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 Goyanka Maths Study 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.

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,

sin2 θ + cos2 θ = 1

Taking,

L.H.S = tan2 θ cos2 θ

= (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,

sin2 θ + cos2 θ = 1  ⇒ sin2 θ = 1 – cos2 θ

Taking L.H.S,

L.H.S = cosec θ √(1 – cos2 θ)

= cosec θ √( sin2 θ)

= 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

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

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,

= 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,

= sin2 A + cos2 A

= 1

= R.H.S

– Hence Proved

11.

Solution:

We know that, sin2 θ + cos2 θ = 1

Taking the L.H.S,

= 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.H.S

– Hence Proved

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

Solution:

Taking L.H.S,

= 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,

= (sec θ – tan θ)2

= R.H.S

– Hence Proved

15. 

Solution:

Taking L.H.S,

= 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 

= 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)

= (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.

Solution:

(i) Taking the L.H.S and using sin2 θ + cos2 θ = 1, we have

L.H.S = cot θ – tan θ

= 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.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

= – sin θ + sin θ

= 0

= R.H.S

• Hence proved

25.

Solution:

Taking L.H.S,

= 2 sec2 A

= R.H.S

• Hence proved

26. 

Solution:

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

= 2/ cos θ

= 2 sec θ

= R.H.S

• Hence proved

27.

Solution:

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

= R.H.S

• Hence proved

28.

Solution:

Taking L.H.S,

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

= R.H.S

• Hence proved

29.

Solution:

Taking L.H.S and using sin2 θ + cos2 θ = 1, we have

= R.H.S

• Hence proved

30.

Solution:

Taking LHS, we have

= 1 + tan θ + cot θ

= R.H.S

• Hence proved

31. sec6 θ = tan6 θ + 3 tan2 θ 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

33.

Solution:

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

= R.H.S

• Hence proved

34.

Solution:

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

sin2A = 1 − cos2A

⇒ sin2A = (1 – cos A)(1 + cos A)

• Hence proved

35.

Solution:

We have,

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

= R.H.S

• Hence proved

36.

Solution:

We have,

On multiplying 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.H.S

• Hence proved

(ii)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

= 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

= 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.H.S

• Hence proved

(iii)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

= 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.H.S

• Hence proved

40.

Solution:

Taking L.H.S and rationalizing the numerator and denominator with (1 – cos A), we get

= (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 on simplifying we have,

= 2 cosec A cot A = RHS

• Hence proved

42.

Solution:

Taking LHS, we have

= cos A + sin A

= RHS

• Hence proved

43.

Solution:

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

= 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.

Solution:

Given,

tan θ = 1/√2

By using sec2 θ − tan2 θ = 1,

4.

Solution:

Given,

tan θ = 3/4

By using sec2 θ − tan2 θ = 1,

sec θ = 5/4

Since, sec θ = 1/ cos θ

⇒ cos θ = 1/ sec θ

= 1/ (5/4)

= 4/5

So,

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,

= 25/ 1

= 25

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

= 3/5

7.

Solution:

Given,

cosec A = √2

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

= 4/2

= 2