RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities

RD Sharma Solutions Class 10 Maths Chapter 6 – Free PDF Download

RD Sharma Solutions for Class 10 Maths Chapter 6 – Trigonometric Identities is provided here. The branch of Mathematics which deals with the measurement of the sides and the angles of a triangle is trigonometry. We know that by now this topic would already seem difficult and complicated as its completely new to you. So, in order to make your learning process smooth and hassle-free the RD Sharma Solutions prepared by our expert team at BYJU’S will help students get the correct understanding of various chapters in the book.

Trigonometric Identities is the 6th Chapter of RD Sharma Class 10 which has two exercises and its solved answers with detailed explanations are given here RD Sharma Solutions for Class 10. The previous chapter was about trigonometric ratios and relations between them. But this chapter will be about proving some trigonometric identities and use them to prove other useful trigonometric identities.

Download the PDF of RD Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities here

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Access the RD Sharma Solutions For Class 10 Maths Chapter 6 – Trigonometric Identities

RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.1 Page No: 6.43

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

= cosec θ – cot θ

= 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  cos A)  (1/ cos A sin A)

= 1

= 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 A

= {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

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

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

  • 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 − cos2A

⇒ 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:

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

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

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

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

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 A)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

RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.2 Page No: 6.54

1. If cos θ = 4/5, find all other trigonometric ratios of angle θ. 

Solution:

 

We have,

cos θ = 4/5

And we know that,

sin θ = √(1 – cos2 θ)

⇒ sin θ = √(1 – (4/5)2)

= √(1 – (16/25))

= √[(25 – 16)/25]

= √(9/25)

= 3/5

∴ sin θ = 3/5

Since, cosec θ = 1/ sin θ

= 1/ (3/5)

⇒ cosec θ = 5/3

And, sec θ = 1/ cos θ

= 1/ (4/5)

⇒ cosec θ = 5/4

Now,

tan θ = sin θ/ cos θ

= (3/5)/ (4/5)

⇒ tan θ = 3/4

And, cot θ = 1/ tan θ

= 1/ (3/4)

⇒ cot θ = 4/3

2. If sin θ = 1/√2, find all other trigonometric ratios of angle θ.

Solution:

 

We have,

sin θ = 1/√2

And we know that,

cos θ = √(1 – sin2 θ)

⇒ cos θ = √(1 – (1/√2)2)

= √(1 – (1/2))

= √[(2 – 1)/2]

= √(1/2)

= 1/√2

∴ cos θ = 1/√2

Since, cosec θ = 1/ sin θ

= 1/ (1/√2)

⇒ cosec θ = √2

And, sec θ = 1/ cos θ

= 1/ (1/√2)

⇒ sec θ = √2

Now,

tan θ = sin θ/ cos θ

= (1/√2)/ (1/√2)

⇒ tan θ = 1

And, cot θ = 1/ tan θ

= 1/ (1)

⇒ cot θ = 1

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 1

3.

Solution:

Given,

tan θ = 1/√2

By using sec2 θ − tan2 θ = 1,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 2
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 3

4.

Solution:

Given,

tan θ = 3/4

By using sec2 θ − tan2 θ = 1,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 4

sec θ = 5/4

Since, sec θ = 1/ cos θ

⇒ cos θ = 1/ sec θ

= 1/ (5/4)

= 4/5

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 5

So,

 

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 6

5.

Solution:

Given, tan θ = 12/5

Since, cot θ = 1/ tan θ = 1/ (12/5) = 5/12

Now, by using cosec2 θ − cot2 θ = 1

cosec θ = √(1 + cot2 θ)

= √(1 + (5/12)2 )

= √(1 + 25/144)

= √(169/ 25)

⇒ cosec θ = 13/5

Now, we know that

sin θ = 1/ cosec θ

= 1/ (13/5)

⇒ sin θ = 5/13

Putting value of sin θ in the expression we have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 7

= 25/ 1

= 25

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 8

6.

Solution:

Given,

cot θ = 1/√3

Using cosec2 θ − cot2 θ = 1, we can find cosec θ

cosec θ = √(1 + cot2 θ)

= √(1 + (1/√3)2)

= √(1 + (1/3)) = √((3 + 1)/3)

= √(4/3)

⇒ cosec θ = 2/√3

So, sin θ = 1/ cosec θ = 1/ (2/√3)

⇒ sin θ = √3/2

And, we know that

cos θ = √(1 – sin2 θ)

= √(1 – (√3/2)2)

= √(1 – (3/4))

= √((4 – 3)/4)

= √(1/4)

⇒ cos θ = 1/2

Now, using cos θ and sin θ in the expression, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 9

= 3/5

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 10

7.

Solution:

Given,

cosec A = √2

Using cosec2 A − cot2 A = 1, we find cot A

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 11

= 4/2

= 2

Frequently Asked Questions on RD Sharma Solutions for Class 10 Maths Chapter 6

What are trigonometric ratios according to Chapter 6 of RD Sharma Solutions in the board exam of Class 10 Maths?

Trigonometric Ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle. The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle. The three sides of the right triangle are:
Hypotenuse (the longest side)
Perpendicular (opposite side to the angle)
Base (Adjacent side to the angle)

Why should we download RD Sharma Solutions for Class 10 Maths Chapter 6 from BYJU’S?

BYJU’S provides the most accurate answers for the questions present in the RD Sharma Solutions for Class 10 Maths Chapter 6. These solutions can be viewed online as well as downloaded in the PDF format. The solutions of this chapter are explained by experts very clearly with neat diagrams wherever necessary. Our specialist teachers formulate these exercises to help you with your exam preparation to achieve good marks in Maths. The solutions are stepwise and detailed to make learning easy for students.

Is RD Sharma Solutions for Class 10 Maths Chapter 6 important from an exam point of view?

Yes, all the chapters present in RD Sharma Solutions for Class 10 Maths are important for board exams as well as for higher grades. Students should practice all the questions present in RD Sharma Solutions for Class 10 Maths Chapter 6 to procure high marks.

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