Solution:
Question 10.
If sin θ= cos (θ – 45°), where θ and (θ – 45°) are acute angles, find the degree measure of θ.
Solution:
Question 11.
If A, B, C are the interior angles of a AABC, show that :
(i) sinB+C2cosA2
(ii) cosB+C2sinA2
Solution:
Question 12.
If 2θ + 45° and 30° – θ are acute angles, find the degree measures of θ satisfying sin (20 + 45°) = cos (30° – θ).
Solution:
Question 13.
If θ is a positive acute angle such that sec θ = cosec 60°, And the value of 2 cos2 θ-1.
Solution:
Question 14.
If cos 2 θ – sin 4 θ, where 2 θ and 4 θ are acute angles, find the value of θ.
Solution:
Question 15.
If sin 3 θ = cos (θ – 6°), where 3 θ and θ – 6° are acute angles, find the value of θ.
Solution:
Question 16.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Question 17.
If sec 2A = cosec (A – 42°), where 2A is an acute angle, find the value of A. (C.B.S.E. 2008)
Solution:
We have provided complete details of RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.
FAQs on RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3
What are the benefits of using RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3?
1. Correct answers according to the latest CBSE guidelines and syllabus.
2. The RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3 is written in simple language to assist students in their board examination, & competitive examination preparation.
Where can I download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3 free PDF?
You can download RD Sharma Class 10 Maths Solutions Chapter 5 Exercise 5.3 free PDF from the above article.
Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3?
Yes, learning all of the concepts included in RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.3 is required in order to achieve high scores on the Class 10 board exams. These solutions were created by subject matter specialists who compiled model questions that covered all of the textbook’s exercise questions.