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 known as trigonometry. Students who find difficulty to understand the concepts covered in this chapter can make their learning process smooth and hassle-free using RD Sharma Solutions. These solutions are well structured by our expert team at BYJU’S to help students grasp the in-depth knowledge of concepts which are vital for examinations.

Trigonometric Identities is the 6th Chapter of RD Sharma Solutions Class 10. This chapter consists of two exercises. Students can find the precise answers for these exercises in RD Sharma Solutions for Class 10. The previous chapter was about trigonometric ratios and relations between them. But this chapter explains the trigonometric identities in a comprehensive manner in accordance with the student’s intelligence quotient.

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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

23.

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 14

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

25.

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 18

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

27.

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

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

29.

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

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

30.

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

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

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

L.H.S = R.H.S

– Hence proved

33.

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 37

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

34.

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 39

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

35.

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 41

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

36.

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

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

37. (i)

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

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)

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

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

39.

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

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.

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

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.

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

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

42.

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

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.

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 67

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

3.

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

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

4.
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 3

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

So,

\(\begin{array}{l}\frac{1-cos\theta }{1+cos\theta }=\frac{1-\frac{4}{5}}{1+\frac{4}{5}}=\frac{\frac{1}{5}}{\frac{9}{5}}=\frac{1}{9}\end{array} \)

 

5. 

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

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

6.

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

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

7.

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

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

Why should we follow RD Sharma Solutions for Class 10 Maths Chapter 6 during exam preparation?

RD Sharma Solutions for Class 10 Maths Chapter 6 provides thorough knowledge of concepts to assist students for effective learning. The solutions are designed by our faculty at BYJU’S with the aim to improve academic performance among students. Those who want to clear their doubts quickly are suggested to download the solutions in a PDF format as per their needs. Practising these solutions on a daily basis help students to reduce their stress and boost confidence in solving the problems with ease.

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. Students can download these solutions in PDF format both in online and offline mode as per their requirements. The solutions of this chapter are explained by experts very clearly with neat diagrams wherever necessary. The main objective of developing these solutions is to make learning fun and interesting among students. The accurate solutions help students to clear their confusion which they come across while revising the problems.

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

Yes, RD Sharma Solutions for Class 10 Maths Chapter 6 is an important chapter from an exam point of view as it continues in higher grades. High professional teachers recommend students to follow RD Sharma Solutions in a descriptive manner to improve their skills in the respective subject. Diligent practise of these solutions not only clear doubts but also help to procure high marks in board exams.

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