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RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 Trigonometric Identities

RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 Trigonometric Identities has been provided here. Students can refer to these questions before their examinations for better preparation.

Neha Tanna14 Nov, 2024

RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1: Chapter 6 of RD Sharma’s Class 10 Maths book covers Trigonometric Identities and provides foundational knowledge on trigonometric functions and their identities.

These identities are essential for simplifying complex trigonometric expressions and solving equations. The exercise includes questions that require students to apply these identities to verify equalities or transform expressions, enhancing problem-solving skills in trigonometry. This section forms the basis for more advanced trigonometric concepts in later chapters.

RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 Overview

Chapter 6, Exercise 6.1 of RD Sharma's Class 10 Maths book covers fundamental trigonometric identities, which are crucial for simplifying and solving trigonometric equations. This exercise focuses on the primary identities. Understanding these identities is essential as they serve as building blocks for more complex trigonometric problems in higher mathematics. Mastery of these identities enables students to tackle problems in geometry, physics, and calculus, making them foundational for both academic exams and practical applications in various scientific fields.

RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 PDF

Below, we have provided the RD Sharma Solutions for Class 10 Maths, Chapter 6, Exercise 6.1, covering Trigonometric Identities. This exercise helps students understand and solve problems related to the fundamental trigonometric identities. You can download the PDF for a detailed, step-by-step explanation and solutions to practice and strengthen your concepts in trigonometry.

RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 PDF

RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 Trigonometric Identities

Below is the RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 Trigonometric Identities -

Prove the following trigonometric identities:

1. (1 – cos ² A) cosec ² A = 1

Solution:

Taking the L.H.S, (1 – cos ² A) cosec ² A = (sin ² A) cosec ² A [∵ sin ² A + cos ² A = 1 ⇒1 – sin ² A = cos ² A] = 1 ² = 1 = R.H.S – Hence Proved

2. (1 + cot ² A) sin ² A = 1

Solution:

By using the identity, cosec ² A – cot ² A = 1 ⇒ cosec ² A = cot ² A + 1 Taking, L.H.S = (1 + cot ² A) sin ² A = cosec ² A sin ² A = (cosec A sin A) ² = ((1/sin A) × sin A) ² = (1) ² = 1 = R.H.S – Hence Proved

3. tan ² θ cos ² θ = 1 − cos ² θ

Solution:

We know that, sin ² θ + cos ² θ = 1 Taking, L.H.S = tan ² θ cos ² θ = (tan θ × cos θ) ² = (sin θ) ² = sin ² θ = 1 – cos ² θ = R.H.S – Hence Proved

4. cosec θ √(1 – cos ² θ) = 1

Solution:

Using identity, sin ² θ + cos ² θ = 1 ⇒ sin ² θ = 1 – cos ² θ Taking L.H.S, L.H.S = cosec θ √(1 – cos ² θ) = cosec θ √( sin ² θ) = cosec θ x sin θ = 1 = R.H.S – Hence Proved

5. (sec ² θ − 1)(cosec ² θ − 1) = 1

Solution:

Using identities, (sec ² θ − tan ² θ) = 1 and (cosec ² θ − cot ² θ) = 1 We have, L.H.S = (sec ² θ – 1)(cosec ² θ – 1) = tan ² θ × cot ² θ = (tan θ × cot θ) ² = (tan θ × 1/tan θ) ² = 1 ² = 1 = R.H.S – Hence Proved

6. tan θ + 1/ tan θ = sec θ cosec θ

Solution:

We have, L.H.S = tan θ + 1/ tan θ = (tan ² θ + 1)/ tan θ = sec ² θ / tan θ [∵ sec ² θ − tan ² θ = 1] = (1/cos ² θ) x 1/ (sin θ/cos θ) [∵ tan θ = sin θ / cos θ] = cos θ/ (sin θ x cos ² θ) = 1/ cos θ x 1/ sin θ = sec θ x cosec θ = sec θ cosec θ = R.H.S – Hence Proved

