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

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

Access RD Sharma Solutions for Class 10 Maths 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

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:

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:

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

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 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

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 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.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 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 simplifying, we have,

= 2 sec2 A

= RHS

• Hence proved