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 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 Download PDF
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 + cot2Â A) sin2Â A = 1Â
Solution:Â
By using the identity,
cosec2 A – cot2 A = 1 ⇒ cosec2 A = cot2 A + 1
Taking,
L.H.S = (1 + cot2Â A) sin2Â A
= cosec2Â A sin2Â 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 θÂ
Solution:Â
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) = 1Â
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 θÂ
Solution:
Taking L.H.S,
L.H.S = tan2 θ − sin2 θÂ
= tan2 θ sin2 θ
= R.H.S
– Hence Proved
17. (cosec θ + sin θ)(cosec θ – sin θ) = cot2θ + cos2θÂ
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 θÂ
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) = 1Â
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 AÂ 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 AÂ = 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 tan2 θ sec2 θ + 1
Solution:Â
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 + cos2AÂ = 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