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

RD Sharma Class 10 Solutions Chapter 6 Ex 6.1 PDF Free Download

In this exercise, students will be proving useful trigonometric identities using some basic trigonometric identities. The RD Sharma Solutions Class 10 is the most vital resource for students to clarify doubts and prepare confidently for their exams. For a detailed step by step solution to exercise questions, students can download the RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities Exercise 6.1 PDF below.

RD Sharma Solutions for Class 10 Chapter 6 Trigonometric Identities Exercise 6.1 Download PDF

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Access RD Sharma Solutions for Class 10 Chapter 6 Trigonometric Identities Exercise 6.1

Prove the following trigonometric identities:

1. (1 – cos2 A) cosec2 A = 1

Solution:

Taking the L.H.S,

(1 – cos2 A) cosec2 A

= (sin2 A) cosec2 A [∵ sin2 A + cos2 A = 1 ⇒1 – sin2 A = cos2 A]

= 12

= 1 = R.H.S

– Hence Proved

2. (1 + cot2 A) sin2 A = 1 

Solution: 

By using the identity,

cosec2 A – cot2 A = 1 ⇒ cosec2 A = cot2 A + 1

Taking,

L.H.S = (1 + cot2 A) sin2 A

= cosec2 A sin2 A

= (cosec A sin A)2 

= ((1/sin A) × sin A)2 

= (1)2 

= 1 

= R.H.S

– Hence Proved

3. tan2 θ cos2 θ = 1 − cos2 θ 

Solution: 

We know that,

  sinθ + cosθ = 1 

Taking,

L.H.S = tanθ cosθ

= (tan θ × cos θ)2

= (sin θ)2 

= sin2 θ 

= 1 – cos2 θ 

= R.H.S

– Hence Proved

4. cosec θ √(1 – cos2 θ) = 1

Solution:

Using identity,

sinθ + cosθ = 1  ⇒ sinθ = 1 – cosθ

Taking L.H.S,

L.H.S = cosec θ √(1 – cos2 θ)

= cosec θ √( sinθ)

= cosec θ x sin θ

= 1

= R.H.S

– Hence Proved

5. (sec2 θ − 1)(cosec2 θ − 1) = 1 

Solution:

Using identities, 

(sec2 θ − tan2 θ) = 1 and (cosec2 θ − cot2 θ) = 1

We have,

L.H.S = (sec2 θ – 1)(cosec2θ – 1)

= tan2θ × cot2θ

= (tan θ × cot θ)2

= (tan θ × 1/tan θ)2

= 12

= 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

= (tan2 θ + 1)/ tan θ

= sec2 θ / tan θ [∵ sec2 θ − tan2 θ = 1]

= (1/cos2 θ) x 1/ (sin θ/cos θ) [∵ tan θ = sin θ / cos θ]

= cos θ/ (sin θ x cos2 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, 

sin2 θ + cos2 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, 

sin2 θ + cos2 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. cos2 θ + 1/(1 + cot2 θ) = 1

Solution:

We already know that,

cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3

= cos2 A + sin2 A

= 1

= R.H.S

– Hence Proved

10. sin2 A + 1/(1 + tan 2 A) = 1

Solution:

We already know that,

sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 4

= sin2 A + cos2 A

= 1

= R.H.S

– Hence Proved

11.
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 5

Solution:

We know that, sin2 θ + cos2 θ = 1

Taking the L.H.S,

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

= R.H.S

– Hence Proved

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

Solution:

We know that, 

sin2 θ + cos2 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. R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 10

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. tan2 θ − sin2 θ = tan2 θ sin2 θ 

Solution:

Taking L.H.S,

L.H.S = tan2 θ − sin2 θ 

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 12

= tan2 θ sin2 θ 

= R.H.S

– Hence Proved

17. (cosec θ + sin θ)(cosec θ – sin θ) = cot2θ + cos2θ 

Solution:

Taking L.H.S = (cosec θ + sin θ)(cosec θ – sin θ)

On multiplying we get,

= cosec2 θ – sin2 θ

= (1 + cot2 θ) – (1 – cos2 θ) [Using cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1]

= 1 + cot2 θ – 1 + cos2 θ 

= cot2 θ + cos2 θ 

= R.H.S     

– Hence Proved

18. (sec θ + cos θ) (sec θ – cos θ) = tan2 θ + sin2 θ 

Solution:

Taking L.H.S = (sec θ + cos θ)(sec θ – cos θ)

On multiplying we get,

= sec2 θ – sin2 θ

= (1 + tan2 θ) – (1 – sin2 θ) [Using sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1]

= 1 + tan2 θ – 1 + sin2 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 – sin2 A / cos2 A [After taking L.C.M]

= cos2 A / cos2 A [∵ 1 – sin2 A = cos2 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)