7. cos θ/ (1 – sin θ) = (1 + sin θ)/ cos θ

Solution:

We know that, sin ² θ + cos ² θ = 1 So, by multiplying both the numerator and the denominator by (1+ sin θ), we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 1

L.H.S = = R.H.S – Hence Proved 8. cos θ/ (1 + sin θ) = (1 – sin θ)/ cos θ

Solution:

We know that, sin ² θ + cos ² θ = 1 So, by multiplying both the numerator and the denominator by (1- sin θ), we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 2

L.H.S = = R.H.S – Hence Proved

9. cos ² θ + 1/(1 + cot ² θ) = 1

Solution:

We already know that, cosec ² θ − cot ² θ = 1 and sin ² θ + cos ² θ = 1 Taking L.H.S,

= cos ² A + sin ² A = 1 = R.H.S – Hence Proved

10. sin ² A + 1/(1 + tan ² A) = 1

Solution:

We already know that, sec ² θ − tan ² θ = 1 and sin ² θ + cos ² θ = 1 Taking L.H.S, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 4

= sin ² A + cos ² A = 1 = R.H.S – Hence Proved

11.

Solution:

We know that, sin ² θ + cos ² θ = 1 Taking the L.H.S, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 6

= cosec θ – cot θ = R.H.S – Hence Proved

12. 1 – cos θ/ sin θ = sin θ/ 1 + cos θ

Solution:

We know that, sin ² θ + cos ² θ = 1 So, by multiplying both the numerator and the denominator by (1+ cos θ), we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 7

= R.H.S – Hence Proved

13. sin θ/ (1 – cos θ) = cosec θ + cot θ

Solution:

Taking L.H.S, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 8

= cosec θ + cot θ = R.H.S – Hence Proved

14. (1 – sin θ) / (1 + sin θ) = (sec θ – tan θ) ²

Solution:

Taking the L.H.S, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 9

= (sec θ – tan θ) ² = R.H.S – Hence Proved

15.

Solution:

Taking L.H.S, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 11

= cot θ = R.H.S – Hence Proved

16. tan ² θ − sin ² θ = tan ² θ sin ² θ

Solution:

Taking L.H.S, L.H.S = tan ² θ − sin ² θ R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 12

= tan ² θ sin ² θ = R.H.S – Hence Proved

17. (cosec θ + sin θ)(cosec θ – sin θ) = cot ² θ + cos ² θ

Solution:

Taking L.H.S = (cosec θ + sin θ)(cosec θ – sin θ) On multiplying, we get, = cosec ² θ – sin ² θ = (1 + cot ² θ) – (1 – cos ² θ) [Using cosec ² θ − cot ² θ = 1 and sin ² θ + cos ² θ = 1] = 1 + cot ² θ – 1 + cos ² θ = cot ² θ + cos ² θ = R.H.S – Hence Proved

18. (sec θ + cos θ) (sec θ – cos θ) = tan ² θ + sin ² θ

Solution:

Taking L.H.S = (sec θ + cos θ)(sec θ – cos θ) On multiplying, we get, = sec ² θ – sin ² θ = (1 + tan ² θ) – (1 – sin ² θ) [Using sec ² θ − tan ² θ = 1 and sin ² θ + cos ² θ = 1] = 1 + tan ² θ – 1 + sin ² θ = tan ² θ + sin ² θ = R.H.S – Hence Proved

19. sec A(1- sin A) (sec A + tan A) = 1

Solution:

Taking L.H.S = sec A(1 – sin A)(sec A + tan A) Substituting sec A = 1/cos A and tan A =sin A/cos A in the above, we have, L.H.S = 1/cos A (1 – sin A)(1/cos A + sin A/cos A) = 1 – sin ² A / cos ² A [After taking L.C.M] = cos ² A / cos ² A [∵ 1 – sin ² A = cos ² A] = 1 = R.H.S – Hence Proved

20. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1

Solution:

Taking L.H.S = (cosec A – sin A)(sec A – cos A)(tan A + cot A)

= (cos ² A/ sin A) (sin ² A/ cos A) (1/ sin A cos A) [∵ sin ² θ + cos ² θ = 1] = (sin A cos A ) (1/ cos A sin A) = 1 = R.H.S – Hence Proved

21. (1 + tan ² θ)(1 – sin θ)(1 + sin θ) = 1

Solution:

Taking L.H.S = (1 + tan ² θ)(1 – sin θ)(1 + sin θ) And, we know sin ² θ + cos ² θ = 1 and sec ² θ – tan ² θ = 1 So, L.H.S = (1 + tan ² θ)(1 – sin θ)(1 + sin θ) = (1 + tan ² θ){(1 – sin θ)(1 + sin θ)} = (1 + tan ² θ)(1 – sin ² θ) = sec ² θ (cos ² θ) = (1/ cos ² θ) x cos ² θ = 1 = R.H.S – Hence Proved

22. sin ² A cot ² A + cos ² A tan ² A = 1

Solution:

We know that, cot ² A = cos ² A/ sin ² A and tan ² A = sin ² A/cos ² A Substituting the above in L.H.S, we get L.H.S = sin ² A cot ² A + cos ² A tan ² A = {sin ² A (cos ² A/ sin ² A)} + {cos ² A (sin ² A/cos ² A)} = cos ² A + sin ² A = 1 [∵ sin ² θ + cos ² θ = 1] = R.H.S – Hence Proved R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 14

23.

Solution:

(i) Taking the L.H.S and using sin ² θ + cos ² θ = 1, we have L.H.S = cot θ – tan θ R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 15

= R.H.S – Hence Proved (ii) Taking the L.H.S and using sin ² θ + cos ² θ = 1, we have L.H.S = tan θ – cot θ R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 16

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 16

= R.H.S – Hence Proved

24. (cos ² θ/ sin θ) – cosec θ + sin θ = 0

Solution:

Taking L.H.S and using sin ² θ + cos ² θ = 1, we have R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 17

= – sin θ + sin θ = 0 = R.H.S

Hence proved

25.

Solution:

Taking L.H.S,

= 2 sec ² A = R.H.S

Hence proved

26.

Solution:

Taking the LHS and using sin ² θ + cos ² θ = 1, we have

= 2/ cos θ = 2 sec θ = R.H.S

Hence proved

27.

Solution:

Taking the LHS and using sin ² θ + cos ² θ = 1, we have R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 23

= R.H.S

Hence proved

28.

Solution:

Taking L.H.S, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 25

Using sec ² θ − tan ² θ = 1 and cosec ² θ − cot ² θ = 1 R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 26

= R.H.S

29.

Solution:

Taking L.H.S and using sin ² θ + cos ² θ = 1, we have R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 34

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 33

= R.H.S

Hence proved

30.

Solution:

Taking LHS, we have R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 36

= 1 + tan θ + cot θ = R.H.S

Hence proved

31. sec ⁶ θ = tan ⁶ θ + 3 tan ² θ sec ² θ + 1

Solution:

From trig. Identities we have, sec ² θ − tan ² θ = 1 On cubing both sides, (sec ² θ − tan ² θ) ³ = 1 sec ⁶ θ − tan ⁶ θ − 3sec ² θ tan ² θ(sec ² θ − tan ² θ) = 1 [Since, (a – b) ³ = a ³ – b ³ – 3ab(a – b)] sec ⁶ θ − tan ⁶ θ − 3sec ² θ tan ² θ = 1 ⇒ sec ⁶ θ = tan ⁶ θ + 3sec ² θ tan ² θ + 1 Hence, L.H.S = R.H.S

Hence proved

32. cosec ⁶ θ = cot ⁶ θ + 3cot ² θ cosec ² θ + 1

Solution:

From trig. Identities we have, cosec ² θ − cot ² θ = 1 On cubing both sides, (cosec ² θ − cot ² θ) ³ = 1 cosec ⁶ θ − cot ⁶ θ − 3cosec ² θ cot ² θ (cosec ² θ − cot ² θ) = 1 [Since, (a – b) ³ = a ³ – b ³ – 3ab(a – b)] cosec ⁶ θ − cot ⁶ θ − 3cosec ² θ cot ² θ = 1 ⇒ cosec ⁶ θ = cot ⁶ θ + 3 cosec ² θ cot ² θ + 1 Hence, L.H.S = R.H.S

Hence proved

33.