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 13

= (cos2 A/ sin A) (sin2 A/ cos A) (1/ sin A cos A) [∵ sin2 θ + cos2 θ = 1]

= sin A x cos A x (1/ cos A sin A)

= R.H.S

– Hence Proved

21. (1 + tan2 θ)(1 – sin θ)(1 + sin θ) = 1

Solution:

Taking L.H.S = (1 + tan2θ)(1 – sin θ)(1 + sin θ)

And, we know sin2 θ + cos2 θ = 1 and sec2 θ – tan2 θ = 1

So,

L.H.S = (1 + tan2 θ)(1 – sin θ)(1 + sin θ)

= (1 + tan2 θ){(1 – sin θ)(1 + sin θ)}

= (1 + tan2 θ)(1 – sin2 θ)

= sec2 θ (cos2 θ)

= (1/ cos2 θ) x cos2 θ

= 1

= R.H.S

– Hence Proved

22. sin2 A cot2 A + cos2 A tan2 A = 1

Solution:

We know that,

cot2 A = cos2 A/ sin2 A and tan2 A = sin2 A/cos2 A

Substituting the above in L.H.S, we get

L.H.S = sin2 A cot2 A + cos2 A tan2

= {sin2 A (cos2 A/ sin2 A)} + {cos2 A (sin2 A/cos2 A)}

= cos2 A + sin2 A

= 1 [∵ sin2 θ + cos2 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 sin2 θ + cos2 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 sin2 θ + cos2 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.H.S

– Hence Proved

24. (cos2 θ/ sin θ) – cosec θ + sin θ = 0

Solution:

Taking L.H.S and using sin2 θ + cos2 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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 1825.

Solution:

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 19

= 2 sec2 A

= R.H.S

  • Hence proved

26. R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 20

Solution:

Taking the LHS and using sin2 θ + cos2 θ = 1, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 21

= 2/ cos θ

= 2 sec θ

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 22

27.

Solution:

Taking the LHS and using sin2 θ + cos2 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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 24

28.

Solution:

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 25

Using sec2 θ − tan2 θ = 1 and cosec2 θ − cot2 θ = 1

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 26

= R.H.S

And, taking

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 27
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 28

=
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 29[Using sec2 θ − tan2 θ = 1 and cosec2 θ − cot2 θ = 1]

=
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 30

=
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 31

= tan2 θ = R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 32

29.

Solution:

Taking L.H.S and using sin2 θ + cos2 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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 35

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. sec6 θ = tan6 θ + 3 tanθ sec2 θ + 1

Solution: 

From trig. Identities we have,

sec2 θ − tan2 θ = 1 

On cubing both sides,

(sec2θ − tan2θ)3 = 1

sec6 θ − tan6 θ − 3sec2 θ tan2 θ(sec2 θ − tan2 θ) = 1 

[Since, (a – b)3 = a3 – b3 – 3ab(a – b)] 

sec6 θ − tan6 θ − 3sec2 θ tan2 θ = 1

⇒ sec6 θ = tan6 θ + 3sec2 θ tan2 θ + 1

Hence, L.H.S = R.H.S

  • Hence proved

32. cosec6 θ = cot6 θ + 3cot2 θ cosec2 θ + 1

Solution:

From trig. Identities we have,

cosec2 θ − cot2 θ = 1 

On cubing both sides,

(cosec2 θ − cot2 θ)3 = 1

cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ (cosec2 θ − cot2 θ) = 1

[Since, (a – b)3 = a3 – b3 – 3ab(a – b)]

cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ = 1

⇒ cosec6 θ = cot6 θ + 3 cosec2 θ cot2 θ + 1 

Hence, L.H.S = R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3733.

Solution:

Taking L.H.S and using sec2 θ − tan2 θ = 1 ⇒ 1 + tan2 θ = sec2 θ

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 38

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3934.

Solution:

Taking L.H.S and using the identity sin2A + cos2A = 1, we get

sin2A = 1 − cos2

⇒ sin2A = (1 – cos A)(1 + cos A)

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 40

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 4135.

Solution:

We have,


R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 42

Rationalizing the denominator and numerator with (sec A + tan A) and using sec2 θ − tan2 θ = 1 we get,

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

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 44

36.

Solution:

We have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 45

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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 45

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 47

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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 49

(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:

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 51(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

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 53(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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 55(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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 57(iv)

Solution:

Taking L.H.S, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 58

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 59

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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 61

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)2 

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 63

41.

Solution:

Considering L.H.S and taking L.C.M and on 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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 65

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

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 6743.

Solution:

Considering L.H.S and taking L.C.M and on simplifying we have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 68

= 2 sec2 A

= RHS

  • Hence proved

Also, access other exercise solutions of RD Sharma Class 10 Maths Chapter 6 Trigonometric Identities

Exercise 6.2 Solutions

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