Solution:

Taking L.H.S and using sec ² θ − tan ² θ = 1 ⇒ 1 + tan ² θ = sec ² θ R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 38

= R.H.S

Hence proved

34.

Solution:

Taking L.H.S and using the identity sin ² A + cos ² A = 1, we get sin ² A = 1 − cos ² A ⇒ sin ² A = (1 – cos A)(1 + cos A) R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 40

Hence proved

35.

Solution:

We have,

Rationalizing the denominator and numerator with (sec A + tan A) and using sec ² θ − tan ² θ = 1 we get, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 43

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 43

= R.H.S

Hence proved

36.

Solution:

We have,

On multiplying the numerator and denominator by (1 – cos A), we get

= R.H.S

Hence proved

37. (i)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with √(1 + sin A), we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 48

= R.H.S

Hence proved

(ii)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 50

= 2 cosec A = R.H.S

Hence proved

38. Prove that:

(i)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 52

= 2 cosec θ = R.H.S

Hence proved

(ii)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 54

= R.H.S

Hence proved

(iii)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 56

= 2 cosec θ = R.H.S

Hence proved

(iv)

Solution:

Taking L.H.S, we have

= R.H.S

Hence proved

39.

Solution:

Taking LHS = (sec A – tan A) ² , we have R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 60

= R.H.S

Hence proved

40.

Solution:

Taking L.H.S and rationalizing the numerator and denominator with (1 – cos A), we get R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 62

= (cosec A – cot A) ² = (cot A – cosec A) ² = R.H.S

Hence proved

41.

Solution:

Considering L.H.S and taking L.C.M and simplifying, we have, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 64

= 2 cosec A cot A = RHS

Hence proved

42.

Solution:

Taking LHS, we have R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 66

= cos A + sin A = RHS

Hence proved

43.

Solution:

Considering L.H.S and taking L.C.M and simplifying, we have, R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 68

= 2 sec ² A = RHS

Hence proved

Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1

Solving RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.1 on Trigonometric Identities offers several benefits for students:

Strengthens Conceptual Understanding : The exercise provides problems based on fundamental trigonometric identities. Practicing these problems helps reinforce the foundational concepts, enabling students to better understand how identities work and how they can be applied.

Improves Problem-Solving Skills : Working through various questions on trigonometric identities hones problem-solving abilities. Students learn techniques to simplify complex trigonometric expressions, which is beneficial not only in trigonometry but also in calculus and other advanced math topics.

Prepares for Competitive Exams : Trigonometry is a significant part of competitive exams and various state-level engineering entrance exams. RD Sharma's problems provide the rigor required to build confidence for these exams.

Boosts Calculation Speed and Accuracy : Practicing RD Sharma exercises improves calculation speed and accuracy, helping students become more efficient at handling trigonometric expressions and equations under time constraints.

Enhances Analytical Thinking : Trigonometric identities require analytical thinking to recognize patterns and apply the right identity. Regular practice with these exercises enhances students' ability to think analytically, which is useful across math and other subjects.

RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1 FAQs

What are trigonometric identities used for?

Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Trigonometric Identities are true for every value of variables occurring on both sides of an equation.

What is the trick for trigonometric identities?

Always Start from the More Complex Side

What are the real world applications of trigonometric identities?

Applications of trigonometry are applied in areas such as architecture, celestial mechanics, surveying, etc. The most common fields are astronomy and physics where it helps in finding the distance between the stars and planets, the path in motion, and analyzing the waves.

What is one important consideration when proving trigonometric identities?

The general method of proving trigonometric identities is to work on each side of the equation separately, and simplify or manipulate each side until you reach the same expression on both sides.

